# ResoluÇÃo halliday (9 ediÇÃo), Exercícios de Física. Universidade Federal do Maranhão (UFMA)

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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 Rπ 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 3V 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 = πr2/2, where r is the radius. Therefore, the volume is

CHAPTER 1 4

2

2 V r zπ=

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 .2V 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 5C B B A

t t t t= + = −

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

( )33 495 s 40B B A A

t t t t′ ′− = − =

when t'AtA = 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( ) 2d 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 2d 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

tθ =

° .

This yields

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

θ °= = °

Using tand r θ= , we have 2 2 2tan 2d r rhθ= = , 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 = πr2. 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 rπ

= = × =

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 = ρV (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 Vρ = , so that the mass can be found from m = ρV, where ρ = 2600 kg/m3. Thus, using V = 4πr3/3, 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 Vρ = , 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 = 4πR3/3, 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 20.00m 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 st = 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 st = 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 Ahρ= where

2(14.0 cm)(17.0 cm) 238 cmA = = 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

Vdoll 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,

Vminiature 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

VΔ = − = −

= ≈ (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

VΔ = − = −

= ≈ × 34. The customer expects a volume V1 = 20 × 7056 in3 and receives V2 = 20 × 5826 in.3, the difference being 31 2 24600 in.V V VΔ = − = , 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: Atotal = 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 ftV = × × = . 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 = ρV:

( ) ( )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 10A

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 hπ 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 dπ ⎛ ⎞

= = = =⎜ ⎟ ⎝ ⎠

Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, 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

vavg 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 Δx1 and Δx2, and the corresponding time intervals be Δt1 and Δt2, respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is Δx = Δx1 + Δx2, and the total time for the trip is Δt = Δt1 + Δt2. 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 Δt 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 Δx1 = 40 km and the time interval is

t1 40 133= =( . km)

(30 km / h) h.

Similarly, during the second part the displacement is Δx2 = 40 km and the time interval is

t2 40 0 67= =( . km)

(60 km / h) h.

The total displacement is Δx = Δx1 + Δx2 = 40 km + 40 km = 80 km, and the total time elapsed is Δt = Δt1 + Δt2 = 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 (Δt1, Δx1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (Δt1, Δx1) to (Δt, Δx) = (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 vup = 40 km/h and vdown = 60 km/h, yields 48 km/h for the average speed. 5. Using x = 3t – 4t2 + t3 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)t2 + (1 m/s3)t3. 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 Δx = 12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement Δx = 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

v0 = (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 v1 = (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 vΔ = . 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 mD = is

5.0 m 10 s 0.50 m/s

Dt R

= = =

9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 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 L1 ≈ 1000 m in the last step. Thus, if L1 and L2 are no different than about 1.4 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter than L2 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 1t , the car moves in the same direction as the wind. Then the effective speed of the car is given by ,1eff 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 t2, the car moves in the opposite direction of the wind and the effective speed would be

,2eff 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

andc w c w d dv v v v t t

+ = − =

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

1 2c

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 10c 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 sst L v= = = in which the slow car has moved a distance of 12.0 mL = , 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 0t = 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 mx x LΔ = − = − = 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/hD 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/hD D D Ds T

= = = +

which should be rounded to 68 km/h. (c) The total distance traveled (2D) 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 (t1, x1) = (T/2, 55T/2) and the second having a slope of 90 and connecting (t1, x1) 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 bxexp( ) 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 = 16te–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 + 6t, 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 – 6.0t2 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 20t 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 x2 = 21.75 cm and x3 = 50.25 cm, respectively. The average velocity during the time interval 2.00 ≤ t ≤ 3.00 s is

v x tavg

cm cm s s

= = − −

Δ Δ

50 25 2175 3 00 2 00 . .

. .

which yields vavg = 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 x2 and x3 (that is, when the particle is at xm = (x2 + x3)/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) = 12t2 – 2t3 we obtain the velocity and the acceleration functions:

v(t) = 24t – 6t2 and a(t) = 24 – 12t

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

(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 vmax = 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,

vavg = 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 2t 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 v1 = +18 m/s at 1 0t = and v2 = –30 m/s at t2 = 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

tb 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 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0). (c) It is clear that a(t) = – 30t is negative for t > 0. (d) The acceleration a(t) = – 30t 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 Δx v t= ' when the velocity is constant during a time interval Δt ' . (a) The entire interval considered is Δt = 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,

vavg 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,

aavg 2 m / s

s m / s= − =2 2 0

360 0 00611. . .

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

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

vavg 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, aavg = 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 aavg).

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 ct2 is that of length, the unit of c must be that of length/time2, or m/s2 in the SI system. (b) Since bt3 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 = 2ct – 3bt2, 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 = x0 = 0 and for t = 1.0 s, x = 1.0 m > x0. 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 Δx = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, 80 mxΔ = − . The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2. (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 = 2c – 6b = 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/sv = × ,

65.70 10 m/sv = × , x0 = 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, v2 = v20 + 2a(xx0). (a) Given that 0 0v = , 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 v0 = 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 v0 = 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 t1 (the duration of part 1) from v = v0 + at. In this way,

120 0 2.0t= + yields t1 = 10 s. We obtain the duration t2 of part 2 from the same equation. Thus, 0 = 20 + (–1.0)t2 leads to t2 = 20 s, and the total is t = t1 + t2 = 30 s. (b) For part 1, taking x0 = 0, we use the equation v2 = v20 + 2a(xx0) 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 x0 = 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 = v0 + 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 v0 = +24.6 m/s and a = – 4.92 m/s2. We also take x0 = 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 2v t at+ (the x graph, shown next, on the left) and v0 + at (the v graph, on the right) over 0 ≤ t ≤ 5 s, with SI units understood.

29. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s2 and a2 = –1.22 m/s2. We use SI units so the velocity at t = t1 is v = 305/60 = 5.08 m/s. (a) We denote Δx as the distance moved during t1, 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 t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances

CHAPTER 2

36

traveled during t1 and t2 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

t3 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 v0 = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = –5.2 m/s2 into v = v0 + 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/s2x 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 v0 to v. 31. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. (a) We solve v = v0 + 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 x0 = 0. The result is

( )2 6 2 131 9.8 m/s (3.1 10 s) 4.6 10 m .2x = × = × 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/sv 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 x0 = 0, and solve x = v0t + 12 at

2 (Eq. 2-15) for the acceleration: a = 2(xv0t)/t2. Substituting x = 24.0 m, v0 = 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/sa = . 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 = v0 + 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 v1 be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h =100/9 m/s speed (corresponding to a passing point of x2 = 76.6 m) of the red car. We have two equations (based on Eq. 2-17):

dx1 = vo t1 + 12 at1 2 where t1 = x1 ⁄ v1

dx2 = vo t2 + 12 at2

2 where t2 = x2 ⁄ v2

CHAPTER 2

38

We simultaneously solve these equations and obtain the following results: (a) The initial velocity of the green car is vo = − 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 st = 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: Δx1 = vo1 t1 + 12 a1 t1

2 where a1 = 2.25 m/s2 and Δx1 = 9004 m Δx2 = v2 t2 − 12 a2 t2

2 where a2 = −0.75 m/s2 and Δx2 = 3(900)4 m In addition, vo1 = v2 = 0. Solving these equations for the times and adding the results gives t = t1 + t2 = 56.6 s. (b) Equation 2-16 is used for part 1 of the trip:

v2 = (vo1)2 + 2ax1 = 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 x0 = –2.0 m. From Table 2-1, we can apply

xx0 = v0t + 12 at 2

with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 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 v0 = 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 ( v0 0= and x0 0= ) at a1

2134= + . m / s until the midway point and then decelerates at a2 2134= − . m / s

until it comes to a stop v2 0=b g at the next station. The velocity at the midpoint is v1, which occurs at x1 = 806/2 = 403m. (a) Equation 2-16 leads to

( )( )2 2 21 0 1 1 12 2 1.34 m/s 403 mv v a x v= + ⇒ = 32.9 m/s.= (b) The time t1 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

vavg 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 vA = 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 aB t 2 where t = 4.0 s .

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

20 + 2t = 12t + 12 aB t 2

where aB = –5/2. There are two distinct roots unless the discriminant

102 − 2(−20)(aB) 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 (xA) tangent to a parabola (xB) at t = 4.

(d) We only care about real roots, which means 102 − 2(−20)(aB) ≥ 0. If |aB| > 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)(aB) > 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 v0 = 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 = v0T – (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/sb 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 v0, 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 (Δx) 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 t0) the gap between the cars is 10.6 m. The speed of the police car at t0 is 30.56 – 5(2.4) = 18.56 m/s. Collision occurs at time t when xyou = xpolice (we choose coordinates such that your position is x = 0 and the police car’s position is x = 10.6 m at t0). Eq. 2-15 becomes, for each car: xpolice – 10.6 = 18.56(t t0) – 12 (5)(t t0)

2

xyou = 30.56(t t0) – 12 (5)(t t0) 2 .

CHAPTER 2

42

Subtracting equations, we find

10.6 = (30.56 – 18.56)(t t0) ⇒ 0.883 s = t t0. At that time your speed is 30.56 + a(t t0) = 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 Δx 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 Δy replacing Δx) 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 2v 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 Δy replacing Δx) 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 y0 = 0. Upon reaching the maximum height y, the speed of the ball is momentarily zero (v = 0). Therefore, we can relate its initial speed v0 to y via the equation 2 200 2v v gy= = − . The time it takes for the ball to reach maximum height is given by 0 0v 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 gy0 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 v0, 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

0t T= > , the ball returns to its original position ( 0y = ). 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 Δy replacing Δx). (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/sv 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 Δy replacing Δx). (a) Starting the clock at the moment the wrench is dropped (v0 = 0), then

2 2 0 2v 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 = v0 – 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 0a 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 Δy replacing Δx). (a) Noting that Δy = yy0 = –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 v0 = –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 Δy replacing Δx) 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, v0 = +12 m/s, and that its initial coordinate is y0 = +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/sv v gt= − = − = −

Its final speed is about 41 m/s. 50. The y coordinate of Apple 1 obeys y – yo1 = – 12 g t

2 where y = 0 when t = 2.0 s. This allows us to solve for yo1, and we find yo1 = 19.6 m. The graph for the coordinate of Apple 2 (which is thrown apparently at t = 1.0 s with velocity v2) is

y – yo2 = v2(t – 1.0) – 12 g (t – 1.0) 2

where yo2 = yo1 = 19.6 m and where y = 0when t = 2.25 s. Thus, we obtain |v2| = 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 = vo + at ⇒ 0 = vo – (9.8 m/s2)(2.0 s) which leads to vo = 19.6 m/s. Now we use Eq. 2-15:

Δy = (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:

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

or | | 59 myΔ = .

47

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

y – y0 = –12 g t 2

where yy0 = –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 = −gτ. 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 Δy replacing Δx) 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 –v0, so that v0 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 Δy’ = Δy = –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 s2v− = − − which leads to v0 = 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 tavg

= Δ Δ

where Δv is the change in its velocity during contact with the ground and 320.0 10 st −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 my = , we obtain

2 2 0 02 ( ) 0 2(9.8 m/s )(0 15 m) 17.15 m/sv 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,

vB2 = vA2 + 2a(yB – yA), with a = –9.8 m/s2, yB – yA = 0.40 m, and vB = 13 vA. It is then straightforward to solve: vA = 3.0 m/s, approximately. 57. The average acceleration during contact with the floor is aavg = (v2 – v1) / Δt, where v1 is its velocity just before striking the floor, v2 is its velocity just as it leaves the floor, and Δt 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 2v v gy= − . With v0 = 0 and y = – 4.00 m, the result is

2

1 2 2(9.8 m/s ) ( 4.00 m) 8.85 m/sv 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 y0 = – 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 tt' where t is the value of time when it lands and t' is one second prior to that. The corresponding distance is yy' = 0.50h, where y denotes the location of the ground. In these terms, y is the same as h, so we have hy' = 0.50h or 0.50h = y' . (a) We find t' and t from Eq. 2-15 (with v0 = 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' = 0.50h, 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 Δy replacing Δx) 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 t1 can be computed from Eq. 2-15, with v0 = 0 and y1 = –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 t2 = t1 – 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 t3 = t1 –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 sv v gt v= − ⇒ = − so that v0 = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the tower (taking y0 = 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 2y 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 v1) the top of the window (which is at y1) at time t1, and passes the bottom (which is at y2) at time t2. We are told y2 – y1 = 1.20 m and t2 – t1 = 0.125 s. Using Eq. 2-15 we have

y y v t t g t t2 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 v0 = 0) reveals the value of y1:

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 (y3 = H) at t3. Because of the symmetry expressed in the problem (“upward flight is a reverse of the fall’’) we know that t3 – t2 = 2.00/2 = 1.00 s. And this means t3 – t1 = 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 y3 = 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 v0 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 y1 = 0.76 m – 0.15 m = 0.61 m, his speed v1 satisfies v v gy0

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 t1 that the player spends ascending in the top Δy1 = 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 tΔ = + ⇒ = =

+

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 t2 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) t2 = 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 2p 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, v0 = 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 (v0 = 0) at t0 = 110 ms and reaches a maximum value of 90 m/s2 at t1= 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 v1, the speed at t1. (b) To compute the speed of the torso at t1=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 v0 = 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 x0 = 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 t1 = 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 x0 = 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 t1 = 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 v0=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 v0 = 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 Δx 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 Δx = 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 3.00dx 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 2v v= implies

2 212.0 3.00 20.0 4.00 4.00 12.0 17.0 0t 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 = v0 – gt yields v0 = 15.7 m/s. (b) One equation that is not dependent on our result from part (a) is yy0 = vt + 12gt

2; this readily gives ymax – y0 = 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 yy0 = v0t + 12gt

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

0 – y0 = (15.68 m/s)(6.00 s) – 12 (9.8 m/s

2)(6.00 s)2.

Therefore, y0 (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

Δx at vt= FHG I KJ =

1 2

2

car truckb g

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 vtΔ = = =

(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

Δx 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 = vavg 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 v0) type. Then the position of the car is given by

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

2

where v0 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 = v0 + 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 ar1 01 01 21

2 = −

and

x v t v ar2 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 x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m and v02 = 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 x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m, and v02 = 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 v02 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 23D , 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 aΔ = = The car then moves at a constant speed pv over a distance 12 0D 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 (v0 = 0) is taken to be x0 = 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/sa 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/sv′ = ) 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 x0 = +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 v0 = –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 0v′ = ) 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 – y01 = – 12 g t

2 where y = 0 when t = 3.0 s. This allows us to solve for yo1, and we find y01 = 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 – y02 = v1(t–1.0) – 12 g (t – 1.0) 2

where y02 = y01 + 10 = 54.1 m and where (again) y = 0 when t = 3.0 s. Thus we obtain |v1| = 17 m/s, approximately. 80. We take +x in the direction of motion. We use subscripts 1 and 2 for the data. Thus, v1 = +30 m/s, v2 = +50 m/s, and x2 – x1 = +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 x2 – x1 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 (v0 = 0) when the clock starts, we find the value of t1 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 x0 = 0). Thus, we are asked to find the value of x1. 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

− + ∫ . 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 (v0 = 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 Δy D gt= − = − 12

2 , or D = gt2/2. Thus, for 1 50.0 mst = ,

D1 3 29 8 50 0 10

2 0 0123=

× =

−. . .

m / s s m = 1.23 cm.

2c h c h

(b) For t2 = 100 ms, ( ) ( )22 3

2 1

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

2 D D

−× = =

(c) For t3 = 150 ms, ( ) ( )22 3

3 1

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

2 D D

−× = =

(d) For t4 = 200 ms, ( ) ( )22 3

4 1

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

2 D D

−× = =

(e) For t4 = 250 ms, D D5 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 x0 = 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 ⎛ ⎞

= =⎜ ⎟ ⎝ ⎠

( ) ( )( )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 = 2D

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 v0 = 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/sv = . (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 x0 = 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 xx0 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 v1 is 1 1/t d v= . Similarly, with a speed v2 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 md = = × , 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: v0 = 5.00 m/s. (b) Substituting v = 15.0 m/s, v0 = 5.00 m/s, and t = 6.00 s into a = (vv0)/t (Eq. 2-11), we find a = 1.67 m/s2. (c) Substituting v = 0 in 2 20 2v 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 mx = . (d) The graphs require computing the time when v = 0, in which case, we use v = v0 + 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 Δy replacing Δx) 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 Δy = 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 = 2ta we have

H gt t H g

= ⇒ = 1 8

2 22 .

We consider two throws, one to height H1 for total time t1 and another to height H2 for total time t2, 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 t2 = 2t1 (as is required by the problem) then H2 = 22H1 = 4H1. 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 x0 = 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 y0 = 0, we have y v t gt= +0 12

2 , which

(with v0 = 0) yields t y g= 2 / . (a) For this part of the motion, y1 = 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 y2 = 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 1t t tΔ = − = 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 x0 = 0 (where v0 = 0), the end of the speeding-up motion occurs at x1 = 1100/2 = 550 m, and the subway train comes to a halt (v2 = 0) at x2 = 1100 m. (a) Using Eq. 2-15, the subway train reaches x1 at

( )1 1 2

1

2 550 m2 30.3 s . 1.2 m/s

xt a

= = =

The time interval t2 – t1 turns out to be the same value (most easily seen using Eq. 2-18 so the total time is t2 = 2(30.3) = 60.6 s. (b) Its maximum speed occurs at t1 and equals

v v a t1 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 Δx replaced by y) because the ball has constant acceleration motion (and we choose y0 = 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 v0

2 and obtain

1 4

2 2 3 2 2 3 3 4

2 2 2v 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 yyA = v2/2g = 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 Δy replacing Δx) 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, v0 = 0, and y0 = 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 v0 = 0. 95. (a) The wording of the problem makes it clear that the equations of Table 2-1 apply, the challenge being that v0, v, and a are not explicitly given. We can, however, apply xx0 = v0t + 12at

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

16 m – 0 = v0(2.0 s) + 1 2 a(2.0 s)

2

27 m – 0 = v0(3.0 s) + 1 2 a(3.0 s)

2

lead to the values v0 = 6.0 m/s and a = 2.0 m/s2. (b) From Table 2-1,

xx0 = vt – 1 2at

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 xx0 = v0t + 12at

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

Δx = (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 = 2gh (from Eq.

69

2-16), and the time for the ball to fall from the board to the lake surface is t1 = 2h g/ (from Eq. 2-15). Now, the time it spends descending in the lake (at constant

velocity v) is

t D v

D gh2 2

= = .

Thus, T = t1 + t2 = 2h g

+ D gh2

, 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 vavg 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 v0 from 210 2y v t gtΔ = + with t = T and Δy = 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, v0 = 0, and y0 = 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 y2

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 y1

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 gt1 12

2= − and the location of diamond 2 is given by y g t2 12

21= − −b g . We are starting the clock when the first object is dropped. We want the time for which y2 – y1 = 10 m. Therefore,

− − + = ⇒ = + = 1 2

1 1 2

10 10 05 152 2g t gt t gb g b g/ . . s. 99. With +y upward, we have y0 = 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 Δy gt= − 12

2 where Δy 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 t1 during which the parachutist is in free fall is (using Eq. 2-15) given by

d gt t1 1 2

1 250 1

2 9 80= =m = 1

2 m / s2.c h

which yields t1 = 3.2 s. The speed of the parachutist just before he opens the parachute is given by the positive root 21 12v gd= , or

v gh1 12 2 9 80 50 31= = =b gc hb g. m / s m m / s.2 If the final speed is v2, then the time interval t2 between the opening of the parachute and the arrival of the parachutist at the ground level is

t v v a2

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 t1 + t2 = 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 Δy into the positive d on the left-hand side, and switches the order of v1 and v2 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 Δy replacing Δx) because this is constant acceleration motion. The ground level is taken to correspond to y = 0. (a) With y0 = h and v0 replaced with –v0, 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 v0 replaced with –v0,

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 y0 = 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 +v0 in the equation where we had put in –v0 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 y0 = h, we obtain

t v gh v

g =

+ +0 2

02 .

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

Δ Δx 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 , sinx ya a a aθ 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 mxa 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 mx ya a+ = − + − =

CHAPTER 3 74

and ( )1 1tan / tan [( 6.86 m) /( 2.50 m)] 250y 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 = rcosθ = (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 2x 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 , sinx yA A A Aθ θ= =

75

Substituting the results calculated above, we obtain

(47.2 m)cos122 25.0 m, (47.2 m)sin122 40.0 mx 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.

° = ° = °

.

° = ° = °

.

° = ° = °

.

° = ° .

° = ° .

° = ° .

5. The vector sum of the displacements dstorm and dnew must give the same result as its originally intended displacement o ˆ(120 km)jd = 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 od d d+ = readily yields A = –100 km and B = 120 km. The

magnitude of dnew is therefore equal to 2 2

new| | 156 kmd 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 md 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)kd = + + . 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 minL , 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) jr A B C= + + = − + −

whose magnitude is

( ) ( )2 22.4 km 2.1 km 3.2 kmr = − + − ≈ . (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 mx x xr c d= + = + = , (b) 3.8 m 2.0 m 5.8 my 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) jr = − + . (b) The magnitude of r is 2 2 2 2| | ( 9.0 m) (10 m) 13 mx 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) jr 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) ma 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) md 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 = CxAx = (15.0 m) cos 200° – (12.0 m) cos 40° = –23.3 m, By =CyAy = (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)jA = , 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), 50B = ° , or

ˆ ˆ ˆ ˆ(7.8 km)(cos50 i sin50 j) (5.01 km)i (5.98 km) jB = ° + ° = + and the unknown C . Thus, A B C= + . (a) The desired displacement is given by ˆ ˆ( 5.01 km) i (0.38 km) jC 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 Δx 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 20.0yc− + − = − 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

→ = (3A) 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 = (3A)2 + A2 ⇒ 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) jR = − + . (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

xθ − ⎛ ⎞= = − ° −⎜ ⎟−⎝ ⎠

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) d1

→ = 4 d3

→ = 8 i^ + 16 j^ , and

(b) d2

→ = d3

→ = 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)jC = ° + ° = + (a) We find ˆ ˆ(0.16 m )i ( 0.83 m ) jD 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 cml = 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 cmAd = + = . (b) The angle of Ad is

1 1 , ,tan ( / ) tan ( ) 90A y A xd dθ

− −= = ∞ = ° . (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 cmBd 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 ma 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 ai 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 aj k+ . Thus, the vector along the line is the difference ˆ ˆ ˆi j ka 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 ka a+ . Thus, the vector along the line is the difference ˆ ˆ ˆi j ka a a− + . (d) If the starting point is (a, a, 0) with the corresponding position vector ˆ ˆ i ja 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 ka 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 2a a a+ = and the magnitude of the parallel component is a, ( )tan 2 / 2a aθ = = . 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 = ( ) kx 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) = 46a 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( dplane

→ · dcross

→ ) where dplane

→ = d1

→ +

d2 →

and in the plane of d1 →

and d2 →

, and dcross →

= d1 →

× d2 →

. Noting that dcross →

is perpendicular to the plane of d1

→ and d2

→ , 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 6.0kb 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 , cosA B AB θ⋅ = , we have

cos A B AB

θ ⋅=

With 6.00A = , 7.00B = and 14.0A 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 1d and 2d 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 1d and 2d 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 1d and 2d 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 bφ 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 jx 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 kx 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 d1 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 d1 vector along the direction of the d2 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 jx 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 kx 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 5A 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 kA 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) 8.54A = + + = .

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 mx x xc a b= + − and = 0 + 3.28 m = 3.28 my 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 bto . 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 mx x xc b a= − = − − = − 3.28 m.y y yc b a= − = The magnitude of c is 2 2 8.0mx 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) 3.58a a= = + =

and 2 2| | (0.50) (4.5) 4.53b 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.

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