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**VOLUME 1.**

**PART 1.**

1 Measurement. 2 Motion Along a Straight Line. 3 Vectors. 4 Motion in Two and Three Dimensions. 5 Force and Motion — I. 6 Force and Motion — II. 7 Kinetic Energy and Work. 8 Potential Energy and Conservation of Energy. 9 Center of Mass and Linear Momentum. 10 Rotation. 11 Rolling, Torque, and Angular Momentum.

**PART 2.**

12 Equilibrium and Elasticity. 13 Gravitation. 14 Fluids. 15 Oscillations. 16 Waves — I. 17 Waves — II. 18 Temperature, Heat, and the First Law of Thermodynamics. 19 The Kinetic Theory of Gases. 20 Entropy and the Second Law of Thermodynamics.

**VOLUME 2.**

**PART 3.**

21 Electric Charge. 22 Electric Fields. 23 Gauss’ Law. 24 Electric Potential. 25 Capacitance. 26 Current and Resistance. 27 Circuits. 28 Magnetic Fields. 29 Magnetic Fields Due to Currents. 30 Induction and Inductance. 31 Electromagnetic Oscillations and Alternating Current. 32 Maxwell’s Equations; Magnetism of Matter.

**PART 4.**

33 Electromagnetic Waves. 34 Images. 35 Interference. 36 Diffraction. 37 Relativity.

**PART 5.**

38 Photons and Matter Waves. 39 More About Matter Waves. 40 All About Atoms. 41 Conduction of Electricity in Solids. 42 Nuclear Physics. 43 Energy from the Nucleus. 44 Quarks, Leptons, and the Big Bang.

1

**Chapter 1
**
1. Various geometric formulas are given in Appendix E.
(a) Expressing the radius of the Earth as

( )( )6 3 36.37 10 m 10 km m 6.37 10 km,*R *−= × = ×
its circumference is 3 42 2 (6.37 10 km) 4.00 10 km.*s R*π π= = × = ×
(b) The surface area of Earth is ( )22 3 8 24 4 6.37 10 km 5.10 10 km .*A R*= π = π × = ×

(c) The volume of Earth is ( )33 3 12 34 4 6.37 10 km 1.08 10 km .3 3*V R
*π π

= = × = ×

2. The conversion factors are: 1 gry 1/10 line= , 1 line 1/12 inch= and 1 point = 1/72 inch. The factors imply that

1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 2 20.50 gry = 0.18 point . 3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2). (a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 μm,

( ) ( )3 3 6 91km 10 m 10 m 10 m m 10 m.= = =μ μ The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μm. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m,

( ) ( )2 2 6 41cm = 10 m = 10 m 10 m m 10 m.− − =μ μ

We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10−4. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,

*CHAPTER 1 *2

( ) ( )6 51.0 yd = 0.91m 10 m m 9.1 10 m.= ×μ μ

4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain

( ) 1 inch 6 picas0.80 cm = 0.80 cm 1.9 picas. 2.54 cm 1 inch

⎛ ⎞⎛ ⎞ ≈⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

(b) With 12 points = 1 pica, we have

( ) 1 inch 6 picas 12 points0.80 cm = 0.80 cm 23 points. 2.54 cm 1 inch 1 pica

⎛ ⎞⎛ ⎞⎛ ⎞ ≈⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

5. Given that 1 furlong 201.168 m= , 1 rod 5.0292 m= and 1chain 20.117 m= , we find the relevant conversion factors to be

1 rod1.0 furlong 201.168 m (201.168 m ) 40 rods, 5.0292 m

= = =

and 1 chain1.0 furlong 201.168 m (201.168 m ) 10 chains

20.117 m = = = .

Note the cancellation of m (meters), the unwanted unit. Using the given conversion
factors, we find
(a) the distance *d in rods* to be

( ) 40 rods4.0 furlongs 4.0 furlongs 160 rods, 1 furlong

*d *= = =

(b) and that distance *in chains* to be

( )10 chains4.0 furlongs 4.0 furlongs 40 chains. 1 furlong

*d *= = =

6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1 fanega = 112 cahiz, or 8.33 × 10

−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the

already completed part) implies that 1 cuartilla = 148 cahiz, or 2.08 × 10 −2 cahiz.

Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and 33.47 10−× .

3

(b) In the second (“fanega”) column, we find 0.250, 8.33 × 10−2, and 4.17 × 10−2 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get 12 = 0.500 for the last entry. (e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94 × 10−3 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 × 10−2 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m3 or 55501 cm3. Thus, 7.00 almudes = 7.0012 fanega =

7.00 12 (55501 cm

3) = 3.24 × 104 cm3.
7. We use the conversion factors found in Appendix D.
2 31 acre ft = (43,560 ft ) ft = 43,560 ft⋅ ⋅
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
2 2 2 7 3(26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) 4.66 10 ft .*V *= = = ×
Thus,

*V *= ×
× ⋅

= × ⋅ 4 66 10

4 3560 10 11 10

7

4 3.

. .ft

ft acre ft acre ft.

3

3

8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have

( ) 258 W50.0 S 50.0 S 60.8 W 212 S

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

(b) In units of Z, we have

( ) 156 Z50.0 S 50.0 S 43.3 Z 180 S

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

9. The volume of ice is given by the product of the semicircular surface area and the
thickness. The area of the semicircle is *A* = π*r*2/2, where *r* is the radius. Therefore, the
volume is

*CHAPTER 1 *4

2

2
*V r z*π=

where *z* is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

( ) 3 2

510 m 10 cm2000 km 2000 10 cm. 1km 1m

*r
*⎛ ⎞ ⎛ ⎞

= = ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

In these units, the thickness becomes

( ) 2

210 cm3000 m 3000 m 3000 10 cm 1m

*z
*⎛ ⎞

= = = ×⎜ ⎟ ⎝ ⎠

which yields ( ) ( )25 2 22 32000 10 cm 3000 10 cm 1.9 10 cm .2*V
*π

= × × = ×

10. Since a change of longitude equal to 360° corresponds to a 24 hour change, then one expects to change longitude by360 / 24 15° = ° before resetting one's watch by 1.0 h. 11. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43. (b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864. 12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so

37 10 14 86400

31 6.

. . m m m

day s day m s

b gc h b gb g

μ μ=

13. The time on any of these clocks is a straight-line function of that on another, with
slopes ≠ 1 and *y*-intercepts ≠ 0. From the data in the figure we deduce

2 594 33 662, .
7 7 40 5*C B B A
*

*t t t t*= + = −

These are used in obtaining the following results. (a) We find

( )33 495 s
40*B B A A
*

*t t t t*′ ′− = − =

when *t'A* − *tA* = 600 s.

5

(b) We obtain ′ − = ′ − = =*t t t tC C B B
*2
7

2 7

495 141b g b g s.
(c) Clock *B* reads *tB* = (33/40)(400) − (662/5) ≈ 198 s when clock *A* reads *tA* = 400 s.
(d) From *tC* = 15 = (2/7)*tB* + (594/7), we get *tB* ≈ −245 s.
14. The metric prefixes (micro (μ), pico, nano, …) are given for ready reference on the
inside front cover of the textbook (also Table 1–2).

(a) ( )6 100 y 365 day 24 h 60 min1 century 10 century 52.6 min.1 century 1 y 1 day 1 hμ − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(b) The percent difference is therefore

52.6 min 50 min 4.9%. 52.6 min

− =

15. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600
seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix,
this is roughly 1.21 × 1012 μs.
16. We denote the pulsar rotation rate *f* (for frequency).

3

1 rotation 1.55780644887275 10 s

*f *−= ×

(a) Multiplying *f* by the time-interval *t* = 7.00 days (which is equivalent to 604800 s, if
we ignore *significant figure* considerations for a moment), we obtain the number of
rotations:

( )3 1 rotation 604800 s 388238218.4

1.55780644887275 10 s
*N *−

⎛ ⎞ = =⎜ ⎟×⎝ ⎠

which should now be rounded to 3.88 × 108 rotations since the time-interval was
specified in the problem to three significant figures.
(b) We note that the problem specifies the *exact* number of pulsar revolutions (one
million). In this case, our unknown is *t*, and an equation similar to the one we set up in
part (a) takes the form *N *= *ft*, or

6 3

1 rotation1 10 1.55780644887275 10 s

*t*−
⎛ ⎞

× = ⎜ ⎟×⎝ ⎠

*CHAPTER 1 *6

which yields the result *t* = 1557.80644887275 s (though students who do this calculation
on their calculator might not obtain those last several digits).
(c) Careful reading of the problem shows that the time-uncertainty *per revolution* is

173 10 s−± × . We therefore expect that as a result of one million revolutions, the uncertainty should be 17 6 11 ( 3 10 )(1 10 )= 3 10 s− −± × × ± × . 17. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning.

Sun. Mon. Tues. Wed. Thurs. Fri. CLOCK -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat.

A −16 −16 −15 −17 −15 −15 B −3 +5 −10 +5 +6 −7 C −58 −58 −58 −58 −58 −58 D +67 +67 +67 +67 +67 +67 E +70 +55 +2 +20 +10 +10

Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E. 18. The last day of the 20 centuries is longer than the first day by

( ) ( )20 century 0.001 s century 0.02 s.=
The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day.
Since the increase occurs uniformly, the cumulative effect *T* is

7

( ) ( )

( )

average increase in length of a day number of days

0.01 s 365.25 day 2000 y day y

7305 s

*T *=

⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ =

or roughly two hours.
19. When the Sun first disappears while lying down, your line of sight to the top of the
Sun is tangent to the Earth’s surface at point *A* shown in the figure. As you stand,
elevating your eyes by a height *h*, the line of sight to the Sun is tangent to the Earth’s
surface at point *B*.

Let *d* be the distance from point *B* to your eyes. From the Pythagorean theorem, we have
2 2 2 2 2( ) 2*d r r h r rh h*+ = + = + +
or 2 22 ,*d rh h*= + where *r* is the radius of the Earth. Since *r h *, the second term can be
dropped, leading to 2 2*d rh*≈ . Now the angle between the two radii to the two tangent
points *A* and *B* is θ, which is also the angle through which the Sun moves about Earth
during the time interval *t* = 11.1 s. The value of θ can be obtained by using

360 24 h

*t*θ
=

° .

This yields

(360 )(11.1 s) 0.04625 . (24 h)(60 min/h)(60 s/min)

θ °= = °

Using tan*d r *θ= , we have 2 2 2tan 2*d r rh*θ= = , or

2 2

tan
*hr
*

θ =

Using the above value for θ and *h* = 1.7 m, we have 65.2 10 m.*r *= ×

*CHAPTER 1 *8

20. (a) We find the volume in cubic centimeters

( ) 33

5 3231 in 2.54 cm193 gal = 193 gal 7.31 10 cm 1gal 1in

⎛ ⎞ ⎛ ⎞ = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

and subtract this from 1 × 106 cm3 to obtain 2.69 × 105 cm3. The conversion gal → in3 is given in Appendix D (immediately below the table of Volume conversions). (b) The volume found in part (a) is converted (by dividing by (100 cm/m)3) to 0.731 m3, which corresponds to a mass of

1000 kg m 0.731 m = 731 kg3 2c h c h using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in

5731kg 4.06 10 min = 0.77 y 0.0018 kg min

= ×

after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h).
21. If *ME* is the mass of Earth, *m* is the average mass of an atom in Earth, and *N* is the
number of atoms, then *ME = Nm* or *N = ME/m*. We convert mass *m* to kilograms using
Appendix D (1 u = 1.661 × 10−27 kg). Thus,

*N M
m
*

*E*= =
×

× = ×

−

598 10 40 1661 10

9 0 10 24

27 49.

. . .kg

u kg ub g c h 22. The density of gold is

3 3

19.32 g 19.32 g/cm . 1 cm

*m
V
*

ρ = = =

(a) We take the volume of the leaf to be its area *A* multiplied by its thickness *z*. With
density ρ = 19.32 g/cm3 and mass *m* = 27.63 g, the volume of the leaf is found to be

*V m*= =
ρ

1430. .cm3

We convert the volume to SI units:

9

( ) 3

3 6 31 m1.430 cm 1.430 10 m . 100 cm

*V *−
⎛ ⎞

= = ×⎜ ⎟ ⎝ ⎠

Since *V* = *Az* with *z* = 1 × 10-6 m (metric prefixes can be found in Table 1–2), we obtain

*A *= ×
×

= −

−

1430 10 1 10

1430 6

6

. . .m m

m 3

2

(b) The volume of a cylinder of length is *V A*= where the cross-section area is that of
a circle: *A* = π*r*2. Therefore, with *r* = 2.500 × 10−6 m and *V* = 1.430 × 10−6 m3, we obtain

4 2 7.284 10 m 72.84 km.

*V
r*π

= = × =

23. We introduce the notion of density:

ρ = *m
V
*

and convert to SI units: 1 g = 1 × 10−3 kg. (a) For volume conversion, we find 1 cm3 = (1 × 10−2m)3 = 1 × 10−6m3. Thus, the density in kg/m3 is

3 3 3 3 3

3 6 3

1 g 10 kg cm1 g cm 1 10 kg m . cm g 10 m

−

−

⎛ ⎞ ⎛ ⎞⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Thus, the mass of a cubic meter of water is 1000 kg.
(b) We divide the mass of the water by the time taken to drain it. The mass is found from
*M* = ρ*V* (the product of the volume of water and its density):

( ) ( )3 3 3 65700 m 1 10 kg m 5.70 10 kg.*M *= × = ×
The time is *t* = (10h)(3600 s/h) = 3.6 × 104 s, so the *mass flow rate R* is

6

4

5.70 10 kg 158 kg s. 3.6 10 s

*MR
t
*

× = = =

×

24. The metric prefixes (micro (μ), pico, nano, …) are given for ready reference on the
inside front cover of the textbook (see also Table 1–2). The surface area *A* of each grain
of sand of radius *r* = 50 μm = 50 × 10−6 m is given by *A* = 4π(50 × 10−6)2 = 3.14 × 10−8
m2 (Appendix E contains a variety of geometry formulas). We introduce the notion of

*CHAPTER 1 *10

density, /*m V*ρ = , so that the mass can be found from *m* = ρ*V*, where ρ = 2600 kg/m3.
Thus, using *V* = 4π*r*3/3, the mass of each grain is

( )363 9 3

4 50 10 m4 kg2600 1.36 10 kg.
3 m 3
*rm V
*

ππρ ρ −

− ×⎛ ⎞ ⎛ ⎞= = = = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

We observe that (because a cube has six equal faces) the indicated surface area is 6 m2.
The number of spheres (the grains of sand) *N* that have a total surface area of 6 m2 is
given by

2 8

8 2

6 m 1.91 10 . 3.14 10 m

*N *−= = ××

Therefore, the total mass *M* is ( ) ( )8 91.91 10 1.36 10 kg 0.260 kg.*M Nm *−= = × × =
25. The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0 × 106 m3. Letting “*d*”
stand for the thickness of the mud after it has (uniformly) distributed in the valley, then
its volume there would be (400 m)(400 m)*d*. Requiring these two volumes to be equal,
we can solve for *d*. Thus, *d* = 25 m. The volume of a small part of the mud over a patch
of area of 4.0 m2 is (4.0)*d* = 100 m3. Since each cubic meter corresponds to a mass of
1900 kg (stated in the problem), then the mass of that small part of the mud is

51.9 10 kg× . 26. (a) The volume of the cloud is (3000 m)π(1000 m)2 = 9.4 × 109 m3. Since each cubic meter of the cloud contains from 50 × 106 to 500 × 106 water drops, then we conclude that the entire cloud contains from 4.7 × 1018 to 4.7 × 1019 drops. Since the volume of each drop is 43 π(10 × 10

− 6 m)3 = 4.2 × 10−15 m3, then the total volume of water in a cloud

is from 32 10× to 42 10× m3.
(b) Using the fact that 3 3 3 31 L 1 10 cm 1 10 m−= × = × , the amount of water estimated in
part (a) would fill from 62 10× to 72 10× bottles.
(c) At 1000 kg for every cubic meter, the mass of water is from 62 10× to 72 10× kg.
The coincidence in numbers between the results of parts (b) and (c) of this problem is due
to the fact that each liter has a mass of one kilogram when water is at its normal density
(under standard conditions).
27. We introduce the notion of density, /*m V*ρ = , and convert to SI units: 1000 g = 1 kg,
and 100 cm = 1 m.
(a) The density ρ of a sample of iron is

11

( ) 3

3 31 kg 100 cm7.87 g cm 7870 kg/m . 1000 g 1 m

ρ ⎛ ⎞ ⎛ ⎞

= =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

If we ignore the empty spaces between the close-packed spheres, then the density of an
individual iron atom will be the same as the density of any iron sample. That is, if *M* is
the mass and *V* is the volume of an atom, then

26 29 3

3 3

9.27 10 kg 1.18 10 m . 7.87 10 kg m

*MV
*−

−×= = = × ×ρ

(b) We set *V* = 4π*R*3/3, where *R* is the radius of an atom (Appendix E contains several
geometry formulas). Solving for *R*, we find

( ) 1 329 31 3 103 1.18 10 m3 1.41 10 m.
4 4
*VR
*

− −

⎛ ⎞×⎛ ⎞ ⎜ ⎟= = = ×⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

The center-to-center distance between atoms is twice the radius, or 2.82 × 10−10 m. 28. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom (in the cat) as 10 u ≈ 2 × 10−26 kg, then there are roughly (10 kg)/( 2 × 10−26 kg) ≈ 5 × 1026 atoms. This is close to being a factor of a thousand greater than Avogadro’s number. Thus this is roughly a kilomole of atoms. 29. The mass in kilograms is

28 9 100 16 10 10 0 3779. .piculs gin 1picul

tahil 1gin

chee 1tahil

hoon 1 chee

g 1hoon

b g FHG I KJ F HG

I KJ F HG

I KJ F HG

I KJ F HG

I KJ

which yields 1.747 × 106 g or roughly 1.75× 103 kg.
30. To solve the problem, we note that the first derivative of the function with respect to
time gives the rate. Setting the rate to zero gives the time at which an extreme value of
the variable mass occurs; here that extreme value is a maximum.
(a) Differentiating 0.8( ) 5.00 3.00 20.00*m t t t*= − + with respect to *t* gives

0.24.00 3.00.*dm t
dt
*

−= −

The water mass is the greatest when / 0,*dm dt *= or at 1/ 0.2(4.00 / 3.00) 4.21s.*t *= =

*CHAPTER 1 *12

(b) At 4.21s,*t *= the water mass is

0.8( 4.21s) 5.00(4.21) 3.00(4.21) 20.00 23.2 g.*m t *= = − + =
(c) The rate of mass change at 2.00 s*t *= is

0.2

2.00 s

2

g 1 kg 60 s4.00(2.00) 3.00 g/s 0.48 g/s 0.48 s 1000 g 1 min

2.89 10 kg/min.
*t
*

*dm
dt
*

−

=

−

⎡ ⎤= − = = ⋅ ⋅⎣ ⎦

= ×

(d) Similarly, the rate of mass change at 5.00 s*t *= is

0.2

2.00 s

3

g 1 kg 60 s4.00(5.00) 3.00 g/s 0.101g/s 0.101 s 1000 g 1 min

6.05 10 kg/min.
*t
*

*dm
dt
*

−

=

−

⎡ ⎤= − = − = − ⋅ ⋅⎣ ⎦

= − ×

31. The mass density of the candy is

4 3 4 33 0.0200 g 4.00 10 g/mm 4.00 10 kg/cm .

50.0 mm
*m
V
*

ρ − −= = = × = ×

If we neglect the volume of the empty spaces between the candies, then the total mass of
the candies in the container when filled to height *h* is ,*M Ah*ρ= where

2(14.0 cm)(17.0 cm) 238 cm*A *= = is the base area of the container that remains
unchanged. Thus, the rate of mass change is given by

4 3 2( ) (4.00 10 kg/cm )(238 cm )(0.250 cm/s)

0.0238 kg/s 1.43 kg/min.

*dM d Ah dhA
dt dt dt
*

ρ ρ −= = = ×

= =

32. The total volume *V* of the real house is that of a triangular prism (of height *h* = 3.0 m
and base area *A* = 20 × 12 = 240 m2) in addition to a rectangular box (height *h*´ = 6.0 m
and same base). Therefore,

31 1800 m . 2 2

*hV hA h A h A*⎛ ⎞′ ′= + = + =⎜ ⎟
⎝ ⎠

(a) Each dimension is reduced by a factor of 1/12, and we find

*V*doll
3 3m m= FHG
I
KJ ≈1800

1 12

10 3

c h . .

13

(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore,

*V*miniature
3m 6.0 10 m= FHG
I
KJ ≈ ×

−1800 1 144

3 4 3c h .

33. In this problem we are asked to differentiate between three types of tons:
*displacement* ton, *freight* ton and *register* ton, all of which are units of volume. The three
different tons are given in terms of *barrel bulk*, with

31 barrel bulk 0.1415 m 4.0155 U.S. bushels= = using 31 m 28.378 U.S. bushels.= Thus, in terms of U.S. bushels, we have

4.0155 U.S. bushels1 displacement ton (7 barrels bulk) 28.108 U.S. bushels 1 barrel bulk

⎛ ⎞= × =⎜ ⎟ ⎝ ⎠

4.0155 U.S. bushels1 freight ton (8 barrels bulk) 32.124 U.S. bushels 1 barrel bulk

4.0155 U.S. bushels1 register ton (20 barrels bulk) 80.31 U.S. bushels 1 barrel bulk

⎛ ⎞= × =⎜ ⎟ ⎝ ⎠

⎛ ⎞= × =⎜ ⎟ ⎝ ⎠

(a) The difference between 73 “freight” tons and 73 “displacement” tons is

73(freight tons displacement tons) 73(32.124 U.S. bushels 28.108 U.S. bushels)

293.168 U.S. bushels 293 U.S. bushels

*V*Δ = − = −

= ≈ (b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is

3

73(register tons displacement tons) 73(80.31 U.S. bushels 28.108 U.S. bushels)

3810.746 U.S. bushels 3.81 10 U.S. bushels

*V*Δ = − = −

= ≈ ×
34. The customer expects a volume *V*1 = 20 × 7056 in3 and receives *V*2 = 20 × 5826 in.3,
the difference being 31 2 24600 in.*V V V*Δ = − = , or

( ) 3

3 3

2.54cm 1L24600 in. 403L 1 inch 1000 cm

*V
*⎛ ⎞ ⎛ ⎞Δ = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

where Appendix D has been used.
35. The first two conversions are easy enough that a *formal* conversion is not especially
called for, but in the interest of *practice makes perfect* we go ahead and proceed formally:

*CHAPTER 1 *14

(a) ( ) 2 peck11 tuffets = 11 tuffets 22 pecks 1 tuffet

⎛ ⎞ =⎜ ⎟

⎝ ⎠ .

(b) ( ) 0.50 Imperial bushel11 tuffets = 11 tuffets 5.5 Imperial bushels 1 tuffet

⎛ ⎞ =⎜ ⎟

⎝ ⎠ .

(c) ( ) 36.3687 L11 tuffets = 5.5 Imperial bushel 200 L 1 Imperial bushel

⎛ ⎞ ≈⎜ ⎟

⎝ ⎠ .

36. Table 7 can be completed as follows: (a) It should be clear that the first column (under “wey”) is the reciprocal of the first row – so that 910 = 0.900,

3
40 ** = **7.50 × 10

−2, and so forth. Thus, 1 pottle = 1.56 × 10−3 wey and 1 gill = 8.32 × 10−6 wey are the last two entries in the first column. (b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 chaldron (that is, the entries along the “diagonal” in the table must be 1’s). To find out how many chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 112

chaldron = 1 bag. Thus, the next entry in that second column is 112 = 8.33 × 10 −2.

Similarly, 1 pottle = 1.74 × 10−3 chaldron and 1 gill = 9.24 × 10−6 chaldron. (c) In the third column (under “bag”), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1 pottle = 2.08 × 10−2 bag, and 1 gill = 1.11 × 10−4 bag. (d) In the fourth column (under “pottle”), we find 1 chaldron = 576 pottle, 1 bag = 48 pottle, 1 pottle = 1 pottle, and 1 gill = 5.32 × 10−3 pottle. (e) In the last column (under “gill”), we obtain 1 chaldron = 1.08 × 105 gill, 1 bag = 9.02 × 103 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill. (f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since each bag is 0.1091 m3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m3. 37. The volume of one unit is 1 cm3 = 1 × 10−6 m3, so the volume of a mole of them is 6.02 × 1023 cm3 = 6.02 × 1017 m3. The cube root of this number gives the edge length:

5 38.4 10 m× . This is equivalent to roughly 8 × 102 km.
38. (a) Using the fact that the area *A* of a rectangle is (width) × (length), we find

15

( ) ( )( )

( ) ( )( ) total

2

2

3.00acre 25.0 perch 4.00 perch

40 perch 4 perch 3.00 acre 100 perch

1acre

580 perch .

*A *= +

⎛ ⎞ = +⎜ ⎟

⎝ ⎠ =

We multiply this by the perch2 → rood conversion factor (1 rood/40 perch2) to obtain the
answer: *A*total = 14.5 roods.
(b) We convert our intermediate result in part (a):

( ) 2

2 5 2 total

16.5ft580 perch 1.58 10 ft . 1perch

*A
*⎛ ⎞

= = ×⎜ ⎟ ⎝ ⎠

Now, we use the feet → meters conversion given in Appendix D to obtain

( ) 2

5 2 4 2 total

1m1.58 10 ft 1.47 10 m . 3.281ft

*A
*⎛ ⎞

= × = ×⎜ ⎟ ⎝ ⎠

39. This problem compares the U.K gallon with U.S. gallon, two non-SI units for volume.
The interpretation of the type of gallons, whether U.K. or U.S., affects the amount of
gasoline one calculates for traveling a given distance.
If the fuel consumption rate is *R * (in miles/gallon), then the amount of gasoline (in
gallons) needed for a trip of distance *d* (in miles) would be

(miles)(gallon)
(miles/gallon)
*dV
*

*R
*=

Since the car was manufactured in the U.K., the fuel consumption rate is calibrated based on U.K. gallon, and the correct interpretation should be “40 miles per U.K. gallon.” In U.K., one would think of gallon as U.K. gallon; however, in the U.S., the word “gallon” would naturally be interpreted as U.S. gallon. Note also that since 1 U.K. gallon 4.5460900 L= and 1 U.S. gallon 3.7854118 L= , the relationship between the two is

1 U.S. gallon1 U.K. gallon (4.5460900 L) 1.20095 U.S. gallons 3.7854118 L

⎛ ⎞= =⎜ ⎟ ⎝ ⎠

(a) The amount of gasoline actually required is

750 miles 18.75 U. K. gallons 18.8 U. K. gallons 40 miles/U. K. gallon

*V *′ = = ≈

*CHAPTER 1 *16

This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons. (b) Using the conversion factor found above, the actual amount required is equivalent to

( ) 1.20095 U.S. gallons18.75 U. K. gallons 22.5 U.S. gallons 1 U.K. gallon

*V
*⎛ ⎞′ = × ≈⎜ ⎟
⎝ ⎠

.

40. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to simply taking the reciprocal of the number given in Eq. 1-9 and rounding off appropriately. Thus, the answer is 6.0 × 1026. 41. Using the (exact) conversion 1 in = 2.54 cm = 0.0254 m, we find that

0.0254 m1 ft 12 in. (12 in.) 0.3048 m 1in.

⎛ ⎞ = = × =⎜ ⎟

⎝ ⎠

and 3 3 31 ft (0.3048 m) 0.0283 m= = for volume (these results also can be found in
Appendix D). Thus, the volume of a cord of wood is 3(8 ft) (4 ft) (4 ft) 128 ft*V *= × × = .
Using the conversion factor found above, we obtain

3 3 3 3

3

0.0283 m1 cord 128 ft (128 ft ) 3.625 m 1 ft

*V
*⎛ ⎞

= = = × =⎜ ⎟ ⎝ ⎠

which implies that 3 11 m cord 0.276 cord 0.3 cord 3.625

⎛ ⎞= = ≈⎜ ⎟ ⎝ ⎠

.

42. (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eq. 1-9, we find

( ) 27

261.6605402 10 kg18u = 18u 3.0 10 kg. 1u

− −⎛ ⎞× = ×⎜ ⎟

⎝ ⎠

(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules:

21 46

26

1.4 10 5 10 . 3.0 10

*N *−
×

≈ ≈ × ×

43. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 × 106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3 × 106 mg/week)/(604800 s/week) = 3.8 mg/s. 44. The volume of the water that fell is

17

( ) ( ) ( ) ( ) ( ) ( )

2 2 2

6 2

6 3

1000 m 0.0254 m26 km 2.0 in. 26 km 2.0 in. 1 km 1 in.

26 10 m 0.0508 m 1.3 10 m .

*V
*⎛ ⎞ ⎛ ⎞

= = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

= ×

= ×

We write the mass-per-unit-volume (density) of the water as:

3 31 10 kg m .*m
V
*

= = ×ρ

The mass of the water that fell is therefore given by *m* = ρ*V*:

( ) ( )3 3 6 3 91 10 kg m 1.3 10 m 1.3 10 kg.*m *= × × = ×
45. The number of seconds in a year is 3.156 × 107. This is listed in Appendix D and
results from the product

(365.25 day/y) (24 h/day) (60 min/h) (60 s/min). (a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year. (b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by

10 10

10 6

10 4= − u - day,

which may also be expressed as 10 86400 1

8 64− F HG

I KJ =u - day

u - sec u - day

u - sec.c h . 46. The volume removed in one year is

*V* = (75 10 m ) (26 m) 2 10 m4 2 7 3× ≈ ×

which we convert to cubic kilometers: *V *= ×
F
HG

I KJ =2 10

1 0 0207 3

m km 1000 m

km3 3c h . . 47. We convert meters to astronomical units, and seconds to minutes, using

*CHAPTER 1 *18

8

1000 m 1 km 1 AU 1.50 10 km 60 s 1 min .

=

= × =

Thus, 3.0 × 108 m/s becomes

8

8

3.0 10 m 1 km AU 60 s 0.12 AU min. s 1000 m 1.50 10 km min

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞× ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎟⎜⎟ ⎟ ⎟ =⎜ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟⎟⎜ ⎝ ⎠×⎝ ⎠ ⎝ ⎠⎝ ⎠

48. Since one atomic mass unit is 241 u 1.66 10 g−= × (see Appendix D), the mass of one
mole of atoms is about 24 23(1.66 10 g)(6.02 10 ) 1g.*m *−= × × = On the other hand, the mass
of one mole of atoms in the common Eastern mole is

75 g 10 g 7.5

*m*′ = =

Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is

23 23

10 g 1.66 10 g 10 u.
6.02 10*A
*

*m
N
*

−′ = = × = ×

49. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain

1 1

1 97 1

3 88 2ken

m m

m

2

2

2

2= = . . .

(b) Similarly, we find 1 1

197 1

7 653 3ken

m m

m

3 3

3= = . . .

(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus,

( ) ( )22 33.00 5.50 156 ken .*r h*π π= =
(d) If we multiply this by the result of part (b), we determine the volume in cubic meters:
(155.5)(7.65) = 1.19 × 103 m3.
50. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would
be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is
39.4205 km. The difference is 5.95 km.
51. (a) For the minimum (43 cm) case, 9 cubits converts as follows:

19

( ) 0.43m9cubits 9cubits 3.9m. 1cubit

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

And for the maximum (53 cm) case we obtain

( ) 0.53m9cubits 9cubits 4.8m. 1cubit

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

(b) Similarly, with 0.43 m → 430 mm and 0.53 m → 530 mm, we find 3.9 × 103 mm and
4.8 × 103 mm, respectively.
(c) We can convert length and diameter first and then compute the volume, or first
compute the volume and then convert. We proceed using the latter approach (where *d* is
diameter and is length).

( ) 3

2 3 3 3 cylinder, min

0.43m28 cubit 28 cubit 2.2 m . 4 1 cubit

*V d*π
⎛ ⎞

= = = =⎜ ⎟ ⎝ ⎠

Similarly, with 0.43 m replaced by 0.53 m, we obtain *V*cylinder, max = 4.2 m3.
52. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the
ratio 25 wp/11 barn along with appropriate conversion factors:

( ) ( ) ( ) ( ) ( ) ( )

2

28 2 36

100 hide 110 acre 4047 m 1 wp 1acre1 hide

1 10 m 1 barn

25 wp 1 10 .

11 barn −×

≈ ×

53. The objective of this problem is to convert the Earth-Sun distance to parsecs and
light-years. To relate parsec (pc) to AU, we note that when θ is measured in radians, it is
equal to the arc length *s* divided by the radius *R*. For a very large radius circle and small
value of θ, the arc may be approximated as the straight line-segment of length 1 AU.
Thus,

( ) 61 arcmin 1 2 radian1 arcsec 1 arcsec 4.85 10 rad 60 arcsec 60 arcmin 360

θ − ⎛ ⎞⎛ ⎞° π⎛ ⎞= = = ×⎜ ⎟⎜ ⎟⎜ ⎟°⎝ ⎠⎝ ⎠⎝ ⎠

Therefore, one parsec is

5o 6 1 AU1 pc 2.06 10 AU

4.85 10
*sR
*θ −

= = = = × ×

Next, we relate AU to light-year (ly). Since a year is about 3.16 × 107 s, we have

( ) ( )7 121ly 186,000mi s 3.16 10 s 5.9 10 mi= × = × .

(a) Since 51 pc 2.06 10 AU= × , inverting the relationship gives

*CHAPTER 1 *20

( ) 65 1 pc1 AU 1 AU 4.9 10 pc.

2.06 10 AU
*R *−

⎛ ⎞ = = = ×⎜ ⎟×⎝ ⎠

(b) Given that 61AU 92.9 10 mi= × and 121ly 5.9 10 mi= × , the two expressions together lead to

6 6 5 5 12

1 ly1 AU 92.9 10 mi (92.9 10 mi) 1.57 10 ly 1.6 10 ly 5.9 10 mi

− −⎛ ⎞= × = × = × ≈ ×⎜ ⎟×⎝ ⎠ .

Our results can be further combined to give 1 pc 3.2 ly= . 54. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3 = 1.639 × 10−2 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460 ft2/gal becomes

22 2 2460 ft 1 m 1 gal460 ft /gal 11.3 m L.

gal 3.28 ft 3.79 L ⎛ ⎞⎛ ⎞ ⎛ ⎞

= =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

(b) Also, since 1 m3 is equivalent to 1000 L, our result from part (a) becomes

2 2 4 1

3

11.3 m 1000 L11.3 m /L 1.13 10 m . L 1 m

−⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

(c) The inverse of the original quantity is (460 ft2/gal)−1 = 2.17 × 10−3 gal/ft2. (d) The answer in (c) represents the volume of the paint (in gallons) needed to cover a square foot of area. From this, we could also figure the paint thickness [it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part (b)].

21

**Chapter 2
**1.The speed (assumed constant) is *v* = (90 km/h)(1000 m/km) ⁄ (3600 s/h) = 25 m/s.
Thus, in 0.50 s, the car travels a distance *d* = *vt* = (25 m/s)(0.50 s) ≈ 13 m.
2. (a) Using the fact that time = distance/velocity while the velocity is constant, we
find

avg 73.2 m 73.2 m 3.05 m1.22 m/s

73.2 m 73.2 m 1.74 m/s.*v *+= =
+

(b) Using the fact that distance = *vt* while the velocity *v* is constant, we find

*v*avg
m / s)(60 s) m / s)(60 s)

s m / s.= + =( . ( . .122 3 05

120 2 14

(c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.

3. Since the trip consists of two parts, let the displacements during first and second
parts of the motion be Δ*x*1 and Δ*x*2, and the corresponding time intervals be Δ*t*1 and Δ*t*2,
respectively. Now, because the problem is one-dimensional and both displacements
are in the same direction, the total displacement is Δ*x* = Δ*x*1 + Δ*x*2, and the total time
for the trip is Δ*t* = Δ*t*1 + Δ*t*2. Using the definition of average velocity given in Eq. 2-2,
we have

1 2 avg

1 2

.*x xxv
t t t
*

Δ + ΔΔ = =

Δ Δ + Δ

To find the average speed, we note that during a time Δ*t* if the velocity remains a
positive constant, then the speed is equal to the magnitude of velocity, and the
distance is equal to the magnitude of displacement, with | |*d x v t*= Δ = Δ .

*CHAPTER 2
*

22

(a) During the first part of the motion, the displacement is Δ*x*1 = 40 km and the time
interval is

*t*1
40 133= =( . km)

(30 km / h) h.

Similarly, during the second part the displacement is Δ*x*2 = 40 km and the time
interval is

*t*2
40 0 67= =( . km)

(60 km / h) h.

The total displacement is Δ*x* = Δ*x*1 + Δ*x*2 = 40 km + 40 km = 80 km, and the total time
elapsed is Δ*t* = Δ*t*1 + Δ*t*2 = 2.00 h. Consequently, the average velocity is

avg (80 km) 40 km/h. (2.0 h)

*xv
t
*

Δ = = =

Δ

(b) In this case, the average speed is the same as the magnitude of the average
velocity: avg 40 km/h.*s *=
(c) The graph of the entire trip is shown below; it consists of two contiguous line
segments, the first having a slope of 30 km/h and connecting the origin to (Δ*t*1, Δ*x*1) =
(1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (Δ*t*1, Δ*x*1)
to (Δ*t, *Δ*x*) = (2.00 h, 80 km).

4. Average speed, as opposed to average velocity, relates to the total distance, as
opposed to the net displacement. The distance *D* up the hill is, of course, the same as
the distance down the hill, and since the speed is constant (during each stage of the
motion) we have speed = *D*/*t*. Thus, the average speed is

*D D
t t
*

*D
D
v
*

*D
v
*

up down

up down

up down

+

+ =

+

2

which, after canceling *D* and plugging in *v*up = 40 km/h and *v*down = 60 km/h, yields 48
km/h for the average speed.
5. Using *x* = 3*t* – 4*t*2 + *t*3 with SI units understood is efficient (and is the approach we

23

will use), but if we wished to make the units explicit we would write

*x* = (3 m/s)*t* – (4 m/s2)*t*2 + (1 m/s3)*t*3.
We will quote our answers to one or two significant figures, and not try to follow the
significant figure rules rigorously.
(a) Plugging in *t* = 1 s yields *x* = 3 – 4 + 1 = 0.
(b) With *t* = 2 s we get *x* = 3(2) – 4(2)2+(2)3 = –2 m.
(c) With *t* = 3 s we have *x* = 0 m.
(d) Plugging in *t* = 4 s gives *x* = 12 m.
For later reference, we also note that the position at *t* = 0 is *x* = 0.
(e) The position at *t* = 0 is subtracted from the position at *t* = 4 s to find the
displacement Δ*x* = 12 m.
(f) The position at *t *= 2 s is subtracted from the position at *t *= 4 s to give the
displacement Δ*x* = 14 m. Eq. 2-2, then, leads to

avg 14 m 7 m/s.

2 s
*xv
t
*

Δ = = =

Δ

(g) The position of the object for the interval 0 ≤ *t* ≤ 4 is plotted below. The straight
line drawn from the point at (*t, x*) = (2 s , –2 m) to (4 s, 12 m) would represent the
average velocity, answer for part (f).

6. Huber’s speed is

*v*0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h,
where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat
Huber by 19.0 km/h, his speed is *v*1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36
m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2-2, the time through a distance of 200 m
for Whittingham is

*CHAPTER 2
*

24

1

200 m 5.554 s. 36 m/s

*xt
v
*Δ

Δ = = =

7. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h,
the total time that elapses before they crash is *t* = (60 km)/(60 km/h) = 1.0 h. During
this time, the bird travels a distance of *x* = *vt* = (60 km/h)(1.0 h) = 60 km.
8. The amount of time it takes for each person to move a distance *L* with speed *sv * is

/ *st L v*Δ = . With each additional person, the depth increases by one body depth *d*
(a) The rate of increase of the layer of people is

(0.25 m)(3.50 m/s) 0.50 m/s / 1.75 m

*s
*

*s
*

*dvd dR
t L v L
*

= = = = = Δ

(b) The amount of time required to reach a depth of 5.0 m*D *= is

5.0 m 10 s 0.50 m/s

*Dt
R
*

= = =

9. Converting to seconds, the running times are *t*1 = 147.95 s and *t*2 = 148.15 s,
respectively. If the runners were equally fast, then

1 2 avg avg1 2

1 2

.*L Ls s
t t
*

= ⇒ =

From this we obtain

2 2 1 1 1 1

1

148.151 1 0.00135 1.4 m 147.95

*tL L L L L
t
*

⎛ ⎞ ⎛ ⎞− = − = − = ≈⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

where we set *L*1 ≈ 1000 m in the last step. Thus, if *L*1 and *L*2 are no different than
about 1.4 m, then runner 1 is indeed faster than runner 2. However, if *L*1 is shorter
than *L*2 by more than 1.4 m, then runner 2 would actually be faster.
10. Let *wv *be the speed of the wind and *cv * be the speed of the car.
(a) Suppose during time interval 1*t *, the car moves in the same direction as the wind.
Then the effective speed of the car is given by ,1*eff c wv v v*= + , and the distance traveled
is ,1 1 1( )*eff c wd v t v v t*= = + . On the other hand, for the return trip during time interval *t*2,
the car moves in the opposite direction of the wind and the effective speed would be

,2*eff c wv v v*= − . The distance traveled is ,2 2 2( )*eff c wd v t v v t*= = − . The two expressions
can be rewritten as

25

1 2

and*c w c w
d dv v v v
t t
*

+ = − =

Adding the two equations and dividing by two, we obtain 1 2

1
2*c
*

*d dv
t t
*

⎛ ⎞ = +⎜ ⎟

⎝ ⎠ . Thus,

method 1 gives the car’s speed *cv *a in windless situation.
(b) If method 2 is used, the result would be

22 2

1 2 1 2

2 2 1 ( ) / 2

*c w w
c c
*

*c c
*

*c w c w
*

*v v vd d dv vd dt t t t v v
v v v v
*

⎡ ⎤⎛ ⎞−′ ⎢ ⎥= = = = = − ⎜ ⎟+ + ⎢ ⎥⎝ ⎠+ ⎣ ⎦ + −

.

The fractional difference is 2

2 4(0.0240) 5.76 10*c c w
c c
*

*v v v
v v
*

−⎛ ⎞′− = = = ×⎜ ⎟ ⎝ ⎠

.

11. The values used in the problem statement make it easy to see that the first part of
the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour.
Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is
160 km. Thus, a speed *v* = (160 km)/(1.25 h) = 128 km/h is needed.
12. (a) Let the fast and the slow cars be separated by a distance *d* at *t* = 0. If during the
time interval / (12.0 m) /(5.0 m/s) 2.40 s*st L v*= = = in which the slow car has moved
a distance of 12.0 m*L *= , the fast car moves a distance of *vt d L*= + to join the line
of slow cars, then the shock wave would remain stationary. The condition implies a
separation of
(25 m/s)(2.4 s) 12.0 m 48.0 m.*d vt L*= − = − =
(b) Let the initial separation at 0*t *= be 96.0 m.*d *= At a later time *t*, the slow and
the fast cars have traveled *sx v t*= and the fast car joins the line by moving a distance
*d x*+ . From

,
*s
*

*x d xt
v v
*

+ = =

we get 5.00 m/s (96.0 m) 24.0 m,

25.0 m/s 5.00 m/s
*s
*

*s
*

*vx d
v v
*

= = = − −

which in turn gives (24.0 m) /(5.00 m/s) 4.80 s.*t *= = Since the rear of the slow-car
pack has moved a distance of 24.0 m 12.0 m 12.0 m*x x L*Δ = − = − = downstream, the
speed of the rear of the slow-car pack, or equivalently, the speed of the shock wave, is

shock 12.0 m 2.50 m/s. 4.80 s

*xv
t
*

Δ = = =

*CHAPTER 2
*

26

(c) Since *x L*> , the direction of the shock wave is downstream.
13. (a) Denoting the travel time and distance from San Antonio to Houston as *T* and *D*,
respectively, the average speed is

avg1
(55 km/h)( /2) (90 km/h)( / 2) 72.5 km/h*D T Ts
*

*T T
*+

= = =

which should be rounded to 73 km/h. (b) Using the fact that time = distance/speed while the speed is constant, we find

avg2 / 2 / 2 55 km/h 90 km/h

68.3 km/h*D D
D Ds
T
*

= = = +

which should be rounded to 68 km/h.
(c) The total distance traveled (2*D*) must not be confused with the net displacement
(zero). We obtain for the two-way trip

avg 72.5 km/h 68.3 km/h

2 70 km/h.*D D
Ds *= =
+

(d) Since the net displacement vanishes, the average velocity for the trip in its entirety
is zero.
(e) In asking for a *sketch*, the problem is allowing the student to arbitrarily set the
distance *D* (the intent is *not* to make the student go to an atlas to look it up); the
student can just as easily arbitrarily set *T* instead of *D*, as will be clear in the following
discussion. We briefly describe the graph (with kilometers-per-hour understood for
the slopes): two contiguous line segments, the first having a slope of 55 and
connecting the origin to (*t*1, *x*1) = (*T*/2, 55*T*/2) and the second having a slope of 90 and
connecting (*t*1, *x*1) to (*T, D*) where *D* = (55 + 90)*T*/2. The average velocity, from the
graphical point of view, is the slope of a line drawn from the origin to (*T, D*). The
graph (not drawn to scale) is depicted below:

27

14. Using the general property *ddx bx b bx*exp( ) exp( )= , we write

*v dx
dt
*

*d t
dt
*

*e t de
dt
*

*t
t
*

= = FHG I KJ ⋅ + ⋅

F HG I KJ

− −( ) ( )19 19 .

If a concern develops about the appearance of an argument of the exponential (–*t*)
apparently having units, then an explicit factor of 1/*T* where *T* = 1 second can be
inserted and carried through the computation (which does not change our answer).
The result of this differentiation is

*v t e t*= − −16 1( )
with *t* and *v* in SI units (s and m/s, respectively). We see that this function is zero
when *t* = 1 s. Now that we know *when* it stops, we find out *where* it stops by
plugging our result *t* = 1 into the given function *x* = 16*te–t* with *x* in meters. Therefore,
we find *x* = 5.9 m.
15. We use Eq. 2-4 to solve the problem.
(a) The velocity of the particle is

*v dx
dt
*

*d
dt
*

*t t t*= = − + = − + ( ) .4 12 3 12 62

Thus, at *t* = 1 s, the velocity is *v* = (–12 + (6)(1)) = –6 m/s.
(b) Since *v* < 0, it is moving in the –*x* direction at *t* = 1 s.
(c) At *t* = 1 s, the *speed* is |*v*| = 6 m/s.
(d) For 0 < *t* < 2 s, |*v*| decreases until it vanishes. For 2 < *t* < 3 s, |*v*| increases from
zero to the value it had in part (c). Then, |*v*| is larger than that value for *t* > 3 s.
(e) Yes, since *v* smoothly changes from negative values (consider the *t* = 1 result) to
positive (note that as *t* → + ∞, we have *v* → + ∞). One can check that *v* = 0 when

2 s.*t *=
(f) No. In fact, from *v* = –12 + 6*t*, we know that *v* > 0 for *t* > 2 s.
16. We use the functional notation *x*(*t*), *v*(*t*), and *a*(*t*) in this solution, where the latter
two quantities are obtained by differentiation:

*v t
dx t
*

*dt
t a t
*

*dv t
dt
*

b g b g b g b g= = − = = −12 12and
with SI units understood.
(a) From *v*(*t*) = 0 we find it is (momentarily) at rest at *t* = 0.

*CHAPTER 2
*

28

(b) We obtain *x*(0) = 4.0 m.
(c) and (d) Requiring *x*(*t*) = 0 in the expression *x*(*t*) = 4.0 – 6.0*t*2 leads to *t* = ±0.82 s
for the times when the particle can be found passing through the origin.
(e) We show both the asked-for graph (on the left) as well as the “shifted” graph that
is relevant to part (f). In both cases, the time axis is given by –3 ≤ *t* ≤ 3 (SI units
understood).

(f) We arrived at the graph on the right (shown above) by adding 20*t* to the *x*(*t*)
expression.
(g) Examining where the slopes of the graphs become zero, it is clear that the shift
causes the *v* = 0 point to correspond to a larger value of *x* (the top of the second curve
shown in part (e) is higher than that of the first).
17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and
work with distances in centimeters and times in seconds.
(a) We plug into the given equation for *x* for *t* = 2.00 s and *t* = 3.00 s and obtain *x*2 =
21.75 cm and *x*3 = 50.25 cm, respectively. The average velocity during the time
interval 2.00 ≤ *t* ≤ 3.00 s is

*v x
t*avg

cm cm s s

= = − −

Δ Δ

50 25 2175 3 00 2 00 . .

. .

which yields *v*avg = 28.5 cm/s.
(b) The instantaneous velocity is *v tdxdt*= = 4 5

2. , which, at time *t* = 2.00 s, yields *v* =
(4.5)(2.00)2 = 18.0 cm/s.
(c) At *t* = 3.00 s, the instantaneous velocity is *v* = (4.5)(3.00)2 = 40.5 cm/s.
(d) At *t* = 2.50 s, the instantaneous velocity is *v* = (4.5)(2.50)2 = 28.1 cm/s.
(e) Let *tm* stand for the moment when the particle is midway between *x*2 and *x*3 (that is,
when the particle is at *xm* = (*x*2 + *x*3)/2 = 36 cm). Therefore,

*x t tm m m*= + ⇒ =9 75 15 2 596
3. . .

in seconds. Thus, the instantaneous speed at this time is *v* = 4.5(2.596)2 = 30.3 cm/s.

29

(f) The answer to part (a) is given by the slope of the straight line between *t* = 2 and *t*
= 3 in this *x*-vs-*t* plot. The answers to parts (b), (c), (d), and (e) correspond to the
slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate
points.

18. (a) Taking derivatives of *x*(*t*) = 12*t*2 – 2*t*3 we obtain the velocity and the
acceleration functions:

*v*(*t*) = 24*t* – 6*t*2 and *a*(*t*) = 24 – 12*t*

with length in meters and time in seconds. Plugging in the value *t* = 3 yields
(3) 54 m*x *= .

(b) Similarly, plugging in the value *t* = 3 yields *v*(3) = 18 m/s.
(c) For *t* = 3, *a*(3) = –12 m/s2.
(d) At the maximum *x*, we must have *v* = 0; eliminating the *t* = 0 root, the velocity
equation reveals *t* = 24/6 = 4 s for the time of maximum *x*. Plugging *t* = 4 into the
equation for *x* leads to *x* = 64 m for the largest *x* value reached by the particle.
(e) From (d), we see that the *x* reaches its maximum at *t* = 4.0 s.
(f) A maximum v requires *a* = 0, which occurs when *t* = 24/12 = 2.0 s. This, inserted
into the velocity equation, gives *v*max = 24 m/s.
(g) From (f), we see that the maximum of *v* occurs at *t* = 24/12 = 2.0 s.
(h) In part (e), the particle was (momentarily) motionless at *t* = 4 s. The acceleration at
that time is readily found to be 24 – 12(4) = –24 m/s2.
(i) The *average velocity* is defined by Eq. 2-2, so we see that the values of *x* at *t* = 0
and *t* = 3 s are needed; these are, respectively, *x* = 0 and *x* = 54 m (found in part (a)).
Thus,

*v*avg =
54 0
3 0

− −

= 18 m/s.

19. We represent the initial direction of motion as the +*x* direction. The average
acceleration over a time interval 1 2*t t t*≤ ≤ is given by Eq. 2-7:

*CHAPTER 2
*

30

2 1 avg

2 1

( ) ( ) .*v t v tva
t t t
*

−Δ = =

Δ −

Let *v*1 = +18 m/s at 1 0*t *= and *v*2 = –30 m/s at *t*2 = 2.4 s. Using Eq. 2-7 we find

22 1 avg

2 1

( ) ( ) ( 30 m/s) ( 1 m/s) 20 m/s 2.4 s 0

*v t v ta
t t
*

− − − + = = = −

− − .

The average acceleration has magnitude 20 m/s2 and is in the opposite direction to the
particle’s initial velocity. This makes sense because the velocity of the particle is
decreasing over the time interval.
20. We use the functional notation *x*(*t*), *v*(*t*) and *a*(*t*) and find the latter two quantities
by differentiating:

*v t
dx t
*

*t
t a t
*

*dv t
dt
*

*t*b g b g b g b g= = − + = = −15 20 302 and
with SI units understood. These expressions are used in the parts that follow.
(a) From 0 15 202= − +*t *, we see that the only positive value of *t* for which the
particle is (momentarily) stopped is *t *= =20 15 12/ . s .
(b) From 0 = – 30*t*, we find *a*(0) = 0 (that is, it vanishes at *t* = 0).
(c) It is clear that *a*(*t*) = – 30*t* is negative for *t* > 0.
(d) The acceleration *a*(*t*) = – 30*t* is positive for *t* < 0.
(e) The graphs are shown below. SI units are understood.

21. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices, the initial position of the man is taken as the origin and his

31

direction of motion during 5 min ≤ *t* ≤ 10 min is taken to be the positive *x* direction.
We also use the fact that Δ Δ*x v t*= ' when the velocity is constant during a time
interval Δ*t *' .
(a) The entire interval considered is Δ*t* = 8 – 2 = 6 min, which is equivalent to 360 s,
whereas the sub-interval in which he is *moving* is only 8 5 3min 180 s.*t '*Δ = − = =
His position at *t* = 2 min is *x* = 0 and his position at *t* = 8 min is *x v t*′= Δ =
(2.2)(180) 396 m= . Therefore,

*v*avg
m

s m / s= − =396 0

360 110. .

(b) The man is at rest at *t* = 2 min and has velocity *v* = +2.2 m/s at *t* = 8 min. Thus,
keeping the answer to 3 significant figures,

*a*avg
2 m / s

s m / s= − =2 2 0

360 0 00611. . .

(c) Now, the entire interval considered is Δ*t* = 9 – 3 = 6 min (360 s again), whereas the
sub-interval in which he is moving is 9 5 4 min 240 s*t*′Δ = − = = ). His position at

3 min*t *= is *x* = 0 and his position at *t* = 9 min is (2.2)(240) 528 m*x v t*′= Δ = = .
Therefore,

*v*avg
m

s m / s.= − =528 0

360 147.

(d) The man is at rest at *t* = 3 min and has velocity *v* = +2.2 m/s at *t* = 9 min.
Consequently, *a*avg = 2.2/360 = 0.00611 m/s2 just as in part (b).
(e) The horizontal line near the bottom of this *x*-vs-*t* graph represents the man
standing at *x* = 0 for 0 ≤ *t* < 300 s and the linearly rising line for 300 ≤ *t* ≤ 600 s
represents his constant-velocity motion. The lines represent the answers to part (a)
and (c) in the sense that their slopes yield those results.

The graph of *v*-vs-*t* is not shown here, but would consist of two horizontal “steps”
(one at *v* = 0 for 0 ≤ *t* < 300 s and the next at *v* = 2.2 m/s for 300 ≤ *t* ≤ 600 s). The
indications of the average accelerations found in parts (b) and (d) would be dotted
lines connecting the “steps” at the appropriate *t* values (the slopes of the dotted lines
representing the values of *a*avg).

*CHAPTER 2
*

32

22. In this solution, we make use of the notation *x*(*t*) for the value of *x* at a particular *t*.
The notations *v*(*t*) and *a*(*t*) have similar meanings.
(a) Since the unit of *ct*2 is that of length, the unit of *c* must be that of length/time2, or
m/s2 in the SI system.
(b) Since *bt*3 has a unit of length, *b* must have a unit of length/time3, or m/s3.
(c) When the particle reaches its maximum (or its minimum) coordinate its velocity is
zero. Since the velocity is given by *v* = *dx*/*dt* = 2*ct* – 3*bt*2, *v* = 0 occurs for *t* = 0 and
for

*t c
b
*

= = = 2 3

2 30 3 2 0

10( . ) ( . )

. m / s m / s

s . 2

3

For *t* = 0, *x* = *x*0 = 0 and for *t* = 1.0 s, *x* = 1.0 m > *x*0. Since we seek the maximum, we
reject the first root (*t* = 0) and accept the second (*t* = 1s).
(d) In the first 4 s the particle moves from the origin to *x* = 1.0 m, turns around, and
goes back to

*x*( ( . )( . ( . )( .4 30 4 0 2 0 4 0 802 s) m / s s) m / s s) m .2 3 3= − = −
The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m.
(e) Its displacement is Δ*x* = *x*2 – *x*1, where *x*1 = 0 and *x*2 = –80 m. Thus, 80 m*x*Δ = − .
The velocity is given by *v* = 2*ct* – 3*bt*2 = (6.0 m/s2)*t* – (6.0 m/s3)*t*2.
(f) Plugging in *t *= 1 s, we obtain

2 3 2(1 s) (6.0 m/s )(1.0 s) (6.0 m/s )(1.0 s) 0.*v *= − =
(g) Similarly, 2 3 2(2 s) (6.0 m/s )(2.0 s) (6.0 m/s )(2.0 s) 12m/s .*v *= − = −
(h) 2 3 2(3 s) (6.0 m/s )(3.0 s) (6.0 m/s )(3.0 s) 36 m/s .*v *= − = −
(i) 2 3 2(4 s) (6.0 m/s )(4.0 s) (6.0 m/s )(4.0 s) 72 m/s .*v *= − = −

The acceleration is given by *a* = *dv*/*dt* = 2*c* – 6*b* = 6.0 m/s2 – (12.0 m/s3)*t*.
(j) Plugging in *t *= 1 s, we obtain
2 3 2(1 s) 6.0 m/s (12.0 m/s )(1.0 s) 6.0 m/s .*a *= − = −
(k) 2 3 2(2 s) 6.0 m/s (12.0 m/s )(2.0 s) 18 m/s .*a *= − = −

33

(l) 2 3 2(3 s) 6.0 m/s (12.0 m/s )(3.0 s) 30 m/s .*a *= − = −
(m) 2 3 2(4 s) 6.0 m/s (12.0 m/s )(4.0 s) 42 m/s .*a *= − = −
23. Since the problem involves constant acceleration, the motion of the electron can
be readily analyzed using the equations in Table 2-1:

0

2 0 0

2 2 0 0

(2 11)

1 (2 15) 2

2 ( ) (2 16)

*v v at
*

*x x v t at
*

*v v a x x
*

= + −

− = + −

= + − −

The acceleration can be found by solving Eq. (2-16). With 50 1.50 10 m/s*v *= × ,

65.70 10 m/s*v *= × , *x*0 = 0 and *x* = 0.010 m, we find the average acceleration to be

2 2 6 2 5 2

15 20 (5.7 10 m/s) (1.5 10 m/s) 1.62 10 m/s . 2 2(0.010 m)

*v va
x
*

− × − × = = = ×

24. In this problem we are given the initial and final speeds, and the displacement, and
are asked to find the acceleration. We use the constant-acceleration equation given in
Eq. 2-16, *v*2 = *v*20 + 2*a*(*x* – *x*0).
(a) Given that 0 0*v *= , 1.6 m/s,*v *= and 5.0 m,*x *μΔ = the acceleration of the spores
during the launch is

2 2 2 5 2 40

6

(1.6 m/s) 2.56 10 m/s 2.6 10 2 2(5.0 10 m)

*v va g
x *−

− = = = × = ×

×

(b) During the speed-reduction stage, the acceleration is

2 2 2 3 2 20

3

0 (1.6 m/s) 1.28 10 m/s 1.3 10 2 2(1.0 10 m)

*v va g
x *−

− − = = = − × = − ×

×

The negative sign means that the spores are decelerating.
25. We separate the motion into two parts, and take the direction of motion to be
positive. In part 1, the vehicle accelerates from rest to its highest speed; we are
given *v*0 = 0; *v* = 20 m/s and *a* = 2.0 m/s2. In part 2, the vehicle decelerates from its
highest speed to a halt; we are given *v*0 = 20 m/s; *v* = 0 and *a* = –1.0 m/s2 (negative
because the acceleration vector points opposite to the direction of motion).
(a) From Table 2-1, we find *t*1 (the duration of part 1) from *v* = *v*0 + *at*. In this way,

120 0 2.0*t*= + yields *t*1 = 10 s. We obtain the duration *t*2 of part 2 from the same
equation. Thus, 0 = 20 + (–1.0)*t*2 leads to *t*2 = 20 s, and the total is *t* = *t*1 + *t*2 = 30 s.
(b) For part 1, taking *x*0 = 0, we use the equation *v*2 = *v*20 + 2*a*(*x* – *x*0) from Table 2-1

*CHAPTER 2
*

34

and find

2 2 2 2

0 2

(20 m/s) (0) 100 m 2 2(2.0 m/s )

*v vx
a
*

− − = = = .

This position is then the *initial* position for part 2, so that when the same equation is
used in part 2 we obtain

2 2 2 2 0

2

(0) (20 m/s)100 m 2 2( 1.0 m/s )

*v vx
a
*

− − − = =

− .

Thus, the final position is *x* = 300 m. That this is also the total distance traveled
should be evident (the vehicle did not "backtrack" or reverse its direction of motion).
26. The constant-acceleration condition permits the use of Table 2-1.
(a) Setting *v* = 0 and *x*0 = 0 in 2 20 02 ( )*v v a x x*= + − , we find

2 6 2 0

14

1 1 (5.00 10 ) 0.100 m . 2 2 1.25 10

*vx
a
*

× = − = − =

− ×

Since the muon is slowing, the initial velocity and the acceleration must have opposite
signs.
(b) Below are the time plots of the position *x* and velocity *v* of the muon from the
moment it enters the field to the time it stops. The computation in part (a) made no
reference to *t*, so that other equations from Table 2-1 (such as *v v at*= +0 and
*x v t at*= +0 12

2 ) are used in making these plots.

27. We use *v* = *v*0 + *at*, with *t* = 0 as the instant when the velocity equals +9.6 m/s.
(a) Since we wish to calculate the velocity for a time *before t* = 0, we set *t* = –2.5 s.
Thus, Eq. 2-11 gives

*v *= + − =( . . ( . .9 6 32 2 5 16 m / s) m / s s) m / s.2c h

35

(b) Now, *t* = +2.5 s and we find

*v *= + =( . . ( .9 6 32 2 5 18 m / s) m / s s) m / s.2c h
28. We take +*x* in the direction of motion, so *v*0 = +24.6 m/s and *a* = – 4.92 m/s2. We
also take *x*0 = 0.
(a) The time to come to a halt is found using Eq. 2-11:

0 2

24.6 m/s0 5.00 s 4.92 m/s

.*v at t
*−

= + ⇒ = =

(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq. 2-16 (since it does not depend on our answer to part (a)).

( ) 2

2 0 2

(24.6 m/s)0 2 61.5 m 2 4.92 m/s

.*v ax x
*−

= + ⇒ = − =

(c) Using these results, we plot 210 2*v t at*+ (the *x* graph, shown next, on the left) and
*v*0 + *at* (the *v* graph, on the right) over 0 ≤ *t* ≤ 5 s, with SI units understood.

29. We assume the periods of acceleration (duration *t*1) and deceleration (duration *t*2)
are periods of constant *a* so that Table 2-1 can be used. Taking the direction of motion
to be +*x* then *a*1 = +1.22 m/s2 and *a*2 = –1.22 m/s2. We use SI units so the velocity at *t*
= *t*1 is *v* = 305/60 = 5.08 m/s.
(a) We denote Δ*x* as the distance moved during *t*1, and use Eq. 2-16:

2 2 2

0 1 2

(5.08 m/s)2 2(1.22 m/s )

*v v a x x*= + Δ ⇒ Δ = 10.59 m 10.6 m.= ≈

(b) Using Eq. 2-11, we have

0 1 2

1

5.08 m/s 4.17 s. 1.22 m/s

*v vt
a
*−

= = =

The deceleration time *t*2 turns out to be the same so that *t*1 + *t*2 = 8.33 s. The distances

*CHAPTER 2
*

36

traveled during *t*1 and *t*2 are the same so that they total to 2(10.59 m) = 21.18 m. This
implies that for a distance of 190 m – 21.18 m = 168.82 m, the elevator is traveling at
constant velocity. This time of constant velocity motion is

*t*3
16882
508

33 21= =. .

. m m / s

s.

Therefore, the total time is 8.33 s + 33.21 s ≈ 41.5 s.
30. We choose the positive direction to be that of the initial velocity of the car
(implying that *a* < 0 since it is slowing down). We assume the acceleration is constant
and use Table 2-1.
(a) Substituting *v*0 = 137 km/h = 38.1 m/s, *v* = 90 km/h = 25 m/s, and *a* = –5.2 m/s2
into *v* = *v*0 + *at*, we obtain

*t *= −
−

= 25 38

52 2 52

m / s m / s m / s

s .

. .

(b) We take the car to be at *x* = 0 when the brakes
are applied (at time *t* = 0). Thus, the coordinate of
the car as a function of time is given by

( ) ( )2 2138 m/s 5.2 m/s2*x t t*= + −
in SI units. This function is plotted from *t* = 0 to *t*
= 2.5 s on the graph to the right. We have not
shown the *v*-vs-*t* graph here; it is a descending
straight line from *v*0 to *v*.
31. The constant acceleration stated in the problem permits the use of the equations in
Table 2-1.
(a) We solve *v* = *v*0 + *at* for the time:

*t v v
a
*

= −

= ×

= ×0 1

10 8

63 0 10 9 8

31 10( . .

. m / s) m / s

s2

which is equivalent to 1.2 months.
(b) We evaluate *x x v t at*= + +0 0 12

2 , with *x*0 = 0. The result is

( )2 6 2 131 9.8 m/s (3.1 10 s) 4.6 10 m .2*x *= × = ×
Note that in solving parts (a) and (b), we did not use the equation 2 20 02 ( )*v v a x x*= + − .
This equation can be employed for consistency check. The final velocity based on this

37

equation is
2 2 13 7
0 02 ( ) 0 2(9.8 m/s )(4.6 10 m 0) 3.0 10 m/s*v v a x x*= + − = + × − = × ,

which is what was given in the problem statement. So we know the problems have been solved correctly. 32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).

*a v
t
*

= =

F HG

I KJ

= Δ Δ

1020 1000

3600 14

202 4 2 km / h

m / km s / h

s m / s

b g .

. .

In terms of the gravitational acceleration *g*, this is expressed as a multiple of 9.8 m/s2
as follows:

2

2

202.4 m/s 21 . 9.8 m/s

*a g g
*⎛ ⎞

= =⎜ ⎟ ⎝ ⎠

33. The problem statement (see part (a)) indicates that *a* = constant, which allows us
to use Table 2-1.
(a) We take *x*0 = 0, and solve *x* = *v*0*t* + 12 *at
*

2 (Eq. 2-15) for the acceleration: *a* = 2(*x* –
*v*0*t*)/*t*2. Substituting *x* = 24.0 m, *v*0 = 56.0 km/h = 15.55 m/s and *t* = 2.00 s, we find

( ) ( )( ) ( )

20 22

2 24.0m 15.55m/s 2.00s2( ) 3.56m/s , 2.00s

*x v ta
t
*

−− = = = −

or 2| | 3.56 m/s*a *= . The negative sign indicates that the acceleration is opposite to
the direction of motion of the car. The car is slowing down.
(b) We evaluate *v* = *v*0 + *at* as follows:

*v *= − =1555 356 2 00 8 43. . . .m / s m / s s m / s2c h b g

which can also be converted to 30.3 km/h.
34. Let *d* be the 220 m distance between the cars at *t* = 0, and *v*1 be the 20 km/h = 50/9
m/s speed (corresponding to a passing point of *x*1 = 44.5 m) and *v*2 be the 40 km/h
=100/9 m/s speed (corresponding to a passing point of *x*2 = 76.6 m) of the red car.
We have two equations (based on Eq. 2-17):

*d* – *x*1 = *v*o *t*1 + 12 *a**t*1
2 where *t*1 = *x*1 ⁄ *v*1

*d* – *x*2 = *v*o *t*2 + 12 *a**t*2

2 where *t*2 = *x*2 ⁄ *v*2

*CHAPTER 2
*

38

We simultaneously solve these equations and obtain the following results:
(a) The initial velocity of the green car is *v*o = − 13.9 m/s. or roughly − 50 km/h (the
negative sign means that it’s along the –*x* direction).
(b) The corresponding acceleration of the car is *a* = − 2.0 m/s2 (the negative sign
means that it’s along the –*x* direction).
35. The positions of the cars as a function of time are given by

2 2

0

0

1 1( ) ( 35.0 m) 2 2

( ) (270 m) (20 m/s)

*r r r r
*

*g g g
*

*x t x a t a t
*

*x t x v t t
*

= + = − +

= + = −

where we have substituted the velocity and not the speed for the green car. The two
cars pass each other at 12.0 s*t *= when the graphed lines cross. This implies that

21(270 m) (20 m/s)(12.0 s) 30 m ( 35.0 m) (12.0 s)
2 *r
*

*a*− = = − +

which can be solved to give 20.90 m/s .*ra *=
36. (a) Equation 2-15 is used for part 1 of the trip and Eq. 2-18 is used for part 2:
Δ*x*1 = *v*o1 *t*1 + 12 *a*1 *t*1

2 where *a*1 = 2.25 m/s2 and Δ*x*1 = 9004 m
Δ*x*2 = *v*2 *t*2 − 12 *a*2 *t*2

2 where *a*2 = −0.75 m/s2 and Δ*x*2 = 3(900)4 m
In addition, *v*o1 = *v*2 = 0. Solving these equations for the times and adding the results
gives *t = t*1* + t*2* =* 56.6 s.
(b) Equation 2-16 is used for part 1 of the trip:

*v*2 = (*v*o1)2 + 2*a*1Δ*x*1 = 0 + 2(2.25)
900

4 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠

= 1013 m2/s2

which leads to *v* = 31.8 m/s for the maximum speed.
37. (a) From the figure, we see that *x*0 = –2.0 m. From Table 2-1, we can apply

*x* – *x*0 = *v*0*t* + 12 *at
*2

with *t* = 1.0 s, and then again with *t* = 2.0 s. This yields two equations for the two
unknowns, *v*0 and *a*:

39

( ) ( ) ( )

( ) ( ) ( )

2 0

2 0

10.0 2.0 m 1.0 s 1.0 s 2 16.0 m 2.0 m 2.0 s 2.0 s . 2

*v a
*

*v a
*

− − = +

− − = +

Solving these simultaneous equations yields the results *v*0 = 0 and *a* = 4.0 m/s2.
(b) The fact that the answer is positive tells us that the acceleration vector points in
the +*x* direction.
38. We assume the train accelerates from rest ( *v*0 0= and *x*0 0= ) at
*a*1

2134= + . m / s until the midway point and then decelerates at *a*2
2134= − . m / s

until it comes to a stop *v*2 0=b g at the next station. The velocity at the midpoint is *v*1,
which occurs at *x*1 = 806/2 = 403m.
(a) Equation 2-16 leads to

( )( )2 2 21 0 1 1 12 2 1.34 m/s 403 m*v v a x v*= + ⇒ = 32.9 m/s.=
(b) The time *t*1 for the accelerating stage is (using Eq. 2-15)

( )2 1 0 1 1 1 1 2

2 403 m1 24.53 s 2 1.34 m/s

*x v t a t t*= + ⇒ = = .

Since the time interval for the decelerating stage turns out to be the same, we double
this result and obtain *t* = 49.1 s for the travel time between stations.
(c) With a “dead time” of 20 s, we have *T* = *t* + 20 = 69.1 s for the total time between
start-ups. Thus, Eq. 2-2 gives

*v*avg
m
s

m / s .= =806 691

117 .

.

(d) The graphs for *x*, *v* and *a* as a function of *t* are shown below. The third graph, *a*(*t*),
consists of three horizontal “steps” — one at 1.34 m/s2 during 0 < *t* < 24.53 s, and
the next at –1.34 m/s2 during 24.53 s < *t* < 49.1 s and the last at zero during the “dead
time” 49.1 s < *t* < 69.1 s).

*CHAPTER 2
*

40

39. (a) We note that *v*A = 12/6 = 2 m/s (with two significant figures understood).
Therefore, with an initial *x* value of 20 m, car A will be at *x* = 28 m when *t* = 4 s.
This must be the value of *x* for car B at that time; we use Eq. 2-15:

28 m = (12 m/s)*t *+ 12 *a*B *t
*2 where *t* = 4.0 s .

This yields *a*B = – 2.5 m/s2.
(b) The question is: using the value obtained for *a*B in part (a), are there other values
of *t* (besides *t* = 4 s) such that *x*A = *x*B ? The requirement is

20 + 2*t* = 12*t *+ 12 *a*B *t
*2

where *a*B = –5/2. There are two distinct roots unless the discriminant

102 − 2(−20)(*a*B) is zero. In our case, it is zero – which means there is only one root.
The cars are side by side only once at *t* = 4 s.
(c) A sketch is shown below. It consists of a straight line (*x*A) tangent to a parabola (*x*B)
at *t* = 4.

(d) We only care about real roots, which means 102 − 2(−20)(*a*B) ≥ 0. If |*a*B| > 5/2
then there are no (real) solutions to the equation; the cars are never side by side.
(e) Here we have 102 − 2(−20)(*a*B) > 0 ⇒ two real roots. The cars are side by side
at two different times.

41

40. We take the direction of motion as +*x*, so *a* = –5.18 m/s2, and we use SI units, so
*v*0 = 55(1000/3600) = 15.28 m/s.
(a) The velocity is constant during the reaction time *T*, so the distance traveled during
it is

*dr* = *v*0*T* – (15.28 m/s) (0.75 s) = 11.46 m.
We use Eq. 2-16 (with *v* = 0) to find the distance *db* traveled during braking:

( ) 2

2 2 0 2

(15.28 m/s)2
2 5.18 m/s*b b
*

*v v ad d*= + ⇒ = −
−

which yields *db* = 22.53 m. Thus, the total distance is *dr* + *db* = 34.0 m, which means
that the driver *is* able to stop in time. And if the driver were to continue at *v*0, the car
would enter the intersection in *t* = (40 m)/(15.28 m/s) = 2.6 s, which is (barely)
enough time to enter the intersection before the light turns, which many people would
consider an acceptable situation.
(b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than
the distance to the intersection, so the driver cannot stop without the front end of the
car being a couple of meters into the intersection. And the time to reach it at constant
speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The driver is
caught between a rock and a hard place.
41. The displacement (Δ*x*) for each train is the “area” in the graph (since the
displacement is the integral of the velocity). Each area is triangular, and the area of
a triangle is 1/2( base) × (height). Thus, the (absolute value of the) displacement for
one train (1/2)(40 m/s)(5 s) = 100 m, and that of the other train is (1/2)(30 m/s)(4 s) =
60 m. The initial “gap” between the trains was 200 m, and according to our
displacement computations, the gap has narrowed by 160 m. Thus, the answer is
200 – 160 = 40 m.
42. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two-second interval,
you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In
light of the fact that the initial “gap” between cars was 25 m, this means the gap has
narrowed by 10.0 m – that is, to a distance of 15.0 m between cars.
(b) First, we add 0.4 s to the considerations of part (a). During a 2.4 s interval, you
travel 73.33 m. The decelerating police car travels (using Eq. 2-15) 58.93 m during
that time. The initial distance between cars of 25 m has therefore narrowed by 14.4
m. Thus, at the start of your braking (call it *t*0) the gap between the cars is 10.6 m.
The speed of the police car at *t*0 is 30.56 – 5(2.4) = 18.56 m/s. Collision occurs at time
*t* when *x*you = *x*police (we choose coordinates such that your position is *x* = 0 and the
police car’s position is *x* = 10.6 m at *t*0). Eq. 2-15 becomes, for each car:
*x*police – 10.6 = 18.56(*t *−* t*0) – 12 (5)(*t *−* t*0)

2

*x*you = 30.56(*t *−* t*0) – 12 (5)(*t *−* t*0)
2 .

*CHAPTER 2
*

42

Subtracting equations, we find

10.6 = (30.56 – 18.56)(*t *−* t*0) ⇒ 0.883 s = *t *−* t*0.
At that time your speed is 30.56 + *a*(*t *−* t*0) = 30.56 – 5(0.883) ≈ 26 m/s (or 94 km/h).
43. In this solution we elect to wait until the last step to convert to SI units. Constant
acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and
denote the train’s initial velocity as *vt* and the locomotive’s velocity as *v * (which is
also the final velocity of the train, if the rear-end collision is barely avoided). We note
that the distance Δ*x* consists of the original gap between them, *D*, as well as the
forward distance traveled during this time by the locomotive *v t *. Therefore,

*v v x
t
*

*D v t
t
*

*D
t
*

*vt *+ = = + = +
2

Δ .

We now use Eq. 2-11 to eliminate time from the equation. Thus,

*v v D
v v a
*

*vt
t
*

+ =

− +

2 b g / which leads to

*a v v v v v
D D
*

*v vt t t*=
+

−FHG I KJ

−F HG I KJ = − −2

1 2

2 b g . Hence,

*a *= − −FHG
I
KJ = −

1 2 0 676

29 161 12888 2

2

( . km) km h

km h

km / h

which we convert as follows:

*a *= − FHG
I
KJ
F
HG

I KJ = −12888

1000 1

1 3600

0 9942 2

2 km / h m km

h s

m / sc h .
so that its *magnitude* is |*a*| = 0.994 m/s2. A graph is
shown here for the case where a collision is just
avoided (*x* along the vertical axis is in meters and *t*
along the horizontal axis is in seconds). The top
(straight) line shows the motion of the locomotive
and the bottom curve shows the motion of the
passenger train.
The other case (where the collision is not quite
avoided) would be similar except that the slope of
the bottom curve would be greater than that of the
top line at the point where they meet.

44. We neglect air resistance, which justifies setting *a *= –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the motion. We are allowed to use Table 2-1
(with Δ*y* replacing Δ*x*) because this is constant acceleration motion. The ground level

43

is taken to correspond to the origin of the *y* axis.
(a) Using *y v t gt*= −0 12

2 , with *y* = 0.544 m and *t* = 0.200 s, we find

2 2 2

0 / 2 0.544 m (9.8 m/s ) (0.200 s) / 2 3.70 m/s .

0.200 s
*y gtv
*

*t
*+ +

= = =

(b) The velocity at *y* = 0.544 m is

2
0 3.70 m/s (9.8 m/s ) (0.200 s) 1.74 m/s .*v v gt*= − = − =

(c) Using 2 20 2*v v gy*= − (with different values for *y* and *v* than before), we solve for
the value of *y* corresponding to maximum height (where *v* = 0).

2 2 0

2

(3.7 m/s) 0.698 m.
2 2(9.8 m/s )
*vy
g
*

= = =

Thus, the armadillo goes 0.698 – 0.544 = 0.154 m higher.
45. In this problem a ball is being thrown vertically upward. Its subsequent motion is
under the influence of gravity. We neglect air resistance for the duration of the motion
(between “launching” and “landing”), so *a* = –*g* = –9.8 m/s2 (we take downward to be
the –*y* direction). We use the equations in Table 2-1 (with Δ*y* replacing Δ*x*) because
this is *a* = constant motion:

0

2 0 0

2 2 0 0

(2 11)

1 (2 15) 2

2 ( ) (2 16)

*v v gt
*

*y y v t gt
*

*v v g y y
*

= − −

− = − −

= − − −

We set *y*0 = 0. Upon reaching the maximum height *y*, the speed of the ball is
momentarily zero (*v* = 0). Therefore, we can relate its initial speed *v*0 to *y* via the
equation 2 200 2*v v gy*= = − .
The time it takes for the ball to reach maximum height is given by 0 0*v v gt*= − = , or

0 /*t v g*= . Therefore, for the entire trip (from the time it leaves the ground until the
time it returns to the ground), the total flight time is 02 2 /*T t v g*= = .
(a) At the highest point *v* = 0 and *v gy*0 2= . Since *y* = 50 m we find

20 2 2(9.8 m/s )(50 m) 31.3 m/s.*v gy*= = =
(b) Using the result from (a) for *v*0, we find the total flight time to be

*CHAPTER 2
*

44

0 2

2 2(31.3 m/s) 6.39 s 6.4 s 9.8 m/s

*vT
g
*

= = = ≈ .

(c) SI units are understood in the *x* and *v* graphs shown. The acceleration graph is a
horizontal line at –9.8 m/s2.

In calculating the total flight time of the ball, we could have used Eq. 2-15. At

0*t T*= > , the ball returns to its original position ( 0*y *= ). Therefore,

2 0 0

21 0 2

*vy v T gT T
g
*

= − = ⇒ = .

46. Neglect of air resistance justifies setting *a* = –*g* = –9.8 m/s2 (where *down* is our –*y*
direction) for the duration of the fall. This is constant acceleration motion, and we
may use Table 2-1 (with Δ*y* replacing Δ*x*).
(a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward),
we have

2 2
0 2 0 2(9.8 m/s )( 1700 m) 183 m/s*v v g y*= − − Δ = − − − = − .

Its magnitude is therefore 183 m/s.
(b) No, but it is hard to make a convincing case without more analysis. We estimate
the mass of a raindrop to be about a gram or less, so that its mass and speed (from part
(a)) would be less than that of a typical bullet, which is good news. But the fact that
one is dealing with *many* raindrops leads us to suspect that this scenario poses an
unhealthy situation. If we factor in air resistance, the final speed is smaller, of course,
and we return to the relatively healthy situation with which we are familiar.
47. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the fall. This is constant acceleration motion,
which justifies the use of Table 2-1 (with Δ*y* replacing Δ*x*).
(a) Starting the clock at the moment the wrench is dropped (*v*0 = 0), then

2 2
0 2*v v g y*= − Δ leads to

2

2

( 24 m/s) 29.4 m 2(9.8 m/s )

*y *−Δ = − = −

so that it fell through a height of 29.4 m.

45

(b) Solving *v* = *v*0 – *gt* for time, we find:

0 2

0 ( 24 m/s) 2.45 s. 9.8 m/s

*v vt
g
*− − −

= = =

(c) SI units are used in the graphs, and the initial position is taken as the coordinate origin. The acceleration graph is a horizontal line at –9.8 m/s2.

As the wrench falls, with 0*a g*= − < , its speed increases but its velocity becomes
more negative.
48. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the fall. This is constant acceleration motion,
which justifies the use of Table 2-1 (with Δ*y* replacing Δ*x*).
(a) Noting that Δ*y* = *y* – *y*0 = –30 m, we apply Eq. 2-15 and the quadratic formula
(Appendix E) to compute *t*:

Δ Δ

*y v t gt t
v v g y
*

*g
*= − ⇒ =

± − 0

2 0 0 21

2 2

which (with *v*0 = –12 m/s since it is downward) leads, upon choosing the positive root
(so that *t* > 0), to the result:

2 2

2

12 m/s ( 12 m/s) 2(9.8 m/s )( 30 m) 1.54 s.

9.8 m/s
*t
*

− + − − − = =

(b) Enough information is now known that any of the equations in Table 2-1 can be
used to obtain *v*; however, the one equation that does *not* use our result from part (a)
is Eq. 2-16:

*v v g y*= − =0
2 2 271Δ . m / s

where the positive root has been chosen in order to give *speed* (which is the
magnitude of the velocity vector).
49. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the motion. We are allowed to use Table 2-1
(with Δ*y* replacing Δ*x*) because this is constant acceleration motion. We are placing
the coordinate origin on the ground. We note that the initial velocity of the package is

*CHAPTER 2
*

46

the same as the velocity of the balloon, *v*0 = +12 m/s, and that its initial coordinate is
*y*0 = +80 m.
(a) We solve *y y v t gt*= + −0 0 12

2 for time, with *y* = 0, using the quadratic formula
(choosing the positive root to yield a positive value for *t*).

( )( )2 220 0 0 2

12 m/s (12 m/s) 2 9.8 m/s 80 m2 9.8 m/s

5.4 s

*v v gy
t
*

*g
*

+ ++ + = =

=

(b) If we wish to avoid using the result from part (a), we could use Eq. 2-16, but if that is not a concern, then a variety of formulas from Table 2-1 can be used. For instance, Eq. 2-11 leads to

2
0 12 m/s (9.8 m/s )(5.447 s) 41.38 m/s*v v gt*= − = − = −

Its final *speed* is about 41 m/s.
50. The *y* coordinate of Apple 1 obeys *y – y*o1 = – 12 *g t
*

2 where *y* = 0 when *t* = 2.0 s.
This allows us to solve for *y*o1, and we find *y*o1 = 19.6 m.
The graph for the coordinate of Apple 2 (which is thrown apparently at *t* = 1.0 s with
velocity *v*2) is

*y – y*o2 = *v*2(*t – *1.0) – 12 *g* (*t – *1.0)
2

where *y*o2 = *y*o1 = 19.6 m and where *y = *0when *t* = 2.25 s. Thus, we obtain |*v*2*| =* 9.6
m/s, approximately.
51. (a) With upward chosen as the +*y* direction, we use Eq. 2-11 to find the initial
velocity of the package:

*v* = *v*o + *at* ⇒ 0 = *v*o – (9.8 m/s2)(2.0 s)
which leads to *v*o = 19.6 m/s. Now we use Eq. 2-15:

Δ*y* = (19.6 m/s)(2.0 s) + 12 (–9.8 m/s
2)(2.0 s)2 ≈ 20 m .

We note that the “2.0 s” in this second computation refers to the time interval 2 < *t* < 4
in the graph (whereas the “2.0 s” in the first computation referred to the 0 < *t* < 2 time
interval shown in the graph).
(b) In our computation for part (b), the time interval (“6.0 s”) refers to the 2 < *t* < 8
portion of the graph:

Δ*y* = (19.6 m/s)(6.0 s) + 12 (–9.8 m/s
2)(6.0 s)2 ≈ –59 m ,

or | | 59 m*y*Δ = .

47

52. The full extent of the bolt’s fall is given by

*y – y*0 = –12 *g t
*2

where *y* – *y*0 = –90 m (if upward is chosen as the positive *y* direction). Thus the time
for the full fall is found to be *t* = 4.29 s. The first 80% of its free-fall distance is given
by –72 = –*g *τ2/2, which requires time τ = 3.83 s.
(a) Thus, the final 20% of its fall takes *t* – τ = 0.45 s.
(b) We can find that speed using *v* = −*g*τ. Therefore, |*v*| = 38 m/s, approximately.
(c) Similarly, *vfinal* = − *g t *⇒ |*vfinal*| = 42 m/s.
53. The speed of the boat is constant, given by *vb* = *d*/*t*. Here, *d* is the distance of the
boat from the bridge when the key is dropped (12 m) and *t* is the time the key takes in
falling. To calculate *t*, we put the origin of the coordinate system at the point where
the key is dropped and take the *y* axis to be positive in the *downward* direction.
Taking the time to be zero at the instant the key is dropped, we compute the time *t*
when *y* = 45 m. Since the initial velocity of the key is zero, the coordinate of the key
is given by *y gt*= 12

2 . Thus,

*t y
g
*

= = = 2 2 45 303( . m)

9.8 m / s s .2

Therefore, the speed of the boat is

*vb *= =
12 4 0 m
3.03 s

m / s ..

54. (a) We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking
*down* as the –*y* direction) for the duration of the motion. We are allowed to use Eq.
2-15 (with Δ*y* replacing Δ*x*) because this is constant acceleration motion. We use
primed variables (except *t*) with the first stone, which has zero initial velocity, and
unprimed variables with the second stone (with initial downward velocity –*v*0, so that
*v*0 is being used for the initial *speed*). SI units are used throughout.

( )

( )( ) ( )

2

2 0

10 2

11 1 2

*y t gt
*

*y v t g t
*

′Δ = −

Δ = − − − −

Since the problem indicates Δ*y’* = Δ*y* = –43.9 m, we solve the first equation for *t*
(finding *t* = 2.99 s) and use this result to solve the second equation for the initial speed
of the second stone:

*CHAPTER 2
*

48

( ) ( ) ( )( )220 143.9 m 1.99 s 9.8 m/s 1.99 s2*v*− = − −
which leads to *v*0 = 12.3 m/s.
(b) The velocity of the stones are given by

0
( ) ( ), ( 1)*y y
*

*d y d yv gt v v g t
dt dt
*

′Δ Δ′ = = − = = − − −

The plot is shown below:

55. During contact with the ground its average acceleration is given by

*a v
t*avg

= Δ Δ

where Δ*v* is the change in its velocity during contact with the ground and
320.0 10 s*t *−Δ = × is the duration of contact. Thus, we must first find the velocity of the

ball just before it hits the ground (*y* = 0).
(a) Now, to find the velocity just *before* contact, we take *t* = 0 to be when it is dropped.
Using Eq. (2-16) with 0 15.0 m*y *= , we obtain

2 2
0 02 ( ) 0 2(9.8 m/s )(0 15 m) 17.15 m/s*v v g y y*= − − − = − − − = −

where the negative sign is chosen since the ball is traveling downward at the moment of contact. Consequently, the average acceleration during contact with the ground is

2 avg 3

0 ( 17.1 m/s) 857 m/s . 20.0 10 s

*va
t *−

Δ − − = = =

Δ ×

49

(b) The fact that the result is positive indicates that this acceleration vector points upward. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision. 56. We use Eq. 2-16,

*v*B2 = *v*A2 + 2*a*(*y*B – *y*A),
with *a* = –9.8 m/s2, *y*B – *y*A = 0.40 m, and *v*B = 13 *v*A. It is then straightforward to solve:
*v*A = 3.0 m/s, approximately.
57. The average acceleration during contact with the floor is *a*avg = (*v*2 – *v*1) / Δ*t*,
where *v*1 is its velocity just before striking the floor, *v*2 is its velocity just as it leaves
the floor, and Δ*t* is the duration of contact with the floor (12 × 10–3 s).
(a) Taking the *y* axis to be positively upward and placing the origin at the point where
the ball is dropped, we first find the velocity just before striking the floor, using

2 2
1 0 2*v v gy*= − . With *v*0 = 0 and *y* = – 4.00 m, the result is

2

1 2 2(9.8 m/s ) ( 4.00 m) 8.85 m/s*v gy*= − − = − − − = −
where the negative root is chosen because the ball is traveling downward. To find the
velocity just after hitting the floor (as it ascends without air friction to a height of 2.00
m), we use 2 22 02 ( )*v v g y y*= − − with *v* = 0, *y* = –2.00 m (it ends up two meters
*below* its initial drop height), and *y*0 = – 4.00 m. Therefore,

2
2 02 ( ) 2(9.8 m/s ) ( 2.00 m 4.00 m) 6.26 m/s .*v g y y*= − = − + =

Consequently, the average acceleration is

3 22 1 avg 3

6.26 m/s ( 8.85 m/s) 1.26 10 m/s . 12.0 10 s

*v va
t *−

− − − = = = ×

Δ ×

(b) The positive nature of the result indicates that the acceleration vector points
upward. In a later chapter, this will be directly related to the magnitude and direction
of the force exerted by the ground on the ball during the collision.
58. We choose *down* as the +*y* direction and set the coordinate origin at the point
where it was dropped (which is when we start the clock). We denote the 1.00 s
duration mentioned in the problem as *t* – *t'* where *t* is the value of time when it lands
and *t'* is one second prior to that. The corresponding distance is *y* – *y'* = 0.50*h*, where *y*
denotes the location of the ground. In these terms, *y* is the same as *h*, so we have *h* –*y'*
= 0.50*h* or 0.50*h* = *y'* .
(a) We find *t' *and *t* from Eq. 2-15 (with *v*0 = 0):

*CHAPTER 2
*

50

2

2

1 2 2

1 2 . 2

*yy gt t
g
*

*yy gt t
g
*

′ ′= ′ ⇒ ′=

= ⇒ =

Plugging in *y* = *h* and *y'* = 0.50*h*, and dividing these two equations, we obtain

*t
t
*

*h g
h g
*

′ = =

2 0 50 2

050 . /

/ . .b g

Letting *t'* = *t* – 1.00 (SI units understood) and cross-multiplying, we find

*t t t*− = ⇒ =
−

100 0 50 100 1 0 50

. . . .

which yields *t* = 3.41 s.
(b) Plugging this result into *y gt*= 12

2 we find *h* = 57 m.
(c) In our approach, we did not use the quadratic formula, but we did “choose a root”
when we assumed (in the last calculation in part (a)) that 050. = +0.707 instead
of –0.707. If we had instead let 0 50. = –0.707 then our answer for *t* would have
been roughly 0.6 s, which would imply that *t' *= *t* – 1 would equal a negative number
(indicating a time *before* it was dropped), which certainly does not fit with the
physical situation described in the problem.
59. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the motion. We are allowed to use Table 2-1
(with Δ*y* replacing Δ*x*) because this is constant acceleration motion. The ground level
is taken to correspond to the origin of the *y*-axis.
(a) The time drop 1 leaves the nozzle is taken as *t* = 0 and its time of landing on the
floor *t*1 can be computed from Eq. 2-15, with *v*0 = 0 and *y*1 = –2.00 m.

2 1 1 1 2

1 2 2( 2.00 m) 0.639 s . 2 9.8 m/s

*yy gt t
g
*

− − − = − ⇒ = = =

At that moment, the fourth drop begins to fall, and from the regularity of the dripping
we conclude that drop 2 leaves the nozzle at *t* = 0.639/3 = 0.213 s and drop 3 leaves
the nozzle at *t* = 2(0.213 s) = 0.426 s. Therefore, the time in free fall (up to the
moment drop 1 lands) for drop 2 is *t*2 = *t*1 – 0.213 s = 0.426 s. Its position at the
moment drop 1 strikes the floor is

2 2 22 2 1 1 (9.8 m/s )(0.426 s) 0.889 m, 2 2

*y gt*= − = − = −

or about 89 cm below the nozzle.

51

(b) The time in free fall (up to the moment drop 1 lands) for drop 3 is *t*3 = *t*1 –0.426 s
= 0.213 s. Its position at the moment drop 1 strikes the floor is

2 2 2 3 3

1 1 (9.8 m/s )(0.213 s) 0.222 m, 2 2

*y gt*= − = − = −

or about 22 cm below the nozzle. 60. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed is momentarily zero)

( )( )20 00 9.8 m/s 2.5 s*v v gt v*= − ⇒ = −
so that *v*0 = 24.5 m/s (with +*y* up). Now we use Eq. 2-15 to find the height of the
tower (taking *y*0 = 0 at the ground level)

( )( ) ( )( )22 20 0 1 10 24.5 m/s 1.5 s 9.8 m/s 1.5 s .2 2*y y v t at y*− = + ⇒ − = −
Thus, we obtain *y* = 26 m.
61. We choose *down* as the +*y* direction and place the coordinate origin at the top of
the building (which has height *H*). During its fall, the ball passes (with velocity *v*1) the
top of the window (which is at *y*1) at time *t*1, and passes the bottom (which is at *y*2) at
time *t*2. We are told *y*2 – *y*1 = 1.20 m and *t*2 – *t*1 = 0.125 s. Using Eq. 2-15 we have

*y y v t t g t t*2 1 1 2 1 2 1
21

2 − = − + −b g b g

which immediately yields

( )( )2212 1

1.20 m 9.8 m/s 0.125 s 8.99 m/s.

0.125 s
*v
*

− = =

From this, Eq. 2-16 (with *v*0 = 0) reveals the value of *y*1:

2 2 1 1 1 2

(8.99 m/s)2 4.12 m. 2(9.8 m/s )

*v gy y*= ⇒ = =

It reaches the ground (*y*3 = *H*) at *t*3. Because of the symmetry expressed in the
problem (“upward flight is a reverse of the fall’’) we know that *t*3 – *t*2 = 2.00/2 = 1.00
s. And this means *t*3 – *t*1 = 1.00 s + 0.125 s = 1.125 s. Now Eq. 2-15 produces

2 3 1 1 3 1 3 1

2 2 3

1( ) ( ) 2

14.12 m (8.99 m/s) (1.125 s) (9.8 m/s ) (1.125 s) 2

*y y v t t g t t
*

*y
*

− = − + −

− = +

*CHAPTER 2
*

52

which yields *y*3 = *H* = 20.4 m.
62. The height reached by the player is *y* = 0.76 m (where we have taken the origin of
the *y* axis at the floor and +*y* to be upward).
(a) The initial velocity *v*0 of the player is

2
0 2 2(9.8 m/s ) (0.76 m) 3.86 m/s .*v gy*= = =

This is a consequence of Eq. 2-16 where velocity *v* vanishes. As the player reaches *y*1
= 0.76 m – 0.15 m = 0.61 m, his speed *v*1 satisfies *v v gy*0

2 1 2

12− = , which yields

2 2 2
1 0 12 (3.86 m/s) 2(9.80 m/s ) (0.61 m) 1.71 m/s .*v v gy*= − = − =

The time *t*1 that the player spends *ascending* in the top Δ*y*1 = 0.15 m of the jump can
now be found from Eq. 2-17:

( ) ( )1 1 1 1 2 0.15 m1 0.175 s

2 1.71 m/s 0
*y v v t t*Δ = + ⇒ = =

+

which means that the total time spent in that top 15 cm (both ascending and
descending) is 2(0.175 s) = 0.35 s = 350 ms.
(b) The time *t*2 when the player reaches a height of 0.15 m is found from Eq. 2-15:

2 2 2 0 2 2 2 2

1 10.15 m (3.86 m/s) (9.8 m/s ) , 2 2

*v t gt t t*= − = −

which yields (using the quadratic formula, taking the smaller of the two positive roots)
*t*2 = 0.041 s = 41 ms, which implies that the total time spent in that bottom 15 cm
(both ascending and descending) is 2(41 ms) = 82 ms.
63. The time *t* the pot spends passing in front of the window of length *L* = 2.0 m is
0.25 s each way. We use *v* for its velocity as it passes the top of the window (going
up). Then, with *a* = –*g* = –9.8 m/s2 (taking *down* to be the –*y* direction), Eq. 2-18
yields

*L vt gt v L
t
*

*gt*= − ⇒ = −1
2

1 2

2 .

The distance *H* the pot goes above the top of the window is therefore (using Eq. 2-16
with the *final velocity* being zero to indicate the highest point)

( ) ( )22 22 2

2.00 m / 0.25 s (9.80 m/s )(0.25 s) / 2/ / 2 2.34 m.

2 2 2(9.80 m/s )
*L t gtvH
*

*g g
*−−

= = = =

53

64. The graph shows *y* = 25 m to be the highest point (where the speed momentarily
vanishes). The neglect of “air friction” (or whatever passes for that on the distant
planet) is certainly reasonable due to the symmetry of the graph.
(a) To find the acceleration due to gravity *gp* on that planet, we use Eq. 2-15 (with +*y*
up)

( )( ) ( )220
1 125 m 0 0 2.5 s 2.5 s
2 2*p p
*

*y y vt g t g*− = + ⇒ − = +

so that *gp* = 8.0 m/s2.
(b) That same (max) point on the graph can be used to find the initial velocity.

( ) ( ) ( )0 0 0 1 125 m 0 0 2.5 s 2 2

*y y v v t v*− = + ⇒ − = +

Therefore, *v*0 = 20 m/s.
65. The key idea here is that the speed of the head (and the torso as well) at any given
time can be calculated by finding the area on the graph of the head’s acceleration
versus time, as shown in Eq. 2-26:

1 0 0 1

area between the acceleration curve

and the time axis, from o
*v v
*

*t t t
*⎛ ⎞

− = ⎜ ⎟ ⎝ ⎠

(a) From Fig. 2.14a, we see that the head begins to accelerate from rest (*v*0 = 0) at *t*0 =
110 ms and reaches a maximum value of 90 m/s2 at* t*1= 160 ms. The area of this
region is

( )3 21area (160 110) 10 s 90 m/s 2.25 m/s 2

−= − × ⋅ =

which is equal to *v*1, the speed at *t*1.
(b) To compute the speed of the torso at *t*1=160 ms, we divide the area into 4 regions:
From 0 to 40 ms, region A has zero area. From 40 ms to 100 ms, region B has the
shape of a triangle with area

2B 1area (0.0600 s)(50.0 m/s ) 1.50 m/s 2

= = .

From 100 to 120 ms, region C has the shape of a rectangle with area

2 Carea (0.0200 s) (50.0 m/s ) = 1.00 m/s.=

From 110 to 160 ms, region D has the shape of a trapezoid with area

2 D

1area (0.0400 s) (50.0 20.0) m/s 1.40 m/s. 2

= + =

Substituting these values into Eq. 2-26, with *v*0 = 0 then gives

*CHAPTER 2
*

54

1 0 0 1 50 m/s + 1.00 m/s + 1.40 m/s = 3.90 m/s,*v .*− = +
or 1 3 90 m/s.*v .*=
66. The key idea here is that the position of an object at any given time can be
calculated by finding the area on the graph of the object’s velocity versus time, as
shown in Eq. 2-25:

1 0 0 1

area between the velocity curve

and the time axis, from o
*x x .
*

*t t t
*⎛ ⎞

− = ⎜ ⎟ ⎝ ⎠

(a) To compute the position of the fist at *t* = 50 ms, we divide the area in Fig. 2-34
into two regions. From 0 to 10 ms, region A has the shape of a triangle with area

A 1area = (0.010 s) (2 m/s) = 0.01 m. 2

From 10 to 50 ms, region B has the shape of a trapezoid with area

B 1area = (0.040 s) (2 + 4) m/s = 0.12 m. 2

Substituting these values into Eq. 2-25 with *x*0 = 0 then gives
1 0 0 0 01 m + 0.12 m = 0.13 m,*x .*− = +
or 1 0 13 m.*x .*=
(b) The speed of the fist reaches a maximum at *t*1 = 120 ms. From 50 to 90 ms, region
*C* has the shape of a trapezoid with area

C 1area = (0.040 s) (4 + 5) m/s = 0.18 m. 2

From 90 to 120 ms, region D has the shape of a trapezoid with area

D 1area = (0.030 s) (5 + 7.5) m/s = 0.19 m. 2

Substituting these values into Eq. 2-25, with *x*0 = 0 then gives
1 0 0 0 01 m + 0.12 m + 0.18 m + 0.19 m = 0.50 m,*x .*− = +
or 1 0 50 m.*x .*=
67. The problem is solved using Eq. 2-26:

1 0 0 1

area between the acceleration curve

and the time axis, from o
*v v
*

*t t t
*⎛ ⎞

− = ⎜ ⎟ ⎝ ⎠

55

To compute the speed of the unhelmeted, bare head at *t*1 = 7.0 ms, we divide the area
under the *a* vs.* t* graph into 4 regions: From 0 to 2 ms, region A has the shape of a
triangle with area

2A 1area = (0.0020 s) (120 m/s ) = 0.12 m/s. 2

From 2 ms to 4 ms, region B has the shape of a trapezoid with area

2 B

1area = (0.0020 s) (120 + 140) m/s = 0.26 m/s. 2

From 4 to 6 ms, region C has the shape of a trapezoid with area

2 C

1area = (0.0020 s) (140 + 200) m/s = 0.34 m/s. 2

From 6 to 7 ms, region D has the shape of a triangle with area

2 D

1area (0.0010 s) (200 m/s ) 0.10 m/s. 2

= =

Substituting these values into Eq. 2-26, with *v*0=0 then gives
0 12 m/s 0.26 m/s 0.34 m/s 0.10 m/s 0.82 m/s.*unhelmetedv .*= + + + =
Carrying out similar calculations for the helmeted head, we have the following
results: From 0 to 3 ms, region A has the shape of a triangle with area

2A 1area = (0.0030 s) (40 m/s ) = 0.060 m/s. 2

From 3 ms to 4 ms, region B has the shape of a rectangle with area

2 Barea (0.0010 s) (40 m/s ) 0.040 m/s.= =

From 4 to 6 ms, region C has the shape of a trapezoid with area

2 C

1area = (0.0020 s) (40 + 80) m/s = 0.12 m/s. 2

From 6 to 7 ms, region D has the shape of a triangle with area 2

D 1area (0.0010 s) (80 m/s ) 0.040 m/s. 2

= =

Substituting these values into Eq. 2-26, with *v*0 = 0 then gives

helmeted 0 060 m/s 0.040 m/s 0.12 m/s 0.040 m/s 0.26 m/s.*v .*= + + + =

*CHAPTER 2
*

56

Thus, the difference in the speed is
unhelmeted helmeted 0 82 m/s 0.26 m/s 0.56 m/s.*v v v .*Δ = − = − =
68. This problem can be solved by noting that velocity can be determined by the
graphical integration of acceleration versus time. The speed of the tongue of the
salamander is simply equal to the area under the acceleration curve:

2 2 2 2 2 2 21 1 1area (10 s)(100 m/s ) (10 s)(100 m/s 400 m/s ) (10 s)(400 m/s )

2 2 2 5.0 m/s.

*v *− − −= = + + +

=

69. Since /*v dx dt*= (Eq. 2-4), then Δ*x v dt*= z , which corresponds to the area
under the *v* vs *t* graph. Dividing the total area *A* into rectangular (base × height) and
triangular 12 base height×b g areas, we have

*A A A A At t t t*= + + +

= + + +FHG I KJ +

< < < < < < < <0 2 2 10 10 12 12 16

1 2

2 8 8 8 2 4 1 2

2 4 4 4( )( ) ( )( ) ( )( ) ( )( ) ( )( )

with SI units understood. In this way, we obtain Δ*x* = 100 m.
70. To solve this problem, we note that velocity is equal to the time derivative of a
position function, as well as the time integral of an acceleration function, with the
integration constant being the initial velocity. Thus, the velocity of particle 1 can be
written as

( )211 6.00 3.00 2.00 12.0 3.00*dx dv t t tdt dt*= = + + = + .
Similarly, the velocity of particle 2 is

22 20 2 20.0 ( 8.00 ) 20.0 4.00 .*v v a dt t dt t*= + = + − = −∫ ∫
The condition that 1 2*v v*= implies

2 212.0 3.00 20.0 4.00 4.00 12.0 17.0 0*t t t t*+ = − ⇒ + − =

which can be solved to give (taking positive root) ( 3 26) / 2 1.05 s.*t *= − + = Thus,

the velocity at this time is 1 2 12.0(1.05) 3.00 15.6 m/s.*v v*= = + =
71. (a) The derivative (with respect to time) of the given expression for *x* yields the
“velocity” of the spot:

*v*(*t*) = 9 – 94 *t
*2

57

with 3 significant figures understood. It is easy to see that *v* = 0 when *t* = 2.00 s.
(b) At *t* = 2 s, *x* = 9(2) – ¾(2)3 = 12. Thus, the location of the spot when *v* = 0 is 12.0
cm from left edge of screen.
(c) The derivative of the velocity is *a* = – 92 *t*, which gives an acceleration of

29.00 cm/m− (negative sign indicating leftward) when the spot is 12 cm from the
left edge of screen.
(d) Since *v* > 0 for times less than *t* = 2 s, then the spot had been moving rightward.
(e) As implied by our answer to part (c), it moves leftward for times immediately after
*t* = 2 s. In fact, the expression found in part (a) guarantees that for all *t* > 2, *v* < 0
(that is, until the clock is “reset” by reaching an edge).
(f) As the discussion in part (e) shows, the edge that it reaches at some *t* > 2 s cannot
be the right edge; it is the left edge (*x* = 0). Solving the expression given in the
problem statement (with *x* = 0) for positive *t* yields the answer: the spot reaches the
left edge at *t* = 12 s ≈ 3.46 s.
72. We adopt the convention frequently used in the text: that "up" is the positive *y*
direction.
(a) At the highest point in the trajectory *v* = 0. Thus, with *t* = 1.60 s, the equation
*v* = *v*0 – *gt* yields *v*0 = 15.7 m/s.
(b) One equation that is not dependent on our result from part (a) is *y* – *y*0 = *vt* + 12*gt
*

2;
this readily gives *y*max – *y*0 = 12.5 m for the highest ("max") point measured relative to
where it started (the top of the building).
(c) Now we use our result from part (a) and plug into *y* − *y*0 = *v*0*t* + 12*gt
*

2 with *t* = 6.00
s and *y* = 0 (the ground level). Thus, we have

0 – *y*0 = (15.68 m/s)(6.00 s) – 12 (9.8 m/s

2)(6.00 s)2.

Therefore, *y*0 (the height of the building) is equal to 82.3 m.
73. We denote the required time as *t*, assuming the light turns green when the clock
reads zero. By this time, the distances traveled by the two vehicles must be the same.
(a) Denoting the acceleration of the automobile as *a* and the (constant) speed of the
truck as *v* then

Δ*x at vt*= FHG
I
KJ =

1 2

2

car truckb g

which leads to

*CHAPTER 2
*

58

( ) 2

2 9.5 m/s2 8.6 s . 2.2 m/s

*vt
a
*

= = =

Therefore,
( )( )9.5 m/s 8.6 s 82 m .*x vt*Δ = = =

(b) The speed of the car at that moment is

( )( )2car 2.2 m/s 8.6 s 19 m/s .*v at*= = =
74. If the plane (with velocity *v*) maintains its present course, and if the terrain
continues its upward slope of 4.3°, then the plane will strike the ground after traveling

Δ*x h*= =
°

= ≈ tan

. θ

35 4655 m tan 4.3

m 0.465 km.

This corresponds to a time of flight found from Eq. 2-2 (with *v* = *v*avg since it is
constant)

*t x
v
*

= = = ≈ Δ 0 465 0 000358. . km

1300 km / h h 1.3 s.

This, then, estimates the time available to the pilot to make his correction.
75. We denote *tr* as the reaction time and *tb* as the braking time. The motion during *tr*
is of the constant-velocity (call it *v*0) type. Then the position of the car is given by

*x v t v t atr b b*= + +0 0
21

2

where *v*0 is the initial velocity and *a* is the acceleration (which we expect to be
negative-valued since we are taking the velocity in the positive direction and we know
the car is decelerating). *After* the brakes are applied the velocity of the car is given by
*v* = *v*0 + *atb*. Using this equation, with *v* = 0, we eliminate *tb* from the first equation
and obtain

*x v t v
a
*

*v
a
*

*v t v
ar r
*

= − + = − 1

0 0 2

0 2

0 0 21

2 2 .

We write this equation for each of the initial velocities:

*x v t v
ar*1 01
01
21

2 = −

and

*x v t v
ar*2 02
02
21

2 = − .

Solving these equations simultaneously for *tr* and *a* we get

59

*t v x v x
v v v vr
*

= −

− 02 2

1 01 2

2

01 02 02 01b g and

*a v v v v
v x v x
*

= − − −

1 2

02 01 2

01 02 2

02 1 01 2

.

(a) Substituting *x*1 = 56.7 m, *v*01 = 80.5 km/h = 22.4 m/s, *x*2 = 24.4 m and *v*02 = 48.3
km/h = 13.4 m/s, we find

2 2 2 2 02 1 01 2

01 02 02 01

(13.4 m/s) (56.7 m) (22.4 m/s) (24.4 m) ( ) (22.4 m/s)(13.4 m/s)(13.4 m/s 22.4 m/s)

0.74 s.

*r
v x v xt
*

*v v v v
*− −

= = − −

=

(b) Similarly, substituting *x*1 = 56.7 m, *v*01 = 80.5 km/h = 22.4 m/s, *x*2 = 24.4 m, and
*v*02 = 48.3 km/h = 13.4 m/s gives

2 2 2 2 02 01 01 02

02 1 01 2

2

1 1 (13.4 m/s)(22.4 m/s) (22.4 m/s)(13.4 m/s) 2 2 (13.4 m/s)(56.7 m) (22.4 m/s)(24.4 m)

6.2 m/s .

*v v v va
v x v x
*

− − = − = −

− −

= −

The *magnitude* of the deceleration is therefore 6.2 m/s2. Although rounded-off values
are displayed in the above substitutions, what we have input into our calculators are
the “exact” values (such as *v*02 16112= m/s).
76. (a) A constant velocity is equal to the ratio of displacement to elapsed time. Thus,
for the vehicle to be traveling at a constant speed *pv * over a distance 23*D *, the time
delay should be 23 / .*pt D v*=
(b) The time required for the car to accelerate from rest to a cruising speed *pv * is

0 /*pt v a*= . During this time interval, the distance traveled is
2 2

0 0 / 2 / 2 .*px at v a*Δ = =
The car then moves at a constant speed *pv * over a distance 12 0*D x d*− Δ − to reach
intersection 2, and the time elapsed is 1 12 0( ) / *pt D x d v*= − Δ − . Thus, the time delay at
intersection 2 should be set to

2 1212 0

total 0 1

12

( / 2 )

1 2

*p p p
r r r
*

*p p
*

*p
r
*

*p
*

*v v D v a dD x dt t t t t t
a v a v
*

*v D dt
a v
*

− −− Δ − = + + = + + = + +

− = + +

77. Since the problem involves constant acceleration, the motion of the rod can be
readily analyzed using the equations in Table 2-1. We take +*x* in the direction of
motion, so

*CHAPTER 2
*

60

*v *=
F
HG

I KJ = +60

1000 3600

16 7km / h m / km s / h

m / sb g .
and *a* > 0. The location where it starts from rest (*v*0 = 0) is taken to be *x*0 = 0.
(a) Using Eq. 2-7, we find the average acceleration to be

2 20avg 0

16.7 m/s 0 3.09 m/s 3.1 m/s 5.4 s 0

*v vva
t t t
*

−Δ − = = = = ≈

Δ − − .

(b) Assuming constant acceleration 2avg 3.09 m/s*a a*= = , the total distance traveled
during the 5.4-s time interval is

2 2 2 0 0

1 10 0 (3.09 m/s )(5.4 s) 45 m 2 2

*x x v t at*= + + = + + = .

(c) Using Eq. 2-15, the time required to travel a distance of *x* = 250 m is:

( )2

2

2 250 m1 2 13 s 2 3.1 m/s

*xx at t
a
*

= ⇒ = = = .

Note that the displacement of the rod as a function of time can be written as

2 21( ) (3.09 m/s ) 2

*x t t*= . Also we could have chosen Eq. 2-17 to solve for (b):

( ) ( )( )0 1 1 16.7 m/s 5.4 s 45 m. 2 2

*x v v t*= + = =

78. We take the moment of applying brakes to be *t* = 0. The deceleration is constant so
that Table 2-1 can be used. Our primed variables (such as 0 72 km/h = 20 m/s*v*′ = ) refer
to one train (moving in the +*x* direction and located at the origin when *t* = 0) and
unprimed variables refer to the other (moving in the –*x* direction and located at *x*0 =
+950 m when *t* = 0). We note that the acceleration vector of the unprimed train points
in the *positive* direction, even though the train is slowing down; its initial velocity is
*v*0 = –144 km/h = –40 m/s. Since the primed train has the lower initial speed, it should
stop sooner than the other train would (were it not for the collision). Using Eq 2-16, it
should stop (meaning 0*v*′ = ) at

( ) ( )2 2 20 2

0 (20 m/s) 200 m . 2 2 m/s

*v v
x
*

*a
*′ ′− −′ = = =

′ −

The speed of the other train, when it reaches that location, is

( ) ( )( )22 20 2 40 m/s 2 1.0 m/s 200 m 950 m 10 m/s

*v v a x*= + Δ = − + −

=

61

using Eq 2-16 again. Specifically, its velocity at that moment would be –10 m/s since
it is still traveling in the –*x* direction when it crashes. If the computation of *v* had
failed (meaning that a negative number would have been inside the square root) then
we would have looked at the possibility that there was no collision and examined how
far apart they finally were. A concern that can be brought up is whether the primed
train collides before it comes to rest; this can be studied by computing the time it
stops (Eq. 2-11 yields *t* = 20 s) and seeing where the unprimed train is at that moment
(Eq. 2-18 yields *x* = 350 m, still a good distance away from contact).
79. The *y* coordinate of Piton 1 obeys *y – y*01 = – 12 *g t
*

2 where *y* = 0 when *t* = 3.0 s.
This allows us to solve for *y*o1, and we find *y*01 = 44.1 m. The graph for the coordinate
of Piton 2 (which is thrown apparently at *t* = 1.0 s with velocity v1) is

*y – y*02 = *v*1(*t–*1.0) – 12 *g* (*t – *1.0)
2

where *y*02 = *y*01 + 10 = 54.1 m and where (again) *y* = 0 when *t* = 3.0 s. Thus we
obtain |*v*1*| =* 17 m/s, approximately.
80. We take +*x* in the direction of motion. We use subscripts 1 and 2 for the data. Thus,
*v*1 = +30 m/s, *v*2 = +50 m/s, and *x*2 – *x*1 = +160 m.
(a) Using these subscripts, Eq. 2-16 leads to

( ) ( ) 2 2 2 2

22 1

2 1

(50 m/s) (30 m/s) 5.0 m/s . 2 2 160 m

*v va
x x
*

− − = = =

−

(b) We find the time interval corresponding to the displacement *x*2 – *x*1 using Eq. 2-17:

( ) ( )2 1 2 1

1 2

2 2 160 m 4.0 s .

30 m/s 50 m/s
*x x
*

*t t
v v
*

− − = = =

+ +

(c) Since the train is at rest (*v*0 = 0) when the clock starts, we find the value of *t*1 from
Eq. 2-11:

1 0 1 1 2

30 m/s 6.0 s . 5.0 m/s

*v v at t*= + ⇒ = =

(d) The coordinate origin is taken to be the location at which the train was initially at
rest (so *x*0 = 0). Thus, we are asked to find the value of *x*1. Although any of several
equations could be used, we choose Eq. 2-17:

( ) ( )( )1 0 1 1 1 1 30 m/s 6.0 s 90 m . 2 2

*x v v t*= + = =

(e) The graphs are shown below, with SI units understood.

*CHAPTER 2
*

62

81. Integrating (from *t* = 2 s to variable *t* = 4 s) the acceleration to get the velocity and
using the values given in the problem leads to

0 0

2 2 0 0 0 0

1(5.0 ) (5.0)( ) 2

*t t
*

*t t
v v adt v t dt v t t*= + = + = + −∫ ∫ = 17 + 12 (5.0)(42 – 22) = 47 m/s.

82. The velocity *v* at *t* = 6 (SI units and two significant figures understood) is

6

given 2
*v adt
*

− + ∫ . A quick way to implement this is to recall the area of a triangle (12

base × height). The result is *v* = 7 m/s + 32 m/s = 39 m/s.
83. The object, once it is dropped (*v*0 = 0) is in free fall (*a* = –*g* = –9.8 m/s2 if we take
*down* as the –*y* direction), and we use Eq. 2-15 repeatedly.
(a) The (positive) distance *D* from the lower dot to the mark corresponding to a
certain reaction time *t* is given by Δ*y D gt*= − = − 12

2 , or *D* = *gt*2/2. Thus,
for 1 50.0 ms*t *= ,

*D*1
3 29 8 50 0 10

2 0 0123=

× =

−. . .

m / s s m = 1.23 cm.

2c h c h

(b) For *t*2 = 100 ms,
( ) ( )22 3

2 1

9.8 m/s 100 10 s 0.049 m = 4 .

2
*D D
*

−× = =

(c) For *t*3 = 150 ms,
( ) ( )22 3

3 1

9.8 m/s 150 10 s 0.11m = 9 .

2
*D D
*

−× = =

(d) For *t*4 = 200 ms,
( ) ( )22 3

4 1

9.8 m/s 200 10 s 0.196 m =16 .

2
*D D
*

−× = =

(e) For *t*4 = 250 ms, *D D*5
3 29 8 250 10

2 0 306 25=

× =

−. .

m / s s m = .

2

1

c h c h

84. We take the direction of motion as +*x*, take *x*0 = 0 and use SI units, so *v* =
1600(1000/3600) = 444 m/s.

63

(a) Equation 2-11 gives 444 = *a*(1.8) or *a* = 247 m/s2. We express this as a multiple of
*g* by setting up a ratio:

2

2

247 m/s 25 . 9.8 m/s

*a g g
*⎛ ⎞

= =⎜ ⎟ ⎝ ⎠

(b) Equation 2-17 readily yields

( ) ( )( )0 1 1 444 m/s 1.8 s 400 m. 2 2

*x v v t*= + = =

85. Let *D* be the distance up the hill. Then

average speed = total distance traveled

total time of travel =
2*D
*

*D
*20 km/h +

*D
*35 km/h

≈ 25 km/h .

86. We obtain the velocity by integration of the acceleration:

0 0 (6.1 1.2 )

*t
v v t dt*′ ′− = −∫ .

Lengths are in meters and times are in seconds. The student is encouraged to look at
the discussion in the textbook in §2-7 to better understand the manipulations here.
(a) The result of the above calculation is
20 6.1 0.6 ,*v v t t*= + −

where the problem states that *v*0 = 2.7 m/s. The maximum of this function is found by
knowing when its derivative (the acceleration) is zero (*a* = 0 when *t* = 6.1/1.2 = 5.1 s)
and plugging that value of *t* into the velocity equation above. Thus, we find

18 m/s*v *= .
(b) We integrate again to find *x* as a function of *t*:

2 2 30 0 00 0 ( 6.1 0.6 ) 3.05 0.2
*t t
*

*x x v dt v t t dt v t t t*′ ′ ′ ′− = = + − = + −∫ ∫ .
With *x*0 = 7.3 m, we obtain *x* = 83 m for *t* = 6. This is the correct answer, but one has
the right to worry that it might not be; after all, the problem asks for the total distance
traveled (and *x* − *x*0 is just the *displacement*). If the cyclist backtracked, then his total
distance would be greater than his displacement. Thus, we might ask, "did he
backtrack?" To do so would require that his velocity be (momentarily) zero at some
point (as he reversed his direction of motion). We could solve the above quadratic
equation for velocity, for a positive value of *t* where *v* = 0; if we did, we would find
that at *t* = 10.6 s, a reversal does indeed happen. However, in the time interval we
are concerned with in our problem (0 ≤ *t* ≤ 6 s), there is no reversal and the
displacement is the same as the total distance traveled.

*CHAPTER 2
*

64

87. The time it takes to travel a distance *d* with a speed *v*1 is 1 1/*t d v*= . Similarly, with
a speed *v*2 the time would be 2 2/*t d v*= . The two speeds in this problem are

1

2

1609 m/mi55 mi/h (55 mi/h) 24.58 m/s 3600 s/h

1609 m/mi65 mi/h (65 mi/h) 29.05 m/s 3600 s/h

*v
*

*v
*

= = =

= = =

With 5700 km 7.0 10 m*d *= = × , the time difference between the two is

5 1 2

1 2

1 1 1 1(7.0 10 m) 4383 s 73 min 24.58 m/s 29.05 m/s

*t t t d
v v
*

⎛ ⎞ ⎛ ⎞ Δ = − = − = × − = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

or 1 h and 13 min. 88. The acceleration is constant and we may use the equations in Table 2-1. (a) Taking the first point as coordinate origin and time to be zero when the car is there, we apply Eq. 2-17:

( ) ( ) ( )0 0 1 1 15.0 m/s 6.00 s . 2 2

*x v v t v*= + = +

With *x* = 60.0 m (which takes the direction of motion as the +*x* direction) we solve for
the initial velocity: *v*0 = 5.00 m/s.
(b) Substituting *v* = 15.0 m/s, *v*0 = 5.00 m/s, and *t* = 6.00 s into *a* = (*v* – *v*0)/*t* (Eq. 2-11),
we find *a* = 1.67 m/s2.
(c) Substituting *v* = 0 in 2 20 2*v v ax*= + and solving for *x*, we obtain

( ) 2 2 0

2

(5.00 m/s) 7.50m
2 2 1.67 m/s
*vx
a
*

= − = − = − ,

or | | 7.50 m*x *= .
(d) The graphs require computing the time when *v* = 0, in which case, we use *v* = *v*0 +
*at'* = 0. Thus,

0 2

5.00 m/s 3.0s 1.67 m/s

*vt
a
*

− −′ = = = −

indicates the moment the car was at rest. SI units are understood.

65

89. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the motion. We are allowed to use Table 2-1
(with Δ*y* replacing Δ*x*) because this is constant acceleration motion. When something
is thrown straight up and is caught at the level it was thrown from, the time of flight *t*
is half of its time of ascent *ta*, which is given by Eq. 2-18 with Δ*y* = *H* and *v* = 0
(indicating the maximum point).

*H vt gt t H
ga a a
*

= + ⇒ = 1 2

22

Writing these in terms of the total time in the air *t* = 2*ta* we have

*H gt t H
g
*

= ⇒ = 1 8

2 22 .

We consider two throws, one to height *H*1 for total time *t*1 and another to height *H*2 for
total time *t*2, and we set up a ratio:

*H
H
*

*gt
gt
*

*t
t
*

2

1

1 8 2

2

1 8 1

2 2

1

2

= = F HG I KJ

from which we conclude that if *t*2 = 2*t*1 (as is required by the problem) then *H*2 = 22*H*1
= 4*H*1.
90. (a) Using the fact that the area of a triangle is 12 (base) (height) (and the fact that
the integral corresponds to the area under the curve) we find, from *t* = 0 through *t* = 5
s, the integral of *v* with respect to *t* is 15 m. Since we are told that *x*0 = 0 then we
conclude that *x* = 15 m when *t* = 5.0 s.
(b) We see directly from the graph that *v* = 2.0 m/s when *t* = 5.0 s.
(c) Since *a* = *dv*/*dt* = slope of the graph, we find that the acceleration during the
interval 4 < *t* < 6 is uniformly equal to –2.0 m/s2.
(d) Thinking of *x*(*t*) in terms of accumulated area (on the graph), we note that *x*(1) = 1

*CHAPTER 2
*

66

m; using this and the value found in part (a), Eq. 2-2 produces

avg (5) (1) 15 m 1 m 3.5 m/s.

5 1 4 s
*x xv *− −= = =

−

(e) From Eq. 2-7 and the values *v*(*t*) we read directly from the graph, we find

avg (5) (1) 2 m/s 2 m/s 0.

5 1 4 s
*v va *− −= = =

−

91. Taking the +*y* direction *downward* and *y*0 = 0, we have *y v t gt*= +0 12

2 , which

(with *v*0 = 0) yields *t y g*= 2 / .
(a) For this part of the motion, *y*1 = 50 m so that

1 2

2(50 m) 3.2 s . 9.8 m/s

*t *= =

(b) For this next part of the motion, we note that the total displacement is *y*2 = 100 m.
Therefore, the total time is

2 2

2(100 m) 4.5 s . 9.8 m/s

*t *= =

The difference between this and the answer to part (a) is the time required to fall
through that second 50 m distance: 2 1*t t t*Δ = − = 4.5 s – 3.2 s = 1.3 s.
92. Direction of +*x* is implicit in the problem statement. The initial position (when the
clock starts) is *x*0 = 0 (where *v*0 = 0), the end of the speeding-up motion occurs at *x*1 =
1100/2 = 550 m, and the subway train comes to a halt (*v*2 = 0) at *x*2 = 1100 m.
(a) Using Eq. 2-15, the subway train reaches *x*1 at

( )1 1 2

1

2 550 m2 30.3 s . 1.2 m/s

*xt
a
*

= = =

The time interval *t*2 – *t*1 turns out to be the same value (most easily seen using Eq.
2-18 so the total time is *t*2 = 2(30.3) = 60.6 s.
(b) Its maximum speed occurs at *t*1 and equals

*v v a t*1 0 1 1 36 3= + = . .m / s

(c) The graphs are shown below:

67

93. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the stone’s motion. We are allowed to use Table
2-1 (with Δ*x* replaced by *y*) because the ball has constant acceleration motion (and we
choose *y*0 = 0).
(a) We apply Eq. 2-16 to both measurements, with SI units understood.

( ) 2

2 2 2 0 0

2 2 2 2 0 0

12 2 3 2

2 2

*B B A
*

*A A A
*

*v v gy v g y v
*

*v v gy v gy v
*

⎛ ⎞= − ⇒ + + =⎜ ⎟ ⎝ ⎠

= − ⇒ + =

We equate the two expressions that each equal *v*0

2 and obtain

1 4

2 2 3 2 2 3 3 4

2 2 2*v gy g v gy g vA A*+ + = + ⇒ =b g b g
which yields *v g*= =2 4 885b g . m / s.
(b) An object moving upward at *A* with speed *v* = 8.85 m/s will reach a maximum
height *y* – *yA* = *v*2/2*g* = 4.00 m above point *A* (this is again a consequence of Eq. 2-16,
now with the “final” velocity set to zero to indicate the highest point). Thus, the top of
its motion is 1.00 m above point *B*.
94. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking *down*
as the –*y* direction) for the duration of the motion. We are allowed to use Table 2-1
(with Δ*y* replacing Δ*x*) because this is constant acceleration motion. The ground level

*CHAPTER 2
*

68

is taken to correspond to the origin of the *y*-axis. The total time of fall can be
computed from Eq. 2-15 (using the quadratic formula).

Δ Δ

*y v t gt t
v v g y
*

*g
*= − ⇒ =

+ − 0

2 0 0 21

2 2

with the positive root chosen. With *y* = 0, *v*0 = 0, and *y*0 = *h* = 60 m, we obtain

*t
gh
*

*g
h
g
*

= = = 2 2 35. s .

Thus, “1.2 s earlier” means we are examining where the rock is at *t* = 2.3 s:

2 0

1(2.3 s) (2.3 s) 34 m 2

*y h v g y*− = − ⇒ =

where we again use the fact that *h* = 60 m and *v*0 = 0.
95. (a) The wording of the problem makes it clear that the equations of Table 2-1
apply, the challenge being that *v*0, *v*, and *a* are not explicitly given. We can, however,
apply *x* – *x*0 = *v*0*t* + 12*at
*

2 to a variety of points on the graph and solve for the unknowns from the simultaneous equations. For instance,

16 m – 0 = *v*0(2.0 s) +
1
2 *a*(2.0 s)

2

27 m – 0 = *v*0(3.0 s) +
1
2 *a*(3.0 s)

2

lead to the values *v*0 = 6.0 m/s and *a* = 2.0 m/s2.
(b) From Table 2-1,

*x* – *x*0 = *vt* –
1
2*at
*

2 ⇒ 27 m – 0 = *v*(3.0 s) –
1
2 (2.0 m/s

2)(3.0 s)2

which leads to *v* = 12 m/s.
(c) Assuming the wind continues during 3.0 ≤ *t* ≤ 6.0, we apply *x* – *x*0 = *v*0*t* + 12*at
*

2 to this interval (where v0 = 12.0 m/s from part (b)) to obtain

Δ*x* = (12.0 m/s)(3.0 s) +
1
2 (2.0 m/s

2)(3.0 s)2 = 45 m .

96. (a) Let the height of the diving board be *h*. We choose *down* as the +*y* direction
and set the coordinate origin at the point where it was dropped (which is when we
start the clock). Thus, *y* = *h* designates the location where the ball strikes the water.
Let the depth of the lake be *D*, and the total time for the ball to descend be *T*. The
speed of the ball as it reaches the surface of the lake is then *v* = 2*gh * (from Eq.

69

2-16), and the time for the ball to fall from the board to the lake surface is *t*1 =
2*h g*/ (from Eq. 2-15). Now, the time it spends descending in the lake (at constant

velocity *v*) is

*t D
v
*

*D
gh*2 2

= = .

Thus, *T* = *t*1 + *t*2 =
2*h
g
*

+ *D
gh*2

, which gives

( ) ( )( )( ) ( )22 2 4.80 s 2 9.80 m/s 5.20 m 2 5.20 m 38.1 m .*D T gh h*= − = − =
(b) Using Eq. 2-2, the magnitude of the average velocity is

avg 38.1 m 5.20 m 9.02 m/s

4.80 s
*D hv
*

*T
*+ +

= = =

(c) In our coordinate choices, a positive sign for *v*avg means that the ball is going
downward. If, however, upward had been chosen as the positive direction, then this
answer in (b) would turn out negative-valued.
(d) We find *v*0 from 210 2*y v t gt*Δ = + with *t* = *T* and Δ*y* = *h* + *D*. Thus,

( )( )2 0

9.8 m/s 4.80 s5.20 m 38.1 m 14.5 m/s 2 4.80 s 2

*h D gTv
T
*+ +

= − = − =

(e) Here in our coordinate choices the negative sign means that the ball is being
thrown upward.
97. We choose *down* as the +*y* direction and use the equations of Table 2-1 (replacing
*x* with *y*) with *a* = +*g*, *v*0 = 0, and *y*0 = 0. We use subscript 2 for the elevator reaching
the ground and 1 for the halfway point.
(a) Equation 2-16, *v v a y y*2

2 0 2

2 02= + −b g , leads to

( )( )22 22 2 9.8 m/s 120 m 48.5 m/s .*v gy*= = =
(b) The time at which it strikes the ground is (using Eq. 2-15)

( )2 2 2

2 120 m2 4.95 s . 9.8 m/s

*yt
g
*

= = =

(c) Now Eq. 2-16, in the form *v v a y y*1

2 0 2

1 02= + −b g , leads to

*CHAPTER 2
*

70

2
1 12 2(9.8 m/s )(60 m) 34.3m/s.*v gy*= = =

(d) The time at which it reaches the halfway point is (using Eq. 2-15)

1 1 2

2 2(60 m) 3.50 s . 9.8 m/s

*yt
g
*

= = =

98. Taking +*y* to be upward and placing the origin at the point from which the objects
are dropped, then the location of diamond 1 is given by *y gt*1 12

2= − and the location
of diamond 2 is given by *y g t*2 12

21= − −b g . We are starting the clock when the first
object is dropped. We want the time for which *y*2 – *y*1 = 10 m. Therefore,

− − + = ⇒ = + = 1 2

1 1 2

10 10 05 152 2*g t gt t g*b g b g/ . . s.
99. With +*y* upward, we have *y*0 = 36.6 m and *y* = 12.2 m. Therefore, using Eq. 2-18
(the last equation in Table 2-1), we find

2 0

1 22.0 m/s 2

*y y vt gt v*− = + ⇒ = −

at *t* = 2.00 s. The term *speed* refers to the magnitude of the velocity vector, so the
answer is |*v*| = 22.0 m/s.
100. During free fall, we ignore the air resistance and set *a* = –*g* = –9.8 m/s2 where we
are choosing *down* to be the –*y* direction. The initial velocity is zero so that Eq. 2-15
becomes Δ*y gt*= − 12

2 where Δ*y* represents the *negative* of the distance *d* she has
fallen. Thus, we can write the equation as *d gt*= 12

2 for simplicity.
(a) The time *t*1 during which the parachutist is in free fall is (using Eq. 2-15) given by

*d gt t*1 1
2

1 250 1

2 9 80= =m = 1

2 m / s2.c h

which yields *t*1 = 3.2 s. The *speed* of the parachutist just before he opens the parachute
is given by the positive root 21 12*v gd*= , or

*v gh*1 12 2 9 80 50 31= = =b gc hb g. m / s m m / s.2
If the final speed is *v*2, then the time interval *t*2 between the opening of the parachute
and the arrival of the parachutist at the ground level is

*t v v
a*2

1 2 31 30 14= − = − =m / s m / s 2 m / s

s.2 .

71

This is a result of Eq. 2-11 where *speeds* are used instead of the (negative-valued)
velocities (so that final-velocity minus initial-velocity turns out to equal initial-speed
minus final-speed); we also note that the acceleration vector for this part of the motion
is positive since it points upward (opposite to the direction of motion — which makes
it a deceleration). The total time of flight is therefore *t*1 + *t*2 = 17 s.
(b) The distance through which the parachutist falls after the parachute is opened is
given by

*d v v
a
*

= −

= −

≈1 2

2 2 2 2

2 31 3 0

2 2 0 240

m / s m / s m / s

m. 2

b g b g b gc h

. .

In the computation, we have used Eq. 2-16 with both sides multiplied by –1 (which
changes the negative-valued Δ*y* into the positive *d* on the left-hand side, and switches
the order of *v*1 and *v*2 on the right-hand side). Thus the fall begins at a height of *h* = 50
+ *d* ≈ 290 m.
101. We neglect air resistance, which justifies setting *a* = –*g* = –9.8 m/s2 (taking down
as the –*y* direction) for the duration of the motion. We are allowed to use Table 2-1
(with Δ*y* replacing Δ*x*) because this is constant acceleration motion. The ground level
is taken to correspond to *y* = 0.
(a) With *y*0 = *h* and *v*0 replaced with –*v*0, Eq. 2-16 leads to

2 2
0 0 0( ) 2 ( ) 2 .*v v g y y v gh*= − − − = +

The positive root is taken because the problem asks for the speed (the *magnitude* of
the velocity).
(b) We use the quadratic formula to solve Eq. 2-15 for *t*, with *v*0 replaced with –*v*0,

Δ Δ

*y v t gt t
v v g y
*

*g
* = − − ⇒ =

− + − − 0

2 0 0 21

2 2( )

where the positive root is chosen to yield *t* > 0. With *y* = 0 and *y*0 = *h*, this becomes

*t
v gh v
*

*g
*=

+ −0 2

02 .

(c) If it were thrown upward with that speed from height *h* then (in the absence of air
friction) it would return to height *h* with that same downward speed and would
therefore yield the same final speed (before hitting the ground) as in part (a). An
important perspective related to this is treated later in the book (in the context of
energy conservation).
(d) Having to travel up before it starts its descent certainly requires more time than in
part (b). The calculation is quite similar, however, except for now having +*v*0 in the
equation where we had put in –*v*0 in part (b). The details follow:

*CHAPTER 2
*

72

Δ Δ

*y v t gt t
v v g y
*

*g
*= − ⇒ =

+ − 0

2 0 0 21

2 2

with the positive root again chosen to yield *t* > 0. With *y* = 0 and *y*0 = *h*, we obtain

*t
v gh v
*

*g
*=

+ +0 2

02 .

102. We assume constant velocity motion and use Eq. 2-2 (with *v*avg = *v* > 0).
Therefore,

Δ Δ*x v t*= =
F
HG

I KJ

F HG

I KJ × =

−303 1000 100 10 8 43km h

m / km 3600 s / h

s m.c h .

73

**Chapter 3**
1. The *x* and the *y* components of a vector *a * lying on the *xy* plane are given by

cos , sin*x ya a a a*θ θ= =
where | |*a a*= is the magnitude and θ is the angle between *a * and the positive *x* axis.
(a) The *x* component of *a * is given by cos (7.3 m)cos 250 2.50 m*xa a *θ= = ° = − .
(b) Similarly, the *y* component is given by

sin (7.3 m)sin 250 6.86 m 6.9 m.*ya a *θ= = ° = − ≈ −
The results are depicted in the figure below:

In considering the variety of ways to compute these, we note that the vector is 70° below
the – *x* axis, so the components could also have been found from

(7.3 m)cos 70 2.50 m, (7.3 m)sin 70 6.86 m.*x ya a*= − ° = − = − ° = −
Similarly, we note that the vector is 20° to the left from the – *y* axis, so one could also
achieve the same results by using

(7.3 m)sin 20 2.50 m, (7.3 m)cos 20 6.86 m.*x ya a*= − ° = − = − ° = −
As a consistency check, we note that

2 2 2 2( 2.50 m) ( 6.86 m) 7.3 m*x ya a*+ = − + − =

*CHAPTER 3 *74

and
( )1 1tan / tan [( 6.86 m) /( 2.50 m)] 250*y xa a*− −= − − = ° ,

which are indeed the values given in the problem statement.
2. (a) With *r* = 15 m and θ = 30°, the *x* component of *r * is given by

*rx *= *r*cosθ = (15 m) cos 30° = 13 m.
(b) Similarly, the *y* component is given by *ry* = *r *sinθ = (15 m) sin 30° = 7.5 m.
3. A vector *a * can be represented in the *magnitude-angle* notation (*a*, θ), where
2 2*x ya a a*= +
is the magnitude and

1tan *y
x
*

*a
a
*

θ − ⎛ ⎞

= ⎜ ⎟ ⎝ ⎠

is the angle *a * makes with the positive *x* axis.
(a) Given *Ax *= −25.0 m and *Ay *= 40.0 m, 2 2( 25.0 m) (40.0 m) 47.2 m.*A *= − + =
(b) Recalling that tan θ = tan (θ + 180°),

tan–1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°.
Noting that the vector is in the third quadrant (by the signs of its *x* and *y* components) we
see that 122° is the correct answer. The graphical calculator “shortcuts” mentioned above
are designed to correctly choose the right possibility. The results are depicted in the
figure below:

We can check our answers by noting that the *x*- and the *y*- components of *A * can be
written as

cos , sin*x yA A A A*θ θ= =

75

Substituting the results calculated above, we obtain

(47.2 m)cos122 25.0 m, (47.2 m)sin122 40.0 m*x yA A*= ° = − = ° = +
which indeed are the values given in the problem statement.
4. The angle described by a full circle is 360° = 2π rad, which is the basis of our
conversion factor.

(a) ( ) 2 rad20.0 20.0 0.349 rad 360 π

° = ° = °

.

(b) ( ) 2 rad50.0 50.0 0.873 rad 360 π

° = ° = °

.

(c) ( ) 2 rad100 100 1.75 rad 360 π

° = ° = °

.

(d) ( ) 3600.330 rad = 0.330 rad 18.9 2 radπ

° = ° .

(e) ( ) 3602.10 rad = 2.10 rad 120 2 radπ

° = ° .

(f) ( ) 3607.70 rad = 7.70 rad 441 2 radπ

° = ° .

5. The vector sum of the displacements *d*storm and *d*new must give the same result as its
originally intended displacement o ˆ(120 km)j*d *= where east is i , north is j . Thus, we
write

storm new
ˆ ˆ ˆ(100 km) i , i j.*d d A B*= = +

(a) The equation storm new o*d d d*+ = readily yields *A* = –100 km and *B* = 120 km. The

magnitude of *d*new is therefore equal to
2 2

new| | 156 km*d A B*= + = .
(b) The direction is

tan–1 (*B*/*A*) = –50.2° or 180° + ( –50.2°) = 129.8°.
We choose the latter value since it indicates a vector pointing in the second quadrant,
which is what we expect here. The answer can be phrased several equivalent ways:
129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west.
6. (a) The height is *h* = *d* sinθ, where *d* = 12.5 m and θ = 20.0°. Therefore, *h* = 4.28 m.
(b) The horizontal distance is *d* cosθ = 11.7 m.

*CHAPTER 3 *76

7. The displacement of the fly is illustrated in the figure below:

A coordinate system such as the one shown (above right) allows us to express the displacement as a three-dimensional vector. (a)The magnitude of the displacement from one corner to the diagonally opposite corner is

2 2 2| |*d d w l h*= = + + Substituting the values given, we obtain

2 2 2 2 2 2| | (3.70 m) (4.30 m) (3.00 m) 6.42 m*d d w l h*= = + + = + + = .
(b) The displacement vector is along the straight line from the beginning to the end point
of the trip. Since a straight line is the shortest distance between two points, the length of
the path cannot be less than the magnitude of the displacement.
(c) It can be greater, however. The fly might, for example, crawl along the edges of the
room. Its displacement would be the same but the path length would be

11.0 m.*w h*+ + =
(d) The path length is the same as the magnitude of the displacement if the fly flies along
the displacement vector.
(e) We take the *x* axis to be out of the page, the *y* axis to be to the right, and the *z* axis to
be upward. Then the *x* component of the displacement is *w* = 3.70 m, the *y* component of
the displacement is 4.30 m, and the *z* component is 3.00 m. Thus,

ˆ ˆ ˆ(3.70 m) i ( 4.30 m) j (3.00 m)k*d *= + + .
An equally correct answer is gotten by interchanging the length, width, and height.

77

(f) Suppose the path of the fly is as shown by the dotted lines on the upper diagram. Pretend there is a hinge where the front wall of the room joins the floor and lay the wall down as shown on the lower diagram. The shortest walking distance between the lower left back of the room and the upper right front corner is the dotted straight line shown on the diagram. Its length is

( ) ( )2 22 2min 3.70 m 3.00 m (4.30 m) 7.96 m .*L w h*= + + = + + =
To show that the shortest path is indeed given by min*L *, we write the length of the path as
2 2 2 2( )*L y w l y h*= + + − + .
The condition for minimum is given by

2 2 2 2 0

( )
*dL y l y
dy y w l y h
*

− = − =

+ − + .

A little algebra shows that the condition is satisfied when /( )*y lw w h*= + , which gives

2 2 2 2 2 2

min 2 21 1 ( )( ) ( )
*l lL w h w h l
*

*w h w h
*⎛ ⎞ ⎛ ⎞

= + + + = + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ .

Any other path would be longer than 7.96 m.
8. We label the displacement vectors *A *, *B *, and *C * (and denote the result of their vector
sum as *r *). We choose east as the î direction (+x direction) and north as the ĵ direction
(+*y* direction). All distances are understood to be in kilometers.
(a) The vector diagram representing the motion is shown next:

*CHAPTER 3 *78

ˆ(3.1 km) j ˆ( 2.4 km) i ˆ( 5.2 km) j

*A
*

*B
*

*C
*

=

= −

= −

(b) The final point is represented by

ˆ ˆ( 2.4 km )i ( 2.1 km) j*r A B C*= + + = − + −

whose magnitude is

( ) ( )2 22.4 km 2.1 km 3.2 km*r *= − + − ≈ .
(c) There are two possibilities for the angle:

1 2.1 kmtan 41 ,or 221 2.4 km

θ − ⎛ ⎞−

= = ° °⎜ ⎟−⎝ ⎠ .

We choose the latter possibility since *r * is in the third quadrant. It should be noted that
many graphical calculators have polar ↔ rectangular “shortcuts” that automatically
produce the correct answer for angle (measured counterclockwise from the +*x* axis). We
may phrase the angle, then, as 221° counterclockwise from East (a phrasing that sounds
peculiar, at best) or as 41° south from west or 49° west from south. The resultant *r * is
not shown in our sketch; it would be an arrow directed from the “tail” of *A * to the “head”
of *C *.
9. All distances in this solution are understood to be in meters.
(a) ˆ ˆ ˆ ˆ ˆ ˆ[4.0 ( 1.0)] i [( 3.0) 1.0] j (1.0 4.0)k (3.0i 2.0 j 5.0 k) m.*a b*+ = + − + − + + + = − +
(b) ˆ ˆ ˆ ˆ ˆ ˆ[4.0 ( 1.0)]i [( 3.0) 1.0]j (1.0 4.0)k (5.0 i 4.0 j 3.0 k) m.*a b*− = − − + − − + − = − −
(c) The requirement *a b c*− + = 0 leads to *c b a*= − , which we note is the opposite of
what we found in part (b). Thus, ˆ ˆ ˆ( 5.0 i 4.0 j 3.0k) m.*c *= − + +
10. The *x, y*, and *z* components of *r c d*= + are, respectively,
(a) 7.4 m 4.4 m 12 m*x x xr c d*= + = + = ,
(b) 3.8 m 2.0 m 5.8 m*y y yr c d*= + = − − = − , and

79

(c) 6.1 m 3.3 m 2.8 m.*z z zr c d*= + = − + = −
11. We write *r a b*= + . When not explicitly displayed, the units here are assumed to be
meters.
(a) The *x* and the *y* components of *r * are *rx* = *ax* + *bx* = (4.0 m) – (13 m) = –9.0 m and *ry* =
*ay* + *by* = (3.0 m) + (7.0 m) = 10 m, respectively. Thus ˆ ˆ( 9.0m) i (10m) j*r *= − + .
(b) The magnitude of *r *is
2 2 2 2| | ( 9.0 m) (10 m) 13 m*x yr r r r*= = + = − + = .
(c) The angle between the resultant and the +*x* axis is given by

1 1 10.0 mtan tan 48 or 132 9.0 m

*y
*

*x
*

*r
r
*

θ − − ⎛ ⎞ ⎛ ⎞= = = − ° °⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

.

Since the *x* component of the resultant is negative and the *y* component is positive,
characteristic of the second quadrant, we find the angle is 132° (measured
counterclockwise from +*x* axis).
The addition of the two vectors is depicted in the figure below (not to scale). Indeed, we
expect *r * to be in the second quadrant.

12. We label the displacement vectors *A *, *B *, and *C * (and denote the result of their
vector sum as *r *). We choose east as the î direction (+x direction) and north as the ĵ
direction (+*y* direction). We note that the angle between *C * and the *x* axis is 60°. Thus,

*CHAPTER 3 *80

( ) ( )

ˆ(50 km) i ˆ(30 km) j

ˆ ˆ(25 km) cos 60 i + (25 km )sin 60 j

*A
*

*B
*

*C
*

=

=

= ° °

(a) The total displacement of the car from its initial position is represented by

ˆ ˆ(62.5 km) i (51.7 km) j*r A B C*= + + = +
which means that its magnitude is

2 2(62.5km) (51.7 km) 81 km.*r *= + =
(b) The angle (counterclockwise from +*x* axis) is tan–1 (51.7 km/62.5 km) = 40°, which is
to say that it points 40° *north of east*. Although the resultant *r * is shown in our sketch, it
would be a direct line from the “tail” of *A * to the “head” of *C *.
13. We find the components and then add them (as scalars, not vectors). With *d* = 3.40
km and θ = 35.0° we find *d* cos θ + *d* sin θ = 4.74 km.
14. (a) Summing the *x* components, we have

20 m + *bx* – 20 m – 60 m = −140 m,
which gives 80 m.*xb *= −
(b) Summing the *y* components, we have

60 m – 70 m + *cy* – 70 m = 30 m,
which implies *cy* =110 m.
(c) Using the Pythagorean theorem, the magnitude of the overall displacement is given by

2 2 ( 140 m) (30 m) 143 m.− + ≈
(d) The angle is given by 1tan (30 /( 140)) 12− − = − ° , (which would be 12° measured
clockwise from the –*x* axis, or 168° measured counterclockwise from the +*x* axis).

81

15. It should be mentioned that an efficient way to work this vector addition problem is
with the cosine law for general triangles (and since ,*a b *, and *r * form an isosceles
triangle, the angles are easy to figure). However, in the interest of reinforcing the usual
systematic approach to vector addition, we note that the angle *b * makes with the +*x* axis
is 30° +105° = 135° and apply Eq. 3-5 and Eq. 3-6 where appropriate.
(a) The *x* component of *r * is *rx* = (10.0 m) cos 30° + (10.0 m) cos 135° = 1.59 m.
(b) The *y* component of *r * is *ry* = (10.0 m) sin 30° + (10.0 m) sin 135° = 12.1 m.
(c) The magnitude of *r * is 2 2| | (1.59 m) (12.1 m) 12.2 m.*r r*= = + =
(d) The angle between *r * and the +*x* direction is tan–1[(12.1 m)/(1.59 m)] = 82.5°.
16. (a) ˆ ˆ ˆ ˆ ˆ ˆ(3.0 i 4.0 j) m (5.0 i 2.0 j) m (8.0 m) i (2.0 m) j.*a b*+ = + + − = +
(b) The magnitude of *a b*+ is

2 2| | (8.0 m) (2.0 m) 8.2 m.*a b*+ = + =
(c) The angle between this vector and the +*x* axis is

tan–1[(2.0 m)/(8.0 m)] = 14°.
(d) ˆ ˆ ˆ ˆ ˆ ˆ(5.0 i 2.0 j) m (3.0 i 4.0 j) m (2.0 m) i (6.0 m) j .*b a*− = − − + = −
(e) The magnitude of the difference vector *b a*− is

2 2| | (2.0 m) ( 6.0 m) 6.3 m.*b a*− = + − =
(f) The angle between this vector and the +*x* axis is tan-1[( –6.0 m)/(2.0 m)] = –72°. The
vector is 72° clockwise from the axis defined by î .
17. Many of the operations are done efficiently on most modern graphical calculators
using their built-in vector manipulation and rectangular ↔ polar “shortcuts.” In this
solution, we employ the “traditional” methods (such as Eq. 3-6). Where the length unit is
not displayed, the unit meter should be understood.
(a) Using unit-vector notation,

*CHAPTER 3 *82

ˆ ˆ(50 m)cos(30 )i (50 m) sin(30 ) j ˆ ˆ(50 m)cos (195 ) i (50 m)sin (195 ) j ˆ ˆ(50 m)cos (315 ) i (50 m)sin (315 ) j

ˆ ˆ(30.4 m) i (23.3 m) j.

*a
*

*b
*

*c
*

*a b c
*

= ° + °

= ° + °

= ° + °

+ + = −

The magnitude of this result is 2 2(30.4 m) ( 23.3 m) 38 m+ − = .
(b) The two possibilities presented by a simple calculation for the angle between the
vector described in part (a) and the +*x* direction are tan–1[( –23.2 m)/(30.4 m)] = –37.5°,
and 180° + ( –37.5°) = 142.5°. The former possibility is the correct answer since the
vector is in the fourth quadrant (indicated by the signs of its components). Thus, the
angle is –37.5°, which is to say that it is 37.5° *clockwise* from the +*x* axis. This is
equivalent to 322.5° counterclockwise from +*x*.
(c) We find

ˆ ˆ ˆ ˆ[43.3 ( 48.3) 35.4] i [25 ( 12.9) ( 35.4)] j (127 i 2.60 j) m*a b c*− + = − − + − − − + − = +
in unit-vector notation. The magnitude of this result is

2 2 2| | (127 m) (2.6 m) 1.30 10 m.*a b c*− + = + ≈ ×
(d) The angle between the vector described in part (c) and the +*x* axis is

1tan (2.6 m/127 m) 1.2− ≈ ° .
(e) Using unit-vector notation, *d * is given by ˆ ˆ( 40.4 i 47.4 j) m*d a b c*= + − = − + ,

which has a magnitude of 2 2( 40.4 m) (47.4 m) 62 m.− + =
(f) The two possibilities presented by a simple calculation for the angle between the
vector described in part (e) and the +*x* axis are 1tan (47.4 /( 40.4)) 50.0− − = − ° , and
180 ( 50.0 ) 130° + − ° = ° . We choose the latter possibility as the correct one since it
indicates that *d * is in the second quadrant (indicated by the signs of its components).
18. If we wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle
between *C * and the +*x* axis is 180° + 20.0° = 200°.
(a) The *x* and *y* components of *B * are given by

* Bx = Cx* – *Ax* = (15.0 m) cos 200° – (12.0 m) cos 40° = –23.3 m,
* By =Cy* – *Ay* = (15.0 m) sin 200° – (12.0 m) sin 40° = –12.8 m.

83

Consequently, its magnitude is | |*B *= 2 2( 23.3 m) ( 12.8 m) 26.6 m− + − = .
(b) The two possibilities presented by a simple calculation for the angle between *B * and
the +*x* axis are tan–1[( –12.8 m)/( –23.3 m)] = 28.9°, and 180° + 28.9° = 209°. We choose
the latter possibility as the correct one since it indicates that *B * is in the third quadrant
(indicated by the signs of its components). We note, too, that the answer can be
equivalently stated as 151 .− °
19. (a) With i^ directed forward and j^ directed leftward, the resultant is (5.00 i^ + 2.00 j^) m .
The magnitude is given by the Pythagorean theorem: 2 2(5.00 m) (2.00 m)+ = 5.385 m
≈ 5.39 m.
(b) The angle is tan−1(2.00/5.00) ≈ 21.8º (left of forward).
20. The desired result is the displacement vector, in units of km, *A*

→ = (5.6 km), 90º

(measured counterclockwise from the +*x *axis), or ˆ(5.6 km)j*A *= , where ĵ is the unit
vector along the positive *y* axis (north). This consists of the sum of two displacements:
during the whiteout, (7.8 km), 50*B *= ° , or

ˆ ˆ ˆ ˆ(7.8 km)(cos50 i sin50 j) (5.01 km)i (5.98 km) j*B *= ° + ° = +
and the unknown *C *. Thus, *A B C*= + .
(a) The desired displacement is given by ˆ ˆ( 5.01 km) i (0.38 km) j*C A B*= − = − − . The

magnitude is 2 2( 5.01 km) ( 0.38 km) 5.0 km.− + − =
(b) The angle is 1tan [( 0.38 km) /( 5.01 km)] 4.3 ,− − − = ° south of due west.
21. Reading carefully, we see that the (*x, y*) specifications for each “dart” are to be
interpreted as ( , )Δ Δ*x y* descriptions of the corresponding displacement vectors. We
combine the different parts of this problem into a single exposition.
(a) Along the *x* axis, we have (with the centimeter unit understood)

30.0 20.0 80.0 140,*xb*+ − − = −
which gives *bx* = –70.0 cm.
(b) Along the *y* axis we have

*CHAPTER 3 *84

40.0 70.0 70.0 20.0*yc*− + − = −
which yields *cy* = 80.0 cm.
(c) The magnitude of the final location (–140 , –20.0) is 2 2( 140) ( 20.0) 141 cm.− + − =
(d) Since the displacement is in the third quadrant, the angle of the overall displacement
is given by π + 1tan [( 20.0) /( 140)]− − − or 188° counterclockwise from the +*x* axis (or

172− ° counterclockwise from the +*x* axis).
22. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to
write the vectors in unit-vector notation before adding them. However, a very different-
looking approach using the special capabilities of most graphical calculators can be
imagined. Wherever the length unit is not displayed in the solution below, the unit meter
should be understood.
(a) Allowing for the different angle units used in the problem statement, we arrive at

*E
*

*F
*

*G
*

*H
*

*E F G H
*

= +

= −

= +

= − +

+ + + = +

3 73 4 70

1 29 4 83

1 3 73

5 20 3 00

1 28 6 60

. .

. .

.45 .

. .

. .

i j

i j

i j

i j

i j.

(b) The magnitude of the vector sum found in part (a) is 2 2(1.28 m) (6.60 m) 6.72 m+ = .
(c) Its angle measured counterclockwise from the +*x* axis is tan–1(6.60/1.28) = 79.0°.
(d) Using the conversion factor rad = 180π ° , 79.0° = 1.38 rad.
23. The resultant (along the *y* axis, with the same magnitude as *C*

→ ) forms (along with

*C*
→

) a side of an isosceles triangle (with *B*
→

forming the base). If the angle between *C*
→

and the *y* axis is 1tan (3 / 4) 36.87θ −= = ° , then it should be clear that (referring to the
magnitudes of the vectors) 2 sin( / 2)*B C *θ= . Thus (since *C* = 5.0) we find *B* = 3.2.
24. As a vector addition problem, we express the situation (described in the problem
statement) as *A*

→
+ *B*

→
= (3*A*) j^ , where *A*

→
= *A* i^ and *B* = 7.0 m. Since i^ ⊥ j^ we may

use the Pythagorean theorem to express *B* in terms of the magnitudes of the other two
vectors:

85

*B* = (3*A*)2 + *A*2 ⇒ *A* = 1
10

*B* = 2.2 m .

25. The strategy is to find where the camel is ( *C*
→

) by adding the two consecutive displacements described in the problem, and then finding the difference between that

location and the oasis ( *B*
→

). Using the magnitude-angle notation
= (24 15 ) + (8.0 90 ) = (23.25 4.41 )*C *∠ − ° ∠ ° ∠ °
so
(25 0 ) (23.25 4.41 ) (2.5 45 )*B C*− = ∠ ° − ∠ ° = ∠ − °
which is efficiently implemented using a vector-capable calculator in polar mode. The
distance is therefore 2.6 km.
26. The vector equation is *R A B C D*= + + + . Expressing *B * and *D * in unit-vector
notation, we have ˆ ˆ(1.69i 3.63j) m+ and ˆ ˆ( 2.87i 4.10j) m− + , respectively. Where the
length unit is not displayed in the solution below, the unit meter should be understood.
(a) Adding corresponding components, we obtain ˆ ˆ( 3.18 m)i ( 4.72 m) j*R *= − + .
(b) Using Eq. 3-6, the magnitude is
2 2| | ( 3.18 m) (4.72 m) 5.69 m.*R *= − + =
(c) The angle is

1 4.72 mtan 56.0 (with axis). 3.18 m

*x*θ − ⎛ ⎞= = − ° −⎜ ⎟−⎝ ⎠

If measured counterclockwise from +*x*-axis, the angle is then 180 56.0 124° − ° = ° . Thus,
converting the result to polar coordinates, we obtain

− → ∠ °318 4 72 569 124. , . .b g b g
27. Solving the simultaneous equations yields the answers:
(a) *d*1

→
= 4 *d*3

→ = 8 i^ + 16 j^ , and

(b) *d*2

→
= *d*3

→ = 2 i^ + 4 j^.

28. Let *A*

→
represent the first part of Beetle 1’s trip (0.50 m east or ˆ0.5 i ) and *C*

→

represent the first part of Beetle 2’s trip intended voyage (1.6 m at 50º north of east). For

*CHAPTER 3 *86

their respective second parts: *B*
→

is 0.80 m at 30º north of east and *D*
→

is the unknown. The final position of Beetle 1 is

ˆ ˆ ˆ ˆ ˆ(0.5 m)i (0.8 m)(cos30 i sin30 j) (1.19 m) i (0.40 m) j.*A B*+ = + ° + ° = +
The equation relating these is *A B C D*+ = + , where

ˆ ˆ ˆ ˆ(1.60 m)(cos50.0 i sin50.0 j) (1.03 m)i (1.23 m)j*C *= ° + ° = +
(a) We find ˆ ˆ(0.16 m )i ( 0.83 m ) j*D A B C*= + − = + − , and the magnitude is *D* = 0.84 m.
(b) The angle is 1tan ( 0.83/ 0.16) 79− − = − ° , which is interpreted to mean 79º south of
east (or 11º east of south).
29. Let 0 2.0 cm*l *= be the length of each segment. The nest is located at the endpoint of
segment *w*.
(a) Using unit-vector notation, the displacement vector for point *A* is

( ) ( )0 0 0 0 0

ˆ ˆ ˆ ˆ ˆ ˆ(cos 60 i sin60 j) j (cos120 i sin120 j) j

ˆ(2 3) j.

*Ad w v i h l l l l
*

*l
*

= + + + = ° + ° + + ° + ° +

= +

Therefore, the magnitude of *Ad * is | | (2 3)(2.0 cm) 7.5 cm*Ad *= + = .
(b) The angle of *Ad * is

1 1
, ,tan ( / ) tan ( ) 90*A y A xd d*θ

− −= = ∞ = ° .
(c) Similarly, the displacement for point *B* is

( ) ( )0 0 0 0 0 0 0

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ(cos60 i sin 60 j) j (cos60 i sin60 j) (cos30 i sin30 j) i

ˆ ˆ(2 3 / 2) i (3 / 2 3) j.

*Bd w v j p o
*

*l l l l l
*

*l l
*

= + + + +

= ° + ° + + ° + ° + ° + ° +

= + + +

Therefore, the magnitude of *Bd * is
2 20| | (2 3 / 2) (3/ 2 3) (2.0 cm)(4.3) 8.6 cm*Bd l*= + + + = = .
(d) The direction of *Bd * is

87

,1 1 1

,

3 / 2 3tan tan tan (1.13) 48 2 3 / 2

*B y
B
*

*B x
*

*d
d
*

θ − − − ⎛ ⎞ ⎛ ⎞+

= = = = °⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠ .

30. Many of the operations are done efficiently on most modern graphical calculators
using their built-in vector manipulation and rectangular ↔ polar “shortcuts.” In this
solution, we employ the “traditional” methods (such as Eq. 3-6).
(a) The magnitude of *a * is 2 2(4.0 m) ( 3.0 m) 5.0 m.*a *= + − =
(b) The angle between *a * and the +*x* axis is tan–1 [(–3.0 m)/(4.0 m)] = –37°. The vector is
37° *clockwise* from the axis defined by i .
(c) The magnitude of *b * is 2 2(6.0 m) (8.0 m) 10 m.*b *= + =
(d) The angle between *b * and the +*x* axis is tan–1[(8.0 m)/(6.0 m)] = 53°.
(e) ˆ ˆ ˆ ˆ(4.0 m 6.0 m) i [( 3.0 m) 8.0 m]j (10 m)i (5.0 m) j.*a b*+ = + + − + = + The magnitude

of this vector is 2 2| | (10 m) (5.0 m) 11 m;*a b*+ = + = we round to two significant
figures in our results.
(f) The angle between the vector described in part (e) and the +*x* axis is tan–1[(5.0 m)/(10
m)] = 27°.
(g) ˆ ˆ ˆ ˆ(6.0 m 4.0 m) i [8.0 m ( 3.0 m)] j (2.0 m) i (11 m) j.*b a*− = − + − − = + The magnitude

of this vector is 2 2| | (2.0 m) (11 m) 11 m,*b a*− = + = which is, interestingly, the same
result as in part (e) (exactly, not just to 2 significant figures) (this curious coincidence is
made possible by the fact that *a b* ⊥ ).
(h) The angle between the vector described in part (g) and the +*x* axis is tan–1[(11 m)/(2.0
m)] = 80°.
(i) ˆ ˆ ˆ ˆ(4.0 m 6.0 m) i [( 3.0 m) 8.0 m] j ( 2.0 m) i ( 11 m) j.*a b*− = − + − − = − + − The magnitude

of this vector is 2 2| | ( 2.0 m) ( 11 m) 11 m*a b*− = − + − = .
(j) The two possibilities presented by a simple calculation for the angle between the
vector described in part (i) and the +*x* direction are tan–1 [(–11 m)/(–2.0 m)] = 80°, and
180° + 80° = 260°. The latter possibility is the correct answer (see part (k) for a further
observation related to this result).

*CHAPTER 3 *88

(k) Since *a b b a*− = − −( )( )1 , they point in opposite (anti-parallel) directions; the angle
between them is 180°.
31. (a) As can be seen from Figure 3-30, the point diametrically opposite the origin (0,0,0)
has position vector *a a a*i j k+ + and this is the vector along the “body diagonal.”
(b) From the point (*a*, 0, 0), which corresponds to the position vector *a* î, the
diametrically opposite point is (0, *a, a*) with the position vector *a a*j k+ . Thus, the
vector along the line is the difference ˆ ˆ ˆi j k*a a a*− + + .

(c) If the starting point is (0, a, 0) with the corresponding position vector ̂ja , the
diametrically opposite point is (*a*, 0*, a*) with the position vector ˆ ˆi k*a a*+ . Thus, the
vector along the line is the difference ˆ ˆ ˆi j k*a a a*− + .
(d) If the starting point is (*a*, *a*, 0) with the corresponding position vector ˆ ˆ i j*a a*+ , the
diametrically opposite point is (0, 0*, a*) with the position vector k̂a . Thus, the vector
along the line is the difference ˆ ˆ ˆi j k*a a a*− − + .
(e) Consider the vector from the back lower left corner to the front upper right corner. It
is ˆ ˆ ˆ i j k.*a a a*+ + We may think of it as the sum of the vector *a *i parallel to the *x* axis and
the vector *a a* j k+ perpendicular to the *x* axis. The tangent of the angle between the
vector and the *x* axis is the perpendicular component divided by the parallel component.
Since the magnitude of the perpendicular component is 2 2 2*a a a*+ = and the
magnitude of the parallel component is *a*, ( )tan 2 / 2*a a*θ = = . Thus θ = °54 7. . The
angle between the vector and each of the other two adjacent sides (the *y* and *z* axes) is the
same as is the angle between any of the other diagonal vectors and any of the cube sides
adjacent to them.
(f) The length of any of the diagonals is given by 2 2 2 3.*a a a a*+ + =

89

32. (a) With *a* = 17.0 m and θ = 56.0° we find *ax* = *a* cos θ = 9.51 m.
(b) Similarly, *ay* = *a* sin θ = 14.1 m.
(c) The angle relative to the new coordinate system is θ ´ = (56.0° – 18.0°) = 38.0°. Thus,

cos 13.4 m.*xa a *θ′ ′= =
(d) Similarly, *ya*′ = *a* sin θ ´ = 10.5 m.
33. Examining the figure, we see that *a*

→
+ *b*

→
+ *c*

→
= 0, where *a*

→
⊥ *b*

→ .

(a) | *a* → × *b* → | = (3.0)(4.0) = 12 since the angle between them is 90º.
(b) Using the Right-Hand Rule, the vector *a b*× points in the ˆ ˆ ˆi j k× = , or the +*z* direction.
(c) | *a*

→
× *c*

→
| = | *a*

→
× (− *a*

→
− *b*

→
)| = | − ( *a*

→
× *b*

→ )| = 12.

(d) The vector *a b*− × points in the ˆ ˆ ˆi j k− × = − , or the − *z* direction.
(e) | *b*

→
× *c*

→
| = | *b*

→
× (− *a*

→
− *b*

→
)| = | −( *b*

→
× *a*

→
) | = | ( *a*

→
× *b*

→ ) | = 12.

(f) The vector points in the +*z* direction, as in part (a).
34. We apply Eq. 3-30 and Eq. 3-23.
(a) ˆ = ( ) k*x y y xa b a b a b*× − since all other terms vanish, due to the fact that neither *a * nor

*b * have any *z* components. Consequently, we obtain ˆ ˆ[(3.0)(4.0) (5.0)(2.0)]k 2.0k− = .
(b) *x x y ya b a b a b*⋅ = + yields (3.0)(2.0) + (5.0)(4.0) = 26.
(c) ˆ ˆ (3.0 2.0) i (5.0 4.0) j *a b*+ = + + + ⇒ ( + ) = (5.0) (2.0) + (9.0) (4.0) = 46*a b b*⋅ .
(d) Several approaches are available. In this solution, we will construct a *b * unit-vector
and “dot” it (take the scalar product of it) with *a *. In this case, we make the desired unit-
vector by

2 2

ˆ ˆ2.0 i 4.0 jˆ .
| | (2.0) (4.0)
*bb
b
*

+ = =

+

We therefore obtain

*CHAPTER 3 *90

2 2

(3.0)(2.0) (5.0)(4.0)ˆ 5.8. (2.0) (4.0)

*ba a b
*+

= ⋅ = = +

35. (a) The scalar (dot) product is (4.50)(7.30)cos(320º – 85.0º) = – 18.8 .
(b) The vector (cross) product is in the k^ direction (by the right-hand rule) with
magnitude |(4.50)(7.30) sin(320º – 85.0º)| = 26.9 .
36. First, we rewrite the given expression as 4( *d*plane

→
· *d*cross

→
) where *d*plane

→
= *d*1

→ +

*d*2
→

and in the plane of *d*1
→

and *d*2
→

, and *d*cross
→

= *d*1
→

× *d*2
→

. Noting that *d*cross
→

is
perpendicular to the plane of *d*1

→
and *d*2

→ , we see that the answer must be 0 (the scalar

[dot] product of perpendicular vectors is zero).
37. We apply Eq. 3-30 and Eq.3-23. If a vector-capable calculator is used, this makes a
good exercise for getting familiar with those features. Here we briefly sketch the method.
(a) We note that ˆ ˆ ˆ8.0 i 5.0 j 6.0k*b c*× = − + + . Thus,

( ) = (3.0) ( 8.0) (3.0)(5.0) ( 2.0) (6.0) = 21.*a b c*⋅ × − + + − −
(b) We note that ˆ ˆ ˆ + = 1.0 i 2.0 j + 3.0k.*b c *− Thus,

( ) (3.0) (1.0) (3.0) ( 2.0) ( 2.0) (3.0) 9.0.*a b c*⋅ + = + − + − = −

(c) Finally,

ˆ ˆ( + ) [(3.0)(3.0) ( 2.0)( 2.0)] i [( 2.0)(1.0) (3.0)(3.0)] j ˆ[(3.0)( 2.0) (3.0)(1.0)] k

ˆ ˆ ˆ 5i 11j 9k

*a b c*× = − − − + − −

+ − −

= − −

.

38. Using the fact that ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k, j k i, k i j× = × = × = we obtain

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 2 2.00i 3.00j 4.00k 3.00i 4.00j 2.00k 44.0i 16.0j 34.0k.*A B*× = + − × − + + = + +
Next, making use of

ˆ ˆ ˆ ˆ ˆ ˆi i = j j = k k = 1 ˆ ˆ ˆ ˆ ˆ ˆi j = j k = k i = 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

we have

91

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ3 2 3 7.00 i 8.00 j 44.0 i 16.0 j 34.0 k 3[(7.00) (44.0)+( 8.00) (16.0) (0) (34.0)] 540.

*C A B*⋅ × = − ⋅ + +
= − + =

39. From the definition of the dot product between *A * and *B *, cos*A B AB *θ⋅ = , we have

cos *A B
AB
*

θ ⋅=

With 6.00*A *= , 7.00*B *= and 14.0*A B*⋅ = , cos 0.333θ = , or 70.5 .θ = °
40. The displacement vectors can be written as (in meters)

1

2

ˆ ˆ ˆ ˆ(4.50 m)(cos 63 j sin 63 k) (2.04 m) j (4.01 m) k ˆ ˆ ˆ ˆ(1.40 m)(cos30 i sin 30 k) (1.21 m) i (0.70 m) k .

*d
*

*d
*

= ° + ° = +

= ° + ° = +

(a) The dot product of 1*d *and 2*d *is
21 2 ˆ ˆ ˆ ˆ ˆ ˆ(2.04 j 4.01k) (1.21i 0.70 k) = (4.01k) (0.70 k) = 2.81 m .*d d*⋅ = + ⋅ + ⋅
(b) The cross product of 1*d *and 2*d *is

1 2

2

ˆ ˆ ˆ ˆ(2.04 j 4.01k) (1.21i 0.70 k) ˆ ˆ ˆ(2.04)(1.21)( k) + (2.04)(0.70)i (4.01)(1.21) j

ˆ ˆ ˆ(1.43 i 4.86 j 2.48k) m .

*d d*× = + × +

= − +

= + −

(c) The magnitudes of 1*d *and 2*d *are

2 2

1

2 2 2

(2.04 m) (4.01 m) 4.50 m

(1.21 m) (0.70 m) 1.40 m.

*d
*

*d
*

= + =

= + =

Thus, the angle between the two vectors is

2

1 11 2

1 2

2.81 mcos cos 63.5 . (4.50 m)(1.40 m)

*d d
d d
*

θ − − ⎛ ⎞ ⎛ ⎞⋅

= = = °⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

41. Since *ab* cos φ = *axbx* + *ayby* + *azbz*,

*CHAPTER 3 *92

cos .φ =
+ +*a b a b a b
*

*ab
x x y y z z*

The magnitudes of the vectors given in the problem are

2 2 2

2 2 2

| | (3.00) (3.00) (3.00) 5.20

| | (2.00) (1.00) (3.00) 3.74.

*a a
*

*b b
*

= = + + =

= = + + =

The angle between them is found from

(3.00) (2.00) (3.00) (1.00) (3.00) (3.00)cos 0.926. (5.20) (3.74)

φ + += =

The angle is φ = 22°. As the name implies, the scalar product (or dot product) between two vectors is a scalar quantity. It can be regarded as the product between the magnitude of one of the vectors and the scalar component of the second vector along the direction of the first one, as illustrated below (see also in Fig. 3-18 of the text):

cos ( )( cos )*a b ab a b*φ φ⋅ = =

42. The two vectors are written as, in unit of meters,
1 1 1 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ4.0 i+5.0 j i j, 3.0 i+4.0 j i j*x y x yd d d d d d*= = + = − = +
(a) The vector (cross) product gives
1 2 1 2 1 2 ˆ ˆ ˆ( )k [(4.0)(4.0) (5.0)( 3.0)]k=31 k*x y y xd d d d d d*× = − = − −
(b) The scalar (dot) product gives
1 2 1 2 1 2 (4.0)( 3.0) (5.0)(4.0) 8.0.*x x y yd d d d d d*⋅ = + = − + =
(c)
2 2 21 2 2 1 2 2( ) 8.0 ( 3.0) (4.0) 33.*d d d d d d*+ ⋅ = ⋅ + = + − + =

93

(d) Note that the magnitude of the *d*1 vector is 16+25 = 6.4. Now, the dot product is
(6.4)(5.0)cosθ = 8. Dividing both sides by 32 and taking the inverse cosine yields θ =
75.5°. Therefore the component of the *d*1 vector along the direction of the *d*2 vector is
6.4cosθ ≈ 1.6.
43. From the figure, we note that *c b*⊥ , which implies that the angle between *c * and the
+*x* axis is θ + 90°. In unit-vector notation, the three vectors can be written as

î ˆ ˆ ˆ ˆi j ( cos )i ( sin ) j ˆ ˆ ˆ ˆi j [ cos( 90 )]i [ sin( 90 )]j

*x
*

*x y
*

*x y
*

*a a
*

*b b b b b
*

*c c c c c
*

θ θ

θ θ

=

= + = +

= + = + ° + + °

The above expressions allow us to evaluate the components of the vectors.
(a) The *x*-component of *a * is *ax* = *a* cos 0° = *a* = 3.00 m.
(b) Similarly, the *y*-componnet of *a * is *ay* = *a* sin 0° = 0.
(c) The *x*-component of* b *is* bx* = *b* cos 30° = (4.00 m) cos 30° = 3.46 m,
(d) and the *y*-component is *by* = *b* sin 30° = (4.00 m) sin 30° = 2.00 m.
(e) The *x*-component of* c * is* cx* = *c* cos 120° = (10.0 m) cos 120° = –5.00 m,
(f) and the *y*-component is* cy* = *c* sin 30° = (10.0 m) sin 120° = 8.66 m.
(g) The fact that *c pa qb*= + implies

ˆ ˆ ˆ ˆ ˆ ˆ ˆi j ( i) ( i j) ( )i j*x y x x y x x yc c c p a q b b pa qb qb*= + = + + = + +
or

,*x x x y yc pa qb c qb*= + =
Substituting the values found above, we have

5.00 m (3.00 m) (3.46 m) 8.66 m (2.00 m).

*p q
q
*

− = + =

Solving these equations, we find *p* = –6.67.
(h) Similarly, *q* = 4.33 (note that it’s easiest to solve for *q* first). The numbers *p* and *q*
have no units.

*CHAPTER 3 *94

44. Applying Eq. 3-23, *F qv B* = × (where *q* is a scalar) becomes

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆi j k i j k*x y z y z z y z x x z x y y xF F F q v B v B q v B v B q v B v B*+ + = − + − + −
which — plugging in values — leads to three equalities:

4.0 2 (4.0 6.0 )

20 2 (6.0 2.0 ) 12 2 (2.0 4.0 )

*z y
*

*x z
*

*y x
*

*B B
*

*B B
B B
*

= −

− = −

= −

Since we are told that *Bx* = *By*, the third equation leads to *By* = –3.0. Inserting this value
into the first equation, we find *Bz* = –4.0. Thus, our answer is

ˆ ˆ ˆ3.0 i 3.0 j 4.0 k.*B *= − − −
45. The two vectors are given by

ˆ ˆ ˆ ˆ8.00(cos130 i sin130 j) 5.14 i 6.13 j

ˆ ˆ ˆ ˆi j 7.72 i 9.20 j.*x y
*

*A
*

*B B B
*

= ° + ° = − +

= + = − −

(a) The dot product of 5*A B*⋅ is

ˆ ˆ ˆ ˆ5 5( 5.14 i 6.13 j) ( 7.72 i 9.20 j) 5[( 5.14)( 7.72) (6.13)( 9.20)]

83.4.
*A B*⋅ = − + ⋅ − − = − − + −

= −

(b) In unit vector notation
3ˆ ˆ ˆ ˆ ˆ ˆ4 3 12 12( 5.14 i 6.13 j) ( 7.72 i 9.20 j) 12(94.6k) 1.14 10 k*A B A B*× = × = − + × − − = = ×
(c) We note that the azimuthal angle is undefined for a vector along the *z* axis. Thus, our
result is “1.14×103, θ not defined, and φ = 0°.”
(d) Since *A*

→
is in the *xy *plane, and *A B*× is perpendicular to that plane, then the answer is

90°.
(e) Clearly, *A*

→ + 3.00 k^ = –5.14 i^ + 6.13 j^ + 3.00 k^ .

(f) The Pythagorean theorem yields magnitude 2 2 2(5.14) (6.13) (3.00) 8.54*A *= + + = .

The azimuthal angle is θ = 130°, just as it was in the problem statement ( *A*
→

is the

95

projection onto the *xy* plane of the new vector created in part (e)). The angle measured
from the +*z* axis is

φ = cos−1(3.00/8.54) = 69.4°.
46. The vectors are shown on the diagram. The *x* axis runs from west to east and the *y*
axis runs from south to north. Then *ax* = 5.0 m, *ay* = 0,

*bx* = –(4.0 m) sin 35° = –2.29 m, *by* = (4.0 m) cos 35° = 3.28 m.

(a) Let *c a b*= + . Then = 5.00 m 2.29 m = 2.71 m*x x xc a b*= + − and
= 0 + 3.28 m = 3.28 m*y y yc a b*= + . The magnitude of *c* is

( ) ( )2 22 2 2.71m 3.28m 4.2 m.*x yc c c*= + = + =
(b) The angle θ that *c a b*= + makes with the +*x* axis is

1 1 3.28tan tan 50.5 50 . 2.71

*y
*

*x
*

*c
c
*

θ − − ⎛ ⎞ ⎛ ⎞

= = = ° ≈ °⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

The second possibility (θ = 50.4° + 180° = 230.4°) is rejected because it would point in a
direction opposite to *c *.
(c) The vector *b a*− is found by adding −*a b*to . The result is shown on the diagram to
the right. Let .*c b a*= − The components are

2.29 m 5.00 m 7.29 m*x x xc b a*= − = − − = −
3.28 m.*y y yc b a*= − =
The magnitude of *c * is 2 2 8.0m*x yc c c*= + = .

*CHAPTER 3 *96

(d) The tangent of the angle θ that *c * makes with the +*x* axis (east) is

3.28 mtan 4.50. 7.29 m

*y
*

*x
*

*c
c
*

θ = = = − −

There are two solutions: –24.2° and 155.8°. As the diagram shows, the second solution is
correct. The vector *c a b*= − + is 24° north of west.
47. Noting that the given 130° is measured counterclockwise from the +*x* axis, the two
vectors can be written as

ˆ ˆ ˆ ˆ8.00(cos130 i sin130 j) 5.14 i 6.13 j

ˆ ˆ ˆ ˆi j 7.72 i 9.20 j.*x y
*

*A
*

*B B B
*

= ° + ° = − +

= + = − −

(a) The angle between the negative direction of the *y* axis ( ĵ− ) and the direction of *A * is

1 1 1 2 2

ˆ( j) 6.13 6.13cos cos cos 140 . 8.00( 5.14) (6.13)

*A
A
*

θ − − − ⎛ ⎞⎛ ⎞⋅ − − −⎛ ⎞⎜ ⎟= = = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠− +⎝ ⎠ ⎝ ⎠

Alternatively, one may say that the −*y* direction corresponds to an angle of 270°, and the
answer is simply given by 270°−130° = 140°.
(b) Since the *y* axis is in the *xy *plane, and *A B*× is perpendicular to that plane, then the
answer is 90.0°.
(c) The vector can be simplified as

ˆ ˆ ˆ ˆ ˆ ˆ( 3.00k) ( 5.14 i 6.13 j) ( 7.72 i 9.20 j 3.00k)

ˆ ˆ ˆ18.39 i 15.42 j 94.61k

*A B*× + = − + × − − +

= + +

97

Its magnitude is ˆ| ( 3.00k) | 97.6.*A B*× + = The angle between the negative direction of the
*y* axis ( ĵ− ) and the direction of the above vector is

1 15.42cos 99.1 . 97.6

θ − −⎛ ⎞= = °⎜ ⎟ ⎝ ⎠

48. Where the length unit is not displayed, the unit meter is understood.
(a) We first note that the magnitudes of the vectors are 2 2| | (3.2) (1.6) 3.58*a a*= = + =

and 2 2| | (0.50) (4.5) 4.53*b b*= = + = . Now,

cos

(3.2) (0.50) (1.6) (4.5) (3.58) (4.53) cos
*x x y ya b a b a b ab *φ

φ

⋅ = + =

+ =

which leads to φ = 57° (the inverse cosine is double-valued as is the inverse tangent, but
we know this is the right solution since both vectors are in the same quadrant).
(b) Since the angle (measured from +*x*) for *a * is tan–1(1.6/3.2) = 26.6°, we know the
angle for *c * is 26.6° –90° = –63.4° (the other possibility, 26.6° + 90° would lead to a *cx* <
0). Therefore,

*cx* = *c* cos (–63.4° )= (5.0)(0.45) = 2.2 m.
(c) Also, *cy* = *c* sin (–63.4°) = (5.0)( –0.89) = – 4.5 m.
(d) And we know the angle for *d * to be 26.6° + 90° = 116.6°, which leads to

*dx* = *d* cos(116.6°) = (5.0)( –0.45) = –2.2 m.
(e) Finally, *dy* = *d* sin 116.6° = (5.0)(0.89) = 4.5 m.
49. The situation is depicted in the figure below.