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Motion in One Dimension
CHAPTER OUTLINE 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived from Calculus
ANSWERS TO QUESTIONS
* An asterisk indicates an item new to this edition.
*Q2.1 Count spaces (intervals), not dots. Count 5, not 6. The first drop falls at time zero and the last drop at 5 × 5 s = 25 s. The average speed is 600 m25 s = 24 ms, answer (b).
Q2.2 The net displacement must be zero. The object could have moved away from its starting point and back again, but it is at its initial position again at the end of the time interval.
Q2.3 Yes. Yes. If the speed of the object varies at all over the interval, the instantaneous velocity will sometimes be greater than the average velocity and will sometimes be less.
*Q2.4 (a) It speeds up and its acceleration is positive. (b) It slows down overall, since final speed 1 ms is slower than 3 ms. Its acceleration is positive, meaning to the right. (c) It slows down and its acceleration is negative. (d) It speeds up to final speed 7 ms. Its acceleration is negative, mean- ing toward the left or towards increasing-magnitude negative numbers on the track.
Q2.5 No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past to give car B greater acceleration just then.
*Q2.6 (c) A graph of velocity versus time slopes down steadily from an original positive (northward) value. The graph cuts through zero and goes through increasing-magnitude negative values, all with the same constant acceleration.
*Q2.7 (i) none. All of the disks are moving. (ii) (b) shows equal spacing, meaning constant nonzero velocity and constant zero acceleration. (iii) (b) This question has the same physical meaning as question (ii). (iv) (c) shows positive acceleration throughout. (v) (a) shows negative (leftward) acceleration in the last three images.
*Q2.8 Tramping hard on the brake at zero speed on a level road, you do not feel pushed around inside the car. The forces of rolling resistance and air resistance have dropped to zero as the car coasted to a stop, so the car’s acceleration is zero at this moment and afterward. Tramping hard on the brake at zero speed on an uphill slope, you feel thrown backward against your seat. Before, during, and after the zero-speed moment, the car is moving with a downhill acceleration if you do not tramp on the brake. Brian Popp suggested the idea for this question.
15
*Q2.9 With original velocity zero, displacement is proportional to the square of time in (12) at^2. Making the time one-third as large makes the displacement one-ninth as large, answer (c).
Q2.10 No. Constant acceleration only. Yes. Zero is a constant.
Q2.11 They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity of magnitude vi. This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same.
*Q2.12 For the release from rest we have (4 ms)^2 = 02 + 2 gh. For case (i), we have vf^2 = (3 ms)^2 + 2 gh = (3 ms)^2 + (4 ms)^2. Thus answer (d) is true. For case (ii) the same steps give the same answer (d).
*Q2.13 (i) Its speed is zero at b and e. Its speed is equal at a and c, and somewhat larger at d. On the bounce it is moving somewhat slower at f than at d, and slower at g than at c. The assembled answer is d > f > a = c > g > b = e. (ii) The velocity is positive at a, f, and g, zero at b and e, and negative at c and d, with magnitudes as described in part (i). The assembled answer is f > a > g > b = e > c > d. (iii) The acceleration has a very large positive value at e. At all the other points it is −9.8 ms 2. The answer is e > a = b = c = d = f = g.
Q2.14 (b) Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point. So your ball must travel a smaller distance to the passing point that the ball your friend throws.
16 Chapter 2
SOLUTIONS TO PROBLEMS
Section 2.1 Position, Velocity, and Speed
P2.1 (a) vavg x t
m 2 s m s
(b) vavg =^ =
m 4 s
. m s
(c) vavg x x t t
2 1
m m 4 s s
. m s
(d) vavg x x t t
2 1
m m 7 s s
. m s
(e) vavg x x t t
2 1
0 m s
P2.2 (a) (^) vavg = 2 30. m s
(b) v = = m m s = 16.1 m s
x t
(c) vavg
x t
m m 5.00 s m s
18 Chapter 2
P2.7 (a)
(b) At t = 5 0. s, the slope is v ≈ 58 m= 23 2.5 s
m s.
At t = 4 0. s, the slope is v ≈ 54 m= 1 3 s
8 m s.
At t = 3 0. s, the slope is v ≈ 4 9 m= 1 3.4 s
4 m s.
At t = 2 0. s, the slope is v ≈ 3 6 m= 9 4.0 s
.0 m s.
(c) a avg (^) t
v 23 5 0
m s 4 6 s
m s 2 .
(d) Initial velocity of the car was zero.
P2.8 (a) v = ( − ) ( − )
m s
m s
(b) v = ( − ) ( − )
m s
. m s
(c) v = ( − ) ( − )
m m s s
(d) v =
− −( ) ( − )
m s s
m s
P2.9 Once it resumes the race, the hare will run for a time of
t x (^) f xi x
= −^ =
v
1 000 m 800 m 25 8 m s
s.
In this time, the tortoise can crawl a distance x (^) f − xi = ( 0 2. m s (^) )( 25 s) = 5 00. m.
Section 2.3 Acceleration
P2.10 Choose the positive direction to be the outward direction, perpendicular to the wall.
v (^) f = vi + at : a t
= = − −(^ ) ×
∆ − = ×
v 22 0 25 0 3 50 10
m s m s. s
00 4 m s 2
FIG. P2.
Motion in One Dimension 19
P2.11 (a) Acceleration is constant over the first ten seconds, so at the end of this interval
v (^) f = vi + at = 0 + ( 2 00. m s 2 )(10 0. s) = 20 0. m s.
Then a = 0 so v is constant from t =10 0. s to t =15 0. s. And over the last five seconds the velocity changes to
v (^) f = vi + at = 20 0. m s + −( 3 00. m s (^2) )( 5 00. s) = 5 00. mss (^).
(b) In the first ten seconds,
x (^) f = x (^) i + vi t + at = + + (^) ( )( ) =
(^2) 2 00. m s 2 10 0. s (^210) 00 m.
Over the next five seconds the position changes to
x (^) f = x (^) i + vi t + at = + ( )( ) + =
(^2 100) m 20 0. m s 5 00. s 0 20 00 m.
And at t = 20 0. s,
x (^) f = x (^) i + vi t + at = + ( )( ) + −
(^2) m. m s. s ( 3 3 00. m s 2 )(5 00. s ) 2 = 262 m.
P2.12 (a) Acceleration is the slope of the graph of v versus t.
For 0 < t <5 00. s, a = 0.
For 15 0. s < t <20 0. s, a = 0.
For 5 0. s < t <15 0. s, a t t
f i f i
v v .
a =
. m s 2
We can plot a t ( ) as shown.
(b) a t t
f i f i
v v
(i) For 5 00. s < t <15 0. s, t (^) i = 5 00. s, vi = −8 00. m s, t
a t t
f f f i f i
s v m s
v v ))
. m s 2.
(ii) (^) t (^) i = 0 , vi = −8 00. m s, t (^) f = 20 0. s, v (^) f = 8 00. m s
a t t
f i f i
v v (^) 8 00 8 00 20 0 0
. m s 2
FIG. P2.
0 5 10 15
t (s) 20
a (m/s 2 )
Motion in One Dimension 21
Acceleration as a function of time
time
acceleration, arbitrary units
P2.15 (a) At t = 2 00. s, x = ⎡⎣ 3 00 2 00. (. ) − 2 2 00 2 00. (. ) + 3 00. ⎤⎦ m =11 0. m.
At t = 3 00. s, x = ⎡⎣ 3 00 9 00. (. ) −^2 2 00 3 00. (. ) +3 00. ⎤⎦ m =24 0. m
so vavg x t
m m. 3.00 s s
m ss.
(b) At all times the instantaneous velocity is
v =^ d ( − + ) = (^) ( − ) dt
3 00. t^2 2 00. t 3 00. 6 00. t 2 00. m s
At t = 2 00. s, v = ⎡⎣ 6 00 2 00. (^) (. (^) ) − 2 00. ⎤⎦ m s = 10 0. m s.
At t = 3 00. s, v = ⎡⎣ 6 00 3 00. (^) (. (^) ) − 2 00. ⎤⎦ m s = 16 0. m s.
(c) (^) a avg (^) t
v 16 0 10 0 3 00 2 00
m s m s s s m s 2
(d) At all times a d dt
= ( 6 00. t −2 00. ) = 6 00. m s 2. This includes both t = 2 00. s and
t = 3 00. s.
P2.16 (a) (^) a t
v 8 00 6 00
m s. s
m s 2
(b) Maximum positive acceleration is at (^) t = 3 s, and is the slope of the graph, approximately (6 − 2)(4 − 2) = 2 m s 2.
(c) a = 0 at t = 6 s , and also for (^) t > 10 s.
(d) Maximum negative acceleration is at t = 8 s, and is the slope of the graph, approximately −1 5. m s 2.
22 Chapter 2
Section 2.4 Motion Diagrams
- P2.17 (a) The motion is slow at first, then fast, and then slow again.
t
t
0 t
a
x
(b) The motion is constant in speed.
0 t
x
0 t
0 t
a
(c) The motion is speeding up, and we suppose the acceleration is constant.
0 t
x
0 t
0 t
a
P2.18 (a)
(b)
(c)
(d)
(e)
(f ) One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction.
24 Chapter 2
(d) (i) x 1 (^) 0 1 a t 1^2 t^2 2
= + = (^) (3 3. m s (^2) ) or x^ 1 = (1 67.^ m s^2 ) t^2
(ii) x^ 2 t
= ( 15 s) (^) [ 50 m s − (^0) ] + ( 50 m s) ( − 15 s )or x (^) 2 = ( 50 m s) − t 375 m
(iii) For 40 s ≤ t ≤ 50 s,
x
t (^3) t 0
⎝⎜^
area under vs from to 40 s
v (^) 11 2
a 3 ( t − 40 s) + ( 2 50 m s ) ( t − 40 s)
or
x (^) 3 375 1 250 1 t^2 2
= m + m + (^) (− 5 0. m s (^2) ) ( − 40 s) + (( 50 m s) ( t −40 s)
which reduces to
x (^) 3 = ( 250 m s ) − ( t (^) 2 5. m s (^2) ) t 2 −4 375m.
(e) v = total displacement= total elapsed time
1 875 m s
m s 50
P2.23 (a) vi = 100 m s, a = −5 00. m s 2 , v f = vi + at so 0 = 100 − 5 t , v f^2^ = vi^2^ + 2 a x ( f − xi ) so
0 = ( 100 ) −^2 2 5 00(. ) ( x f − 0 ). Thus x f = 1 000 m and t = 20 0. s.
(b) 1 000 m is greater than 800 m. With this acceleration the plane would overshoot the runway: it cannot land.
- P2.24 (a) For the first car the speed as a function of time is v = vi + at = − 3.5 cms + 2.4 cms^2 t. For the second car, the speed is + 5.5 cms + 0. Setting the two expressions equal gives
−3.5 cms + 2.4 cms^2 t = 5.5 cms so t = (9 cms)(2.4 cms 2 ) = 3.75 s.
(b) The first car then has speed −3.5 cms + (2.4 cms^2 )(3.75 s) = 5.50 cms , and this is the constant speed of the second car also.
(c) For the first car the position as a function of time is x (^) i + vi t + (12) at^2 = 15 cm − (3.5 cms) t + (0.5)(2.4 cms 2 ) t^2. For the second car, the position is 10 cm + (5.5 cms) t + 0. At passing, the positions are equal: 15 cm − (3.5 cms) t + (1.2 cms 2 ) t^2 = 10 cm + (5.5 cms) t (1.2 cms 2 ) t^2 − (9 cms) t + 5 cm = 0.
We solve with the quadratic formula:
t = 9 ±^9 −^ 4 1 2^5 = +^ −^ = 2 1 2
and 6 6 90. s and 0.604 s
(d) At 0.604 s, the second and also the first car’s position is 10 cm + (5.5 cms)0.604 s = 13.3 cm. At 6.90 s, both are at position 10 cm + (5.5 cms)6.90 s = 47.9 cm.
continued on next page
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Motion in One Dimension 25
(e) The cars are initially moving toward each other, so they soon share the same position x when their speeds are quite different, giving one answer to (c) that is not an answer to (a). The first car slows down in its motion to the left, turns around, and starts to move toward the right, slowly at first and gaining speed steadily. At a particular moment its speed will be equal to the constant rightward speed of the second car. The distance between them will at that moment be staying constant at its maximum value. The dis- tance between the cars will be far from zero, as the accelerating car will be far to the left of the steadily moving car. Thus the answer to (a) is not an answer to (c). Eventually the accelerating car will catch up to the steadily-coasting car, whizzing past at higher speed than it has ever had before, and giving another answer to (c) that is not an answer to (a). A graph of x versus t for the two cars shows a parabola originally sloping down and then curving upward, intersecting twice with an upward-sloping straight line. The parabola and straight line are running parallel, with equal slopes, at just one point in between their intersections.
P2.25 In the simultaneous equations:
v v
v v
xf xi x
f i xi xf
a t
x x t
we have
v v
v v
xf xi
xi xf
= − ( )( )
= (^) ( +
m s s
m
2
))(^ )
4 20. s
So substituting for vxi gives 62 4 1 2
. m = ⎡⎣ v (^) xf + (5 60. m s (^2) )( 4 20. s) + vxf ⎤⎦( 4 20. s)
. m s = vxf + (^) (5 60. m s (^2) )( 4 20. s). Thus vxf = 3 10. m s.
P2.26 Take any two of the standard four equations, such as
v v
v v
xf xi x
f i xi xf
a t
x x t
Solve one for vxi , and substitute into the other: v (^) xi = vxf − a tx
x f − x i = 1 ( xf − a tx + xf ) t
v v.
Thus
x (^) f − x (^) i = vxf t − a tx
We note that the equation is dimensionally correct. The units are units of length in each term. Like the standard equation x (^) f − x (^) i = vxi t +^1 a tx 2
(^2) , this equation represents that displacement is a quadratic function of time. Our newly derived equation gives us for the situation back in problem 25,
62 4 4 20 1 2
. m = vxf (. s ) − (^) ( −5 60. m s (^2) )( 4 20. s)^2
vxf =
m m 4.20 s m s.
Motion in One Dimension 27
(c) For constant vc and t 0 , ∆ x is minimized by maximizing t (^) m to t (^) m = t 0. Then
∆ x t t t min =^ c − c
v (^) ⎠ = v 0 0 0
(e) This is realized by having the servo motor on all the time. (d) We maximize ∆ x by letting t (^) m approach zero. In the limit ∆ x = vc (^) ( t (^) 0 − (^0) ) = vct 0.
(e) This cannot be attained because the acceleration must be finite.
P2.31 Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit,
x x t a t
t a t t
f i xi x
i d d d d
d
v
v v
v
2
vvi + 1 a td 2
(a) The speed halfway through the photogate in space is given by
v (^) hs^2 v (^) i^2^2 a v (^) i^2 av (^) d td 2
= + ⎛⎝⎜ ^ ⎞⎠⎟ = + ∆.
v (^) hs = v (^) i^2^ + av (^) d ∆ t (^) d and this is not equal to vd unless a = 0.
(b) The speed halfway through the photogate in time is given by v^ ht vi a^ d
t = +
and this is equal to vd as determined above.
P2.32 Take the original point to be when Sue notices the van. Choose the origin of the x -axis at Sue’s car. For her we have x (^) is = 0 , vis = 30 0. m s, a (^) s = −2 00. m s 2 so her position is given by
x (^) s ( ) = t x (^) is + vis t + 1 a ts = ( ) + t − 2
(^2). m s (^) ( 2 00. m s (^22) ) t (^2).
For the van, x (^) iv = 155 m, v (^) iv = 5 00. m s, av =^0 and
x (^) v ( ) = t x (^) i v + vivt + 1 a tv = + ( ) + t 2
(^2 155) 5 00. m s 0.
To test for a collision, we look for an instant t (^) c when both are at the same place:
30 0 155 5 00 0 25 0 155
2 2
t t t t t
c c c c c
From the quadratic formula
t (^) c =
. s or 11 4. s.
The roots are real, not imaginary, so there is a collision. The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position
155 m + (5 00. m s (^) )(11 4. s ) = 212 m.
28 Chapter 2
- P2.33 (a) Starting from rest and accelerating at ab = 13 0. mi h s⋅, the bicycle reaches its maximum speed of vb ,max = 20 0 mi h. in a time
t b a b b ,
,max. .
v (^) mi h mi h s
s.
Since the acceleration ac of the car is less than that of the bicycle, the car cannot catch the bicycle until some time t > t (^) b ,1 (that is, until the bicycle is at its maximum speed and coast- ing). The total displacement of the bicycle at time t is
∆ x (^) b = a tb b + (^) b ( t − tb )
1 2 , ,max , 1 .
v
ft s ⎝⎝⎜^ mi h
⎠⎟^
⎠ (^ ) + (^ )^ −
13 0. mi h 1 54. 2 20 0. s
s mi h t 1 1 54
29 4 22 6
s
ft s ft
⎣⎢^
= ( ) − t
The total displacement of the car at this time is
∆ x (^) c = a tc =
⎝⎜^
⎠⎟^
2.^ ft s 9 00. mi h
mi h s
⎣⎢^
t^2 = ( 6 62. ft s) t^2
At the time the car catches the bicycle ∆ x (^) c = ∆ xb. This gives
( 6 62.^ ft s^2 ) t^^2 = (29 4^.^ ft s^ ) − t^ 22 6. ft or^ t^^2 − (^ 4 44.^ s^ ) + t^ 3 42.^ s^2 =^0
that has only one physically meaningful solution t > t (^) b ,1. This solution gives the total time the bicycle leads the car and is t = 3 45. s.
(b) The lead the bicycle has over the car continues to increase as long as the bicycle is moving faster than the car. This means until the car attains a speed of v (^) c = vb ,max =20 0 mi h.. Thus, the elapsed time when the bicycle’s lead ceases to increase is
t a
b c
v (^) ,max. .
mi h 2 22 mi h s
s
At this time, the lead is
( ∆^ x^ b^ −∆^ x^ c ) (^) max =^ ( ∆^ x^ b −∆ xc ) (^) t =2 22. s =^ ⎡⎣ (^ 29 4.^ ft s^ )^ (^2.^222 s^ ) −22 6.^ ft^ ⎤⎦ −^ ⎡⎣( 6 62.^ ft s^2 )(2 22^.^ s)^2 ⎤⎦⎦
or (^) ( ∆ x (^) b −∆ xc ) (^) max = 10 0 ft..
P2.34 As in the algebraic solution to Example 2.9, we let t represent the time the trooper has been moving. We graph x (^) car = 45 + 45 t
and x (^) trooper = 1 5. t 2.
They intersect at t = 31 s.
FIG. P2.
x (km)
t (s) 10 20 30 40
car
police officer
30 Chapter 2
P2.37 Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a = − g = −9 80. m s 2. During the flight, Goff went 1 mile (1 609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight:
v f^2^ = v i^2^ + 2 a y ( f − yi ) : 0 = vi^2 − 2 9 80(. m s 2 )(1 609 m)
vvi = 178 m s.
His time in the air may be found by considering his motion from just after launch to just before impact: y (^) f − y (^) i = vi t + at
= ( m s) − t (^) ( −9 80. m s (^2) ) t^2.
The root t = 0 describes launch; the other root, t = 36 2. s, describes his flight time. His rate of pay is then
pay rate = $. = ( )( ) = .
s
$ s s h h.
We have assumed that the workman’s flight time, “a mile,” and “a dollar,” were measured to three-digit precision. We have interpreted “up in the sky” as referring to the free fall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away from the job, in order for Goff to survive to resume work.
P2.38 We have y^ f^ = −^ gt^ +^ it + yi
(^2) v
0 = −( 4 90. m s (^2) ) t^2 − (8 00. m s) + t 30 0. m.
Solving for t ,
t = ±^ + −
Using only the positive value for t , we find that t^ =^ 1 79.^ s^.
P2.39 (a) y^ f −^ y^ i =^ vi^ t^ + at
2 : 4 00. = (1 50. ) v − (4 90. )(1 50. ) 2
i and^ vi =^ 10 0.^ m s upward^.
(b) v f = vi + at = 10 0. − ( 9 80. )(1 50. ) = −4 68. m s
v (^) f = 4 68. m s downward
P2.40 The bill starts from rest vi = 0 and falls with a downward acceleration of 9 80. m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of
∆ y = vi t − gt = − ( )( ) = −
(^2 0) 4 90. m s 2 0 20. s 2 0 20. m.
This distance is about twice the distance between the center of the bill and its top edge (≈ 8 cm ). Thus, David will be unsuccessful.
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Motion in One Dimension 31
P2.41 (a) v (^) f = vi − gt : v (^) f = 0 when t = 3 00. s, and g = 9 80. m s 2. Therefore,
vi = gt = ( 9 80. m s (^2) )( 3 00. s ) = 29 4. m s.
(b) y f − y i = 1 ( f + i ) t
v v
y (^) f − yi = (^1) ( )( ) = 2
29 4. m s 3 00. s 44 1. m
*P2.42 We can solve (a) and (b) at the same time by assuming the rock passes the top of the wall and fi nding its speed there. If the speed comes out imaginary, the rock will not reach this elevation. vf^2 = vi^2 + 2 a ( xf − x (^) i ) = (7.4 ms) 2 − 2(9.8 ms^2 )(3.65 m − 1.55 m) = 13.6 m^2 s 2 so the rock does reach the top of the wall with vf = 3.69 ms.
(c) We find the final speed, just before impact, of the rock thrown down: vf^2 = vi^2 + 2 a ( xf − x (^) i ) = (−7.4 ms) 2 − 2(9.8 ms^2 )(1.55 m − 3.65 m) = 95.9 m 2 s^2 vf = −9.79 ms. The change in speed of the rock thrown down is 9.79 − 7.4 = 2.39 ms
(d) The magnitude of the speed change of the rock thrown up is 7.4 − 3.69 = 3.71 ms. This does not agree with 2.39 m s.
The upward-moving rock spends more time in flight, so the planet has more time to change its speed. P2.43 Time to fall 3.00 m is found from the equation describing position as a function of time, with
vi = 0 , thus: 3 00
. m = (^) (9 80. m s (^2) ) t^2 , giving t = 0 782. s.
(a) With the horse galloping at 10 0. m s, the horizontal distance is vt = 7 82. m.
(b) from above t = 0 782. s
P2.44 y^ =^ 3 00.^ t^3 : At t = 2 00. s, y = 3 00 2 00. (^) (. (^) ) =^3 24 0. m and
vy dy dt = = 9 00. t^2 = 36 0. m s ↑.
If the helicopter releases a small mailbag at this time, the mailbag starts its free fall with velocity 36 ms upward. The equation of motion of the mailbag is
y b = ybi + vi t − 1 gt = + t − ( ) t
Setting yb = 0 , 0 = 24 0. + 36 0. t −4 90. t^2. Solving for t , (only positive values of t count), t = 7 96. s.
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Motion in One Dimension 33
P2.47 (a) J da dt
= = constant
da = Jdt
a = J (^) ∫ dt = Jt + c 1
but a = ai when t = 0 so c (^) 1 = ai. Therefore, a = Jt + ai
a d dt d adt adt Jt a (^) i dt Jt a ti c
= (^) ∫ = (^) ∫( + ) = + +
v
v v^1 2
2 2
but (^) v = vi when t = 0 , so c (^) 2 = vi and v = 1 + + v 2
Jt^2 a t i i
v v
v v
∫ ∫
dx dt dx dt
x dt Jt a t dt
x J
i i
2
tt a t t c x x
i i i
3 2 3
v
when t = 0 , so c (^) 3 = x (^) i. Therefore, x = 1 Jt + a ti + (^) it + xi 6
(^3 2) v.
(b) a^2 = (^) ( Jt + a (^) i )^2 = J t^2^2 + a (^) i^2^ + 2 Ja ti a^2 = a (^) i^2^ + (^) ( J t^2^2 + 2 Ja ti )
a^2 a (^) i^2^2 J^1 Jt^2 a ti 2
⎝⎜^
Recall the expression for v : v = 1 + + v 2
Jt^2 a t i i. So^ (^ v^ − vi^ ) =^ Jt^ + a ti
(^2). Therefore,
a^2 = ai^2^ + 2 J ( v − vi (^) ).
34 Chapter 2
P2.48 (a) a d dt
d dt
=^ v^ = ⎡⎣− 5 00. × 10 7 t^2 + 3 00. × 105 t ⎤⎦
a = − (^) ( 10 0. × 10 7 m s (^3) ) + t 3 00. × 105 m s^2
Take x (^) i = 0 at t = 0. Then v = dx dt
x dt t t dt
x
t t − = = (^) (− × + × )
(^0) ∫ ∫ 5 00 10 3 00 10
0
7 2 5 0
v..
7 3 5 2
7 3
× + ×
= − ( × ) +
t t
x m s 3 t ( 1 1 50. × 10 5 m s 2 ) t^2.
(b) The bullet escapes when a = 0 , at − (^) ( 10 0. × 10 7 m s (^3) ) + t 3 00. × 10 5 m s^2 = 0
t = × ×
3 00 10 = 3 00 × 10 −
s. 10.0 10 7 s.
(c) New v = −( 5 00 × (^10 7) ) ( 3 00 × 10 − (^3) ) + (^) ( 3 00 × (^10) ) 3 00 × 10 (^2 )
... (^) (. −−^3 )
v = − 450 m s + 900 m s = 450 m s.
(d) x = − (^) ( 1 67 × (^10 7) ) ( 3 00 × 10 −^3 ) + (^) ( 1 50 × (^10) ) 3 00 × 10 (^3 )
... (^) (. −−^3 ) 2
x = − 0 450. m + 1 35. m = 0 900. m
Additional Problems
- P2.49 (a) The velocity is constant between ti = 0 and t = 4 s. Its acceleration is 0.
(b) a = ( v 9 − v 4 )(9 s − 4 s) = (18 − [−12]) (ms)5 s = 6.0 ms 2.
(c) a = ( v 18 − v 13 )(18 s − 13 s) = (0 − 18) (ms)5 s = −3.6 ms^2.
(d) We read from the graph that the speed is zero at t = 6 s and at 18 s.
(e) and (f) The object moves away from x = 0 into negative coordinates from t = 0 to t = 6 s, but then comes back again, crosses the origin and moves farther into positive coordinates until t = 18 s , then attaining its maximum distance, which is the cumulative distance under the graph line:
(−12 ms)(4 s) + (−6 ms)(2 s) + (9 ms)(3 s) + (18 ms)(4 s) + (9 ms)(5 s) = −60 m + 144 m = 84 m.
(g ) To gauge the wear on the tires, we consider the total distance rather than the resultant dis- placement, by counting the contributions computed in part (f) as all positive:
