CHAPTER 2 Straight-Line Motion Chap 2: Straight-Line Motion Answers to Understanding the Concepts Questions

1. You should be worried about something that might happen to bring the car in front of you to a stop. Your stopping time depends on two factors: Your fixed reaction time, and the time required for your brakes to bring your car to a stop. At a higher initial speed, you travel farther during the time it takes you to react and apply the brakes, and you travel farther in the time it takes your brakes to bring you to a stop. Both factors, then, argue for increasing spacing with increasing speed. 2. The velocity of the chalk is zero at that point, but the acceleration remains g, downward. In fact if the acceleration were zero as well then the chalk would maintain zero velocity there –– i.e., it would “freeze” at the top of its path! 3. We have seen that for a fixed acceleration gx, the relation between fall distance d and fall time t is, 2 1/2 assuming the falling object starts from rest, d = !gxt . Thus for fixed d, t = (2d/gx) . The variation from – l/2 planet to planet for t, that is, with gx, is then (gx) . The larger gx, the smaller the fall time. The speed 1/2 1/2 of the object at the end of the fall is v = gxt = (2dgx) . The speed increases with gx like (gx) . 4. Both you and the bowling balls would be falling at the same rate (g), so there is no reason to worry that any of them would crash onto you. 5. The acceleration of a falling object is equal to g only in a true free fall, which is devoid of any air resistance. In reality, as the object falls, it encounters an air resistance which increases with its speed. Initially, the object is not moving very fast so the air resistance exerted on it is not yet significant, and it falls with an acceleration close to g. As it speeds up, however, it picks up more and more air resistance so its acceleration gradually diminishes, until it reaches a certain speed, at which the upward air resistance equals the downward gravitational pull, whereupon its acceleration is zero and the its speed can no longer increase. This speed is therefore referred to as the terminal speed. So no, the speed of a falling object cannot increase indefinitely. 6. Certainly if the (negative) acceleration has a constant magnitude, the velocity cannot remain positive. Indeed, if the initial velocity has magnitude v0, and the acceleration has the constant magnitude a, then the velocity varies with time according to v = v0 – at, and v = 0 at a time t = v0/a; for times greater than this the velocity is negative. However, the acceleration could steadily decrease in magnitude, while remaining negative, such that the velocity could remain positive. A physical example occurs when a rocket is sent away from Earth with enough initial speed to leave the Solar System--we say that its initial speed exceeds the “escape speed.” If we say that “up” is the positive direction, then the acceleration is negative while the velocity is positive. As the object moves away, the force of gravity on it, and hence its acceleration, decreases in magnitude while remaining negative. For a fast enough start, the object never comes to rest or turns around. Incidentally, the escape speed from Earth is about 11.2 km/s. 7.

2

Treat the jump of the astronaut as a projectile motion. Then the height he can reach is h = v0 /2g, which is inversely proportional to g. Since the astronaut can jump 1.2 m on the surface of Earth, assuming that his 2 initial jumping speed does not change, then he would be able to jump as much as (0.8 m)[(9.8 m/s )/(1.6 2 m/s ) ] = 5 m on the surface of the Moon. Note that we neglected the height of the

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Page 2-1

Chapter 2: Straight-Line Motion

astronaut. To get a more precise result, we need to find how much the center of mass of the astronaut 2 2 can rise on the surface of Earth, and multiply that number by [(9.8 m/s )/(1.6 m/s ) ] = 6.1 to obtain the corresponding value on the Moon. 8.

Common sense tells us that a sudden decrease in speed, i.e., a large deceleration, can cause damage to our body. The airbag prolongs the deceleration process as the object’s speed decreases to zero in a collision. So the magnitude of the deceleration of the object is reduced, lowering the chance of injury or damage.

9. True. This description is consistent with the case when the object is undergoing a uniformly accelerated motion with a pointing to the left, i.e., in the negative x-direction (so a < 0). The velocity of the object as a function of time is v = v0 + at. Since the object starts out moving to the right v0 > 0. But since a <0, v will decrease, and at t = – v0 /a we have v = 0, when the object stops. As t further increases v becomes negative, meaning that the object’s direction of motion is now to the left, while the magnitude of v increase with time so the object speeds up. 10. As long as the motion is in the positive direction, so that the velocity always is positive, there will be no difference between the average speed and the magnitude of the average velocity. This corresponds to Table 2-1. Once negative velocities can occur, even along a straight line, then the average velocity can have any magnitude, including zero, while the average speed is always greater than zero if there is any movement at all. For more complicated motions, the two quantities are not closely related. For example, when a runner goes exactly once around a track, the average velocity is zero (as the net displacement is zero) while the average speed is not. 11. Suppose that the runner does not have a false start, so he or she cannot start running (from rest) until t = 0. So the initial speed at t = 0 should be zero. This is supported by the distance versus time curve, which shows a slope of zero at t = 0. 12. The ancient Greek mathematicians never learned the admittedly subtle notion of a limit –– that the summation of a larger number of smaller and smaller terms, in this case the terms corresponding to the smaller and smaller subdivisions in time and distance traveled, can add to a finite result, in this case the finite time for the runner to catch a tortoise a finite distance ahead of him. Zeno’s paradox certainly doesn’t correspond to our experience! 1/2

2

13. Suppose that the object in question travels x0 from to x. Then v is proportional to (x – x0) , so v is 2 2 proportional to (x – x0). Compare this with the equation v = v0 + 2a (x – x0), and we see that the motion is uniformly accelerated, with zero initial speed (v0 = 0). 14. No. Velocity and acceleration can have different signs. For example, if a car is moving forward, which we choose as the positive direction, then if the driver slams the brake to slow down the car, the acceleration of the car would be negative while its velocity remains positive. In general, if the velocity and acceleration of an object have the same sign, then it must be speeding up; if they have opposite signs it must be slowing down. 15. False. Suppose that the body falls from rest, starting from the origin of the x-axis which points 2 2 downward. Then v = 2gx. As the object has fallen twice as far x doubles to 2x, at which time v = 2g(2x) 2 = 4gx. So v doubles as x does, and v itself increases only by a factor of √2, not four times. 16. Neglecting the effects of air resistance, all three beanbags have exactly the same constant acceleration, namely the acceleration of gravity g, all the time they are in the air. There isn’t much to compare here. 17. Suppose that the three bags are tossed out at about the same time. Then the third one hits the floor first, followed by the second one, then the first one. The first and the third bag will hit the ground with the same speed, which is higher than that of the second one. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 2-2

Fishbane, Gasiorowicz, and Thornton

18. The beanbag is in free fall, so its acceleration is always g, downward, even though its velocity varies. 19. Let the diameter of each wheel be d. Then the distance covered by the bicycle with each turn of its wheels is πd. If the wheels turn at the rate of N turns/s, then the total linear distance covered by the bicycle per second is N(πd). But by definition this is its linear speed: v = N(πd), which gives N = v/πd. So in addition to v we would also need to measure the diameter of each wheel. 20. Let’s assume that the amount of time spent going up equals the amount of time spent coming back down. That means that we can measure the total time of many jumps, an easy thing to do with precision, divide by the total number of jumps to find the time of one jump, and then divide by two to find the time ∆t to fall, say. Our meter stick allows us to measure the height h of a jump; with this information, we can find the 2 2 average acceleration a by applying the formula h = ! a ( t) , or a = 2h/(∆t) . 21. Here are a few examples: A box sliding up or down a straight ramp, a glider on an air track being pulled by a hanging mass, two unequal masses connected by a string hanging over a fixed pulley, a vehicle accelerating uniformly down a straight road. (Note: if an object moves in a circle at uniform speed, its acceleration is not a constant –– the direction of that acceleration is always changing as the object moves. See Chapter 3.) 22. To measure velocity in a straight-line motion we need to know the distance traveled and the time it takes to travel that distance. With a measuring rod we can determine the distance; and since the speed of the film is given (say, 24 frames a second), we also know the time interval between any two frames. For example, we can examine two adjacent frames to determine the distance the person travels, and divide this by the time interval between the two frames. This gives us a reasonably good measurement of the instantaneous velocity of the person, since the time interval between two movie frames is fairly short. Once the velocity is measured, we can then find the difference in velocity between, say, a couple of frames, and divide it by the corresponding time interval. This gives us the approximate value of the instantaneous acceleration. 23. Choose up as positive. The v vs t graph is shown below, where the slope of the line is always equal to –g. The velocity of the ball is zero again at t = 20 s, when it returns to the point where it was initially dropped. 24. As the ball makes contact with the ground it encounters an upward resistance from the ground, quickly causing its downward acceleration to decrease from g to zero. After that the acceleration of the ball becomes upward and the ball eventually acquires an upward speed before leaving contact with the ground. t

t

x

0 (a) object at rest

0

t

x (b) object moving slowly

0

x (c) object moving rapidly

The main disadvantage of switching the axes is that the velocity of the object is no longer equal to the slope of the curve, but rather the reciprocal of the slope.

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Page 2-3

Chapter 2: Straight-Line Motion

Solutions to Problems

1. Displacement = + 32 cm – 27 cm – 23 cm + 39 cm = + 21 cm. The grasshopper is 21 cm from the origin in the positive direction.

ˆ

2. (40 m)i . ˆ

ˆ r d = (40 m)i

3. Distance = (3)(2)(42 m) = 252 m. Displacement = 0, since final position = initial position.

60 Displa cemen t

ˆ (40 m) i

( m )

(80 m) i

40 20 0 0

7

14 21 28 35 42 Time (s)

4. We will choose a coordinate system with the origin at the start and x in the direction of the dash.

0 → 100 m:

r

av

= ∆x /∆t = (50 m – 0 m)i /(5.61 s – 0 s) =

r v

av

r

v

av

r

= ∆x /∆t = (100 m – 50 m)i /(9.86 s – 5.61 s) =

r

= ∆x /∆t = (100 m – 0 m)i / (9.86 s – 0 s) =

(8.9 m/s )i

ˆ

. (11.8 m/s )i

(10.1 m/s )i

ˆ

ˆ

.

.

5. (a) For the first automobile the times for each leg and the elapsed times are ∆t1 = ∆x1/v1 = 15 km/(75 km/h) = 0.20 h = 12 min; t1 = 12 min. ∆t2 = 25 min; t2 = 37 min. ∆t3 = ∆x3/v3 = 40 km/(100 km/h) = 0.4 h = 24 min; t3 = 61 min. ∆t4 = 5 min; t4 = 66 min. ∆t5 = ∆x5/v5 = 55 km/(60 km/h) = 0.92 h = 55 min; t5 = 116 min. (b) From the plot, we see that the two automobiles meet twice. The position of the second auto as a function of time is x2 = [(74 km/h)/(60 min/h)](t – 25 min). At t2 = 37 min, x2 = 15 km = x1; the two autos meet at

140 (min )

50 → 100 m:

r

v

Tim e

0 → 50 m:

120

First auto

100 80 Second auto 60 40 20

t2 = 37 min and x = 15 km. Because they meet during the last leg, we need the position of the first auto during this leg: x1 = [(– 60 km/h)/(60 min/h)](t – 66 min) + 55 km.

0 0

20

40

Position (km)

By setting x1 = x2 , we find that they also meet at t = 68 min and x = 53 km.

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Page 2-4

60

Fishbane, Gasiorowicz, and Thornton

6. 120 car

100

truck x (miles)

80 60 40 20

0

0.5

1

1.5

2

t (hours)

From the plot we see that the truck catches up to and passes the car at around 0.6 hours and then the car catches up to the truck and passes it just short of 1.5 hours. 7. Position (km)

60

60 Time (min)

First auto 40 20 0

20

Second auto 0

40 80 Time (min)

First auto 40

0

120

Second auto 0

40 80 Position (km)

120

First car travels infinitely fast at t = 15 min and at t = 55 min and then travels backward in time, obviously not possible. Second car travels at constant speed. 8. We take north as the y-axis. For the total trip ˆ r r v

av

∆y /∆t = (30 mi + 20 mi) j /[(60 min + 20 min)(1 h/60 min)] =

=

r For the first half: v

av

r ∆y /∆t

=

r For the second half: v

av

ˆ r

= (30 mi) j /[(40 min)(1 h/60 min/h)] = (45 mi/h)j

ˆ

= ∆y /∆t = (20 mi) j /[(40 min)(1 h/60 min/h)] =

9. (a) We take north as the y-axis. The total displacement is r ∆y

(38 mi/h)j

ˆ ˆ

. . ˆ

(30 mi/h)j

.

ˆ

= [(30 mi/h)(2 min) + (45 mi/h)(3 min) + (30 mi/h)(2 min)](1 h/ 60 min) j

= (4.3 mi)j

ˆ

.

(b) The total time for the displacement is ∆t = 2 min + (30 s)(1 min/ 60 s) + 3 min + (3 s)(1 min/ 60 s) + 2 min = 7.55 min, so r r ˆ ˆ . av = ∆y /∆t = (4.25 mi) j /7.55 min = (0.56 mi/min ) j v

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Page 2-5

Chapter 2: Straight-Line Motion

10.

2

Given x = x0 + v0t – !gt , we find the velocity from v = dx/dt = v0 – gt. Because the average velocity is defined as vav = ∆x/∆t: 0 s → 1 s: vav = [x(at t = 1 s) – x(at t = 0 s)]/(1 s – 0 s)

2

2

= {[x0 + v0(1 s) – #!g(1 s) ] – [x0 + v0(0 s) – !g(0 s) ]}/1 s = v0 – !g; 1 s → 2 s: vav = [x(at t = 2 s) – x(at t = 1 s)]/(2 s – 1 s)

2

2

= {[x0 + v0(2 s) – !g(2 s) ] – [x0 + v0(1 s) – !g(1 s) ]}/1 s = v0 – *g; τ s → (τ + 1) s: vav = [x(at t = τ +1) – x(at t = τ)]/[(τ + 1) – τ ] 2

2

= {[x0 + v0(τ + 1) – !g(τ + 1) ] – (x0 + v0 τ – !gτ )}/1

v0 – (τ +!) g, which reduces to previous results with τ = 0 s and τ = 1 s.

=

Because the displacement is the area under the v-t curve, we can approximate the position as xi = xi – 1 + ! (vi + vi – 1)/(ti – ti – 1): t,s 0.0 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5

v,m/s 0.00 0.75 1.75 8.75 21.75 39.75 62.75 103.94 90.75 180.69 122.75 287.44

x, m 0.00 0.19 1.44 6.69 21.94 52.69

v 120

Velocity (m/s)

11.

80

40 t 0

2

4

6

8

Time (s)

12. (a)

y (m) 80 60 40 20 0

0

1 2 3

4 5 6 7 8 9 t (s)

(b) For the average velocity (the 38 m will cancel) r r v av

ˆ

= ∆y /∆t = {(25 m)cos [πt/(6 s)] – (25 m)cos[πt/(6 s)]} j /∆t .

Between 2 s and 3 s, this gives

r

v

r

av

ˆ

ˆ

= ∆y /∆t= (25 m){cos[π(3 s)/(6 s)] – cos[π(2 s)/(6 s)]} j /(1s) =

(−13 m/s )j

.

Between 3 s and 4 s, this gives

r

v

r

av

ˆ

= ∆y /∆t = (25 m){cos[π(4 s)/(6 s)] – cos[π(3 s)/(6 s)]} j /(1 s) =

ˆ (−13 m/s )j

.

(c) From the plot we see that the jumper is closest to the ground at t = 6 s. The instantaneous velocity is

r v

r

ˆ = dy /dt = (25 m)[π/(6 s)]{– sin[πt/(6 s)]} j

ˆ

= – (325m)[π/(6 s)]sin[π(6 s)/(6 s)] j = 0, which agrees with the plot, because the slope of the curve is zero at t = 6 s.

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Page 2-6

Fishbane, Gasiorowicz, and Thornton

∆t = ∆x/v

Displacement (km)

13. If we add the distances for the different trips, we get a total distance of 48 km + 24 km + 5 km + 2.5 km + ... = 79.5+ km. To calculate the distance we can say that the insect will fly at a speed of 240 km/h until the trains collide. The time until the trains collide is relative

= 80 km/(80 km/h + 160 km/h) = @#h. The distance the insect travels is then

80 70 60

Fast train

50 40 Insect

30 20 10 0 0

vinsect∆t = (240 km/h)(@ h) = 80 km.

Slow train 0.05 0.1 0.15 0.2

0.25 0.3 0.35 Time (h)

14. x (m) Light 2

1200 (b) fast

(a) fast

(b) slow

(a) slow Light 1

600

Light 0 0 0 20 40 60 80 120 (c) The timed speed is vtimed = ∆x/∆t = (1200 m – 600 m)/(70 s – 10 s) = 10 m/s (72 km/h). (d) The fastest speed is shown on the plot above as (a) fast:

t (s)

v fastest = ∆x/∆t = (1200 m – 600 m)/(60 s – 20 s) = 15 m/s (108 km/h). 3

2

15. From x = 0.010t – 0.050t + 1.5t cm, we get the speed v 2 = dx/dt = 0.030t – 0.10t + 1.5 cm/s. Then r 2

v

1

v

5

ˆ

= 0.030(1) – 0.10(1) + 1.5 = 1.4 cm/s; v 1 = (1.4 cm/s )i ; r ˆ 2 = 0.030(5) – 0.10(5) + 1.5 = 1.8 cm/s; v 5 = (1.8 cm/s )i ; r ˆ 2 v

v10 = 0.030(10) – 0.10(10) + 1.5 = 3.5 cm/s;

10

=

(3.5 cm/s )i .

The magnitude of the average velocity is v av = ∆x/∆t 3 2 3 2 = {[0.010(10) – 0.050(10) + 1.5(10)] cm} – {[0.010(0) – 0.050(0) + 1.5(0)] cm}/(10 – 0) s = 2.0 cm/s; r ˆ . v av = (2.0 cm/s )i 2

The formula is unrealistic for large times because the t term in v will produce a very large velocity for an ant.

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Page 2-7

Chapter 2: Straight-Line Motion

16. (a) x (m) 3 2 1 t (s) 00 1 2 3 4 5 (b) For the average velocity, we have 2 2 r (2.0 m/s )(tf + 1.0 s) − (2.0 m/s )(ti + 1.0 s) r ∆x

=

tf − ti v av ∆t = Between 1.0 s and 5.0 s, this gives r

r

∆x

(2.0 m/s

2

ˆ

i .

)(5.0 s + 1.0 s) − (2.0 m/s

5.0 s − 1.0 s v av ∆t = Between 2.0 s and 4.0 s, this gives

2

r

r

∆x

(2.0 m/s

2

)(4.0 s + 1.0 s) − (2.0 m/s

ˆ

)(1.0 s + 1.0 s) ˆ

i = (0.366 m/s )i .

=

2

ˆ

)(2.0 s + 1.0 s) ˆ

4.0 s − 2.0 s v av ∆t = i = (0.356 m/s )i . Between 2.8 s and 3.2 s, this gives r 2 2 (2.0 m/s )(3.2 s + 1.0 s) − (2.0 m/s )(2.8 s + 1.0 s) ˆ r ˆ ∆x =

= 3.2 s − 2.8 s v av ∆t = (c) For the instantaneous velocity, we have 2 r 2.0 m/s 1 r dx

2

v = dt = 2 (2.0 m/s )(t + 1.0 s) At 3.0 s this is 2 r 2.0 m/s r 1 dx

i = (0.354 m/s )i .

iˆ .

ˆ

ˆ

i = (0.354 m/s )i .

v = dt = 2 (2.0 m/s 2)(3.0 s + 1.0 s) Note that the smaller the ∆t for ∆t centered at t = 3.0 s, the closer the average velocity approaches the instantaneous velocity. 17. Because the acceleration is constant, we have v = v0 + a(t – t0), or v – v0 = a(t – t0). 2

To reach 35 mi/h: 35 mi/h = 0 + (0.40 m/s )(t – t0), so 2

3

t – t0 = [(35 mi/h)/(0.40 m/s )](1 h/3600 s)(1.609 ⋅ 10 m/mi) = 39 s. 18. Because the acceleration is constant, we have v = v0 + a(t – t0); 2

5 m/s = 10 m/s + (– 0.3 m/s )t, which gives t = 17 s. 19. Because the acceleration is constant, we have v = v0 + a(t – t0); (60 mi/h)(1.61 km/mi)/(3600 s/h) = 0+ a(9.0 s), which gives 2

a= 3.0 m/s = 0.30g.

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Page 2-8

Fishbane, Gasiorowicz, and Thornton 2

+ a∆t and ∆x = v ∆t + !a∆t .

(c) For the separate legs of car A’s motion: I. v = 60 km/h; ∆x = (60 km/h)(1.5 min)(1 h/60 min) = 1.5 km. II. We need the acceleration, which we can get from the velocity equation: 80 km/h = 60 km/h + a(0.25 min), so a = [(80 km/h) – (60 km/h)]/(0.25 min)(60 min/h) 2 = 1.33 km/min . Then 2 ∆x = v0t + !at = (60 km/h)(1 h/60 min)(0.25 min) + 2

0

120 (k m/ h)

0

Car B

40

0

2

2

0

( k m )

!(1.33 km/min )(0.25 min) = 0.29 km.

2

3 Time (min)

4

5 Car A

4 3 2 1 Car B

2

0

!(– 2.67 km/min )(0.5 min) = 0.33 km.

The total distance traveled by A is 1.5 + 0.29 + 2.67 + 0.33 = 4.8 km. For each leg of the motion of car B: vav = ! (v + v0) = ! (120 km/h + 0) = 60 km/h and ∆x = vav∆t = [(60 km/h)/(60 min/h)](1.6 min) = 1.6 km.

1

Di st a n c e

III. v = 80 km/h; ∆x = (80 km/h)(2.0 min)(1 h/60 min) = 2.67 km. IV. We need the acceleration, which we can get from the velocity equation: 0 km/h = 80 km/h + a(0.50 min), so a = [(– 80 km/h)/(0.50 min)](1 h/60 min) 2 = – 2.67 km/min . Then 2 ∆x = v0t + !at = (80 km/h)(1 h/60 min)(0.50 min) +

Car A

80

S p e e d

20. For uniformly accelerated motion we use v = v

0

1

2

3 Time (min)

4

The total distance for B is 3.2 km. 2

2

21. We will take the origin as the location at t = 0 s, so we have x = x0 + v0t + !at = !at . 2 2 At t = 8 s: x = !a(8 s) , at t = 12 s: x = !a(12 s) . 8 12 2 2 Then x – x = 64 m = !a[(12 s) – (8 s) ], which gives 12

8

2

a = 1.6 m/s . 2

22. We choose the origin at the first gate. For constant acceleration we have x = x0 + v0t + !at . For the three gates we have 2 0 = x + v 0 (0.30 s) + !a(0.30 s) ; 0 2 0.60 m = x + v (1.15 s) + !a(1.15 s) ; 0 0 2 1.20 m = x + v (1.70 s) + !a(1.70 s) . 0

0

Thus we have three equations with three unknowns: x0 , v0 , and a. When we solve these, we get x0 = – 0.12 m, v0 = 0.31 m/s, and 2

a = 0.55 m/s . 23. From x = A sin(ωt) we get the velocity and acceleration by differentiation: v = dx/dt = Aω cos(ωt); 2

a = dv/dt = – Aω sin(ωt) .

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Page 2-9

Chapter 2: Straight-Line Motion

24. The average acceleration is found from aav = ∆v/∆t. We will take ∆t = (t + 0.5 s) – (t – 0.5 s) and assume that for small ∆t this is the acceleration at the time midpoint. 2

t,s v,m/s t, s a av, m/s 0.0 0.0 0.5 0.75 1.0 1.0 1.5 1.75 2.0 7.0 2.5 8.75 3.0 13.0 3.5 21.75 4.0 18.0 4.5 39.75 5.0 23.0 5.5 62.75 6.0 28.0 6.5 90.75 7.0 32.0 7.5 122.75 25. The particle never gets farther from the origin than A; it oscillates back and forth through the origin. The magnitude of the velocity is maximum at the origin and zero at x = ±A. The magnitude of the acceleration is maximum at x = ±A and zero at the origin. 2

at

at

26. From x = At + Be we can get the speed: v = dx/dt = 2At + Bae and the acceleration: a = dv/dt = 2A + 2 at Ba e . 2 a(0) Putting the initial conditions into x, we get – 1.5 m = A(0) + Be , or B = – 1.5 m. a(0) Putting the initial conditions into v, we get 0.25 m/s = 2A(0) + Bae ; -1 Ba = 0.25 m/s and thus a = – 0.17 s . –0.1/6, To obtain A, we use the second velocity condition: 0.045 m/s = 2A(0.10 s) + (– 0.25 m/s)e 2 which gives A = –1.0 m/s . At t = 1.0 s the acceleration is 2

-1 2 (–1/6)

a = 2(–1.0 m/s ) + (– 1.5 m/s)(–0.17 s ) e

2

= – 2.0 m/s .

27. We see from a = A – v/t0 that the acceleration is a function of velocity and thus time.

Acceleration

At t = 0, a = A – v0/t0. If this is positive, the velocity will increase, which causes a decrease in the acceleration. The rate of change of the velocity (the slope of the v-t curve) will decrease. Eventually the acceleration becomes zero, at which point the velocity becomes constant (called a terminal velocity). After a long time a = 0, so A – vterminal/t0 = 0, or vterminal = At0 . The graphs assume that A > v0/t0. A t0

O

Velocity

A – (v0/t0)

v0

O

Time

Time 2

2

2

28. For constant acceleration we have x = x0 + v0t + !at and v = v0 + 2a(x – x0). 2 4 2 From the v-equation: (128 mi/h) = 0 + 2a[# mi/ – 0], which gives a = 3.28 ⋅ 10 mi/h . From 4 2 2 the x-equation: # mi = 0 + 0 + ! (3.28 ⋅ 10 mi/h )t , which gives t = 0.00391 h = 14.1 s. 3

29. We convert the speeds to ft/s: (25 mi/h)(5.28 ⋅ 10 ft/mi)/(3600 s/h) = 36.7 ft/s; 50 mi/h = 73.3 ft/s. For constant2 acceleration: 2 2 2 v =v + 2a(x – x ); (73.3 ft/s) = (36.7 ft/s) + 2a(1000 ft – 0), which gives 0

a = 2.02 ft/s

2

0

2

(0.61 m/s ) .

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Page 2-10

Fishbane, Gasiorowicz, and Thornton

30. For constant acceleration the average speed is ! (v0 + v); thus x = vavt = ! (v0+ v)t: 6000 ft = ! [0 + (212 mi/h)(5280 ft/mi)(1 h/3600 s)]t, which gives t = 38.6 s. 31. For constant acceleration, the velocity is given by v = v0 + at: 2 v1 = 0 + (0.50 m/s )(1.0 s) = 0.50 m/s; 2 v2 = 0 + (0.50 m/s )(2.0 s) = 1.0 m/s. For constant acceleration the average velocity is !(v1+ v2): v av = !(0.50 m/s + 1.0 m/s) = 0.75 m/s. 32. For constant acceleration the average speed is !(v0 + v); thus x = vavt = !(v0+ v)t: 3 x = !(0 + 4.2 ⋅ 10 mi/h)(125 s)(1 h/3600 s), which gives x = 73 mi. 2

33. (a) For constant acceleration: v = v0 + at = 8.0 m/s + (– 0.50 m/s )t = 8.0 – 0.50t, easterly with t in s, v in m/s. 2

2

2

(b) For the position: x – x0 = v0t + !at = (8.0 m/s)(5.0 s) + !(– 0.50 m/s )(5.0) = 34 m to the east. 2

34. (a) For constant acceleration: v = v0 + at = 10 m/s + (– 0.2 m/s )t. Plug in t = 1 s and 2 s to obtain v = 9.8 m/s and 9.6 m/s. (b) At t = 2 s, v = 9.6 m/s. So vav = ! (v0 + v) = ! (10 m/s + 9.6 m/s) = 9.8 m/s. 35. The initial speed of the weight just before it touches the surface after falling through a distance h in the 1/2 air is v0 = (2gh) , where h = 6 m. The weight then falls through a distance ∆x in the mud, undergoing 2 2 an acceleration a, reaching final speed v = 0. From v = v0 + 2a∆x = 2gh + 2a∆x = 0 2 4 2 4 2 a = – gh/∆x = – (9.8 m/s )(6 m)/(0.004 m) = – 1 ⋅ 10 m/s , so its magnitude is 1 ⋅ 10 m/s . The average speed of the weight as it decelerates in the mud is vav = !(v0 + v) = !(2gh)

1/2

2

= ![(2(9.8 m/s )(6 m)]

t = ∆x/ vav = 0.004 m/(5.42 m/s) = 7 ⋅ 10

–4

1/2

= 5.42 m/s, so

s.

36. (a) x (m)

v (m/s) 80

400

5 32 t (s) 5 (b) We find the time the airplane rolls from x1 = v0t1; 400 m = (80 m/s)t1, which gives t1 = 5 s. We find the time until the airplane stops from v = v0 + a(t2 – t1);

32 t (s)

2

0 = (80 m/s) + (– 3.0 m/s )(t2 – 5 s), so t2 = 32 s. For the position, we have 2 x2 = x1 + v0(t2 – t1)+ !a(t2 – t1) 2 2 = (400 m) + (80 m/s)(27 s) + !(– 3.0 m/s )(27 s) = 1.5 km.

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Page 2-11

37.

(a) (b) We find the distance before the brakes are applied from x1 = v0t1 = (20 m/s)(1.2 s) = 24 m. 2 We find the acceleration while the brakes are applied from v 2 = v0 + 2a(x2 – x1); 2

Distance (m)

Chapter 2: Straight-Line Motion

80 40 0 0

2

4

2

6 8 Time (s)

Speed (m/s)

0 = (20 m/s) + 2a(80 m – 24 m), which gives a = – 3.6 m/s .

2

2

2

20 10 0

0

2

4

6 8 Time (s)

2

38. Leg I : ∆x1 = v0t + !a1t = 0 + !(3.0 m/s )(4 s) = 24 m; v1 = v0 + at = 0 + (3.0 m/s )(4 s) = 12 2 m/s. Leg II: ∆x2 = v1t + !a2t = (12 m/s)(7 s) + 0 = 84 m. 2 2 2 Leg III: ∆x 3 = v t + !a t = (12 m/s)(15 s) + !(1.0 m/s )(15 s) = 293 m; 1 3 2 v3 = v + a t = 12 m/s + (1.0 m/s )(15 s) = 27 m/s. 2

1

3

2

2

Leg IV: v4 =v3 2 + 2a(∆x4); 0 = (27 m/s) + 2(– 2.5 m/s ) ∆x4 , which gives ∆x4 = 146 m. Total displacement: x = 24 m + 84 m + 293 m + 146 m = 547 m.

39. We will choose a coordinate system with the origin at the release point and take down as the positive 2 direction, so for the penny: x0 = 0; v0 = + 1 m/s; a = + 9.8 m/s . 2 2 2 For the constant acceleration: x – x = v t + !at ; 20 m – 0 = (1 m/s)t + !(9.8 m/s )t . 0

0

Solving the quadratic gives t = 1.92 s, – 2.12 s. Because the penny starts at t = 0, the positive answer is the physically possible answer: t = 1.92 s. 2

40. (a) ∆x1 = v0t + !at ; 2 2 10 m – 0 = 0 + ! (2.8 m/s )t ; the positive answer is t = 2.7 s. 2

(b) ∆x2 = v0t + !at ; 2 2 50 m – 0 = 0 + !(2.8 m/s )t ; the positive answer is t = 6.0 s. 2 The velocity at this time is v = v0 + at = 0 + (2.8 m/s )(6.0 s) = 17 m/s. (c) Because the initial velocity for the second 50 m is the final velocity from the first 50 m, 2 ∆x3 = v0t + !at 2 2 100 m – 50 m = (17 m/s)t + !(2.8 m/s )t ; the positive answer is t = 2.5 s. 2

(d) ∆x4 = v0t + !at ; 2 2 100 m – 0 = 0 + !(2.8 m/s )t ; the positive answer is t = 8.5 s. (e) The early times are longer, but the later times are shorter, because runners cannot maintain a high constant acceleration.

41. From the symmetry of the motion, the initial acceleration lasted for one-half the total time and one-half the total distance traveled. 2 2 2 (a) Thus v2 = v + 2a(x – x ); (144 m/s) = 0 + 2a(3500 m – 0), which gives a = 2.96 m/s . 0

0

2

(b) To find the time we use v = v0 + at; 144 m/s = 0 + (2.96 m/s )t, which gives t = 48.6 s and a total time = 2t = 97.2 s .

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Page 2-12

Fishbane, Gasiorowicz, and Thornton

42. We use a coordinate system with the origin at the corner and t = 0 when you turn the corner. The bus position is given by 2 xb = x0 + v0t + !at = 30 m + 0 + ! (0.6 2 2 m/s )t and its speed is 2

vb = v0 + at = 0 + (0.6 m/s )t. Your speed is vp and your position is given by xp = x0 + vpt = 0 + vpt. For you to catch the bus, your position must be the same as the bus’s position. For you to catch the bus with a minimum speed, your speed must be the same as the bus’s speed when you catch it. This can be seen from the x-t graph; the slope is the speed.

Position

v > minimum

bus v = minimum

O

Time

Thus, xb = xp , or 2 2 30 m + !(0.6 m/s )t = vpt and vb = vp , or 2

(0.6 m/s )t = vp. These equations have two unknowns: t and vp. Solving, we get t = (a)

We choose the origin at the initial passing point and change units of the speeds: 75 mi/h = (75 mi/h)(5280 ft/mi)(1 h/3600 s) = 110 ft/s; 85 mi/h = 125 ft/s. The speeder’s position is xs = x0 + vst = 0 + (110 ft/s)t. For the police car, the distance traveled while accelerating is xp1 = vavt1 = !(125 ft/s + 0)(13 s) = 813 ft. Thereafter its position is xp = xp1 + vp(t – t1) = 813 ft + (125 ft/s)(t – 13 s). The police car will overtake the speeder when xs = xp; (110 ft/s)t = 813 ft + (125 ft/s)(t – 13 s).

10 Distance (1000 ft)

43.

10 s and vp = 6.0 m/s.

8 6 4 2 0 0

20

40 60 80 Time (s)

The solution gives (c) t = 54 s and (b) xs = xp = 5940 ft (1.13 mi). 2

44. With the origin at the point where the brakes are applied, v0 = 35 mi/h = 51 ft/s and a = – 3.0 m/s = 9.8 2 ft/s . We can find the distance she travels before stopping from 2 2 v =v

0

+ 2a(x – x

2

0

);

0 = (51 ft/s) + 2(– 9.8)(x – 0), which gives x = 133 ft . Because this is greater than 90 ft, she does not stop before arriving at the light. 45. We use v = v0 + at; 0 = 58.7 ft/s + (– 0.5g)t, which gives t = 3.65 s.

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Page 2-13

Chapter 2: Straight-Line Motion

46. (a) For the initial acceleration: Car 2

A: aA1 = 5.0 m/s 2 2 To find the speed: vA1 = v0 + 2aA1(x1 – x0); 2 2 vA1 = 0 + 2(5.0 m/s )(200 m – 0), which gives vA1 = 44.7 m/s. 2 To find the time: v A1 = v + a A1 t A1 ; 44.7 m/s = 0 + (5.0 m/s )t A1 , which gives t 2 Car B: a B1 = 4.5 m/s 2 2 = v + 2a (x – x ); To find the speed: v 0

B1

2

B1

0

2

1

A1

= 8.94 s .

0

vB1 = 0 + 2(4.5 m/s )(200 m – 0), which gives vB1 = 42.4 m/s . 2 To find the time: v B1 = v + a B1 t B1 ; 42.4 m/s = 0 + (4.5 m/s )t B1 , which gives t B1 = 9.43 s . 0

(b)

For the second acceleration: Car A: aA2 = 2.5 m/s

2

2

2

To find the speed: vA2 = vA1 + 2aA2(x2 – x1); 2 2 2 vA2 = (44.7 m/s) + 2(2.5 m/s )(400 m – 200 m), which gives vA2 = 54.8 m/s. To find the time: vA2 = vA1 + aA2(tA2 – tA1); 2

54.8 m/s = 44.7 m/s + (2.5 m/s )(tA2 – 8.94 s), which gives tA2 = 13.0 s. 2 Car B: aB2 = 3.0 m/s 2 2 To find the speed: vB2 = vB1 + 2aB2(x2 – x1); 2 2 2 vB2 = (42.4 m/s) + 2(3.0 m/s )(400 m – 200 m), which gives vB2 = 54.8 m/s. To find the time: vB2 = vB1 + aB2(tB2 – tB1); 2

54.8 m/s = 42.4 m/s + (3.0 m/s )(tB2 – 9.43 s), which gives tB2 = 13.6 s.

47. (a) For uniform acceleration: ∆x = vav∆t; 2 ⋅ 10

–2

6

4

m = !(5 ⋅ 10 m/s + 3 ⋅ 10 m/s) ∆t, which gives ∆t = 8.0 ⋅ 10 2

2

–9

s.

(b) To find the acceleration: v = v0 + 2a(x – x0); 6 2 4 2 –2 14 2 (5 ⋅ 10 m/s) = (3 ⋅ 10 m/s) + 2a(2 ⋅ 10 m), which gives a = 6.3 ⋅ 10 m/s . 48. To find the acceleration: v = v0 + at; -4 2 (50 mi/h)(1 h/3600 s) = 0 + a(18 s), which gives a = 7.7 ⋅ 10 mi/s . -4 2 -2 After 36 s: v = v0 + at = 0 + (6.94 ⋅ 10 mi/s )(36 s) = 2.8 ⋅ 10 mi/s = 100 mi/h = 45 m/s. 2

During the first 20 s: x – x0 = v0t + !at = 0 + !(7.7 ⋅ 10 km. During 1 min: x – x0 = 0 + !(7.7 ⋅ 10

-4

2

2

-4

2

2

mi/s )(20 s) = 0.15 mi = 0.25

mi/s )(36 s) = 0.50 mi = 0.80 km.

49. We use a coordinate system with the origin at the ground and up positive. At the instant the object is dropped: 2 2 2 y elev = y0elev + v0elevt1 + !at1 = 0 + 0 + !at1 = !a(3 s) = (4.5 2 s )a; velev = v0elev + at1 = 0 + at1 = a(3 s); 2 Thus for the object’s motion: y0 = (4.5 s )a, v0 = (3 s)a; ao = – 9.8 2 2 m/s : y – y0 = v0t2 + !a0 t2 ; 2

2

2

0 – (4.5 s )a = (3 s)a(3.5 s) + !(– 9.8 m/s )(3.5 s) , which gives a = 4.0 2

2

2

2

m/s . y0 = (4.5 s )a = (4.5 s )(4.0 m/s ) = 18 m.

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Page 2-14

Fishbane, Gasiorowicz, and Thornton

50. We use a coordinate system with the origin at the initial position of B and up positive. The acceleration of A is negative, with the magnitude of the acceleration of B, which is positive. For the motion of the two objects, we have 2

y = y0 + v0t + !at1 ; 2 2 y A = (1.2 m) + 0 + !(– 0.3 m/s )t , 2 2 y B = 0 + 0 + !(+ 0.3 m/s )t . The two objects bump into each other when yA = yB: 2 2

2 2

1.2m + !(– 0.3 m/s )t = !(+ 0.3 m/s )t , which gives t = 2 s. 3

51. For uniform acceleration: x = vavt = !(0 + 60 mi/h)(4 s)(1 h/3600 s)(1.6 ⋅ 10 m/mi) = 54 m. 52. We use a coordinate system with the origin where she slips and down positive. During the 2

2

2

free fall: v1 = v0 + 2a(x1 – x0) = 0 + 2(9.8 m/s )(8 m – 0), or v1 = 12.5 m/s. 2 2 While the rope is stretching: v = v + 2a (x – x ); 2

2

2

1

0 = (12.5 m/s) + 2(– 5)(9.8 m/s )(x

2

2

53. We find the acceleration from v = v 2

0 = (600 m/s) + 2a(20 ⋅ 10 We find the time from v = v0 + at;

2

–2

2 0

2

1

– x ), which gives x 1

2

– x = 1.6 m . 1

+ 2a(x – x ); 0

5

2

5

2

m), which gives a = – 9.0 ⋅ 10 m/s ; |a| = 9.0 ⋅ 10 m/s . 5 2 –4 0 = 600 m/s – (– 9.0 ⋅ 10 m/s )t, we get t = 6.7 ⋅ 10 s .

54. We use a coordinate system with the origin at the top of the tower and down positive. For the motion of the object, we have 2

y = y0 + v0t + !at1 ;

2 2

54.5 m = 0 + 0 + ! (+ 9.8 m/s )t , which gives t = 3.34 s. 55. We take the free fall for 97 stories ≈ 970 ft. The speed at that point is found from v

2

2

= v0 + 2a(x – x0) = 0 + 2(32 ft/s)(970 ft), which gives v ≈ 250 ft/s. If we take this to be the average speed for the rest of the fall, the time to fall the last 3 stories is ∆t ≈ 30 ft/(250 ft/s) ≈ 0.1 s. This is not much time to say anything! 56. We use a coordinate system with the origin at the drum and up positive. For all the sinkers, v0 = 0. We take t = 0 when the string is released. The time for the nth sinker to hit the drum is found from: 2 1/2 y = y + v t + !(– g)t ; 0 = y + 0 + !(– g)t , which gives t = (2y /g) . 0n

0

n

n

0n

n

n

With t1 = 0, equal time intervals requires tn = (n – 1)t2 , n = 3, 4, 5. 1/2 1/2 2 (2y /g) = (n – 1)(2y /g) , or y = (n – 1) y . Thus 0n

y

03

2

= (2) y 2

02

02 =

0n

In terms of the initial positions:

02

0n

4(10 cm) = 40 cm;

y04 = (3) y02 = 9(10 cm) = 90 cm; 2

y 05 = (4) y02 = 16(10 cm) = 160 cm. 57. We use a coordinate system with the origin at the release point and down positive. 2 2 2 From y = y + v0y t + !at ; 1 m = 0 + 0 + ![(9.8 m/s )/6]t , we get t = 1.1 s . 0

58. We use a coordinate system with the origin at the release point and down positive. To find the time for the object to hit the ground, we have 2

y = y0 + v0t + !at ;

2 2

10 m = 0 + 0 + !(+ 25.9 m/s )t , which gives a positive answer of t = 0.88 s.

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Page 2-15

Chapter 2: Straight-Line Motion

59. We use a coordinate system with the origin at the ground and up positive and label the first rock A and the second rock B. To find the time for rock A to hit the ground: 2 yA = y0A + v0AtA + !at ; 2

2

0 = 10 m + (22 m/s) tA + !(– 9.8 m/s ) tA , which gives positive answer of tA = 4.9 s. To find the time for rock B to hit the ground: 2 yB = y0B + v0BtB + !at ; 2

2

0 = 10 m + 0 + !(– 9.8 m/s )tB , which gives positive answer of tB = 1.4 s. Thus rock B must be released ∆t = tA – tB = 3.5 s after rock A. 60. We use a coordinate system with the origin at the release point and up positive, label the first rock A and the second rock B and call the height of the cliff H. For rock A: y = y + v 0A tA + !at 2; A

0

2

A

2

– H = 0 + 0 + !(– g)(4.15 s) , or H = (8.61 s )g.

For rock B to reach the highest point: v 2=v2 + 2(– g)(y top – y ); top 0B 0 1/2 0 = v 2 + 2(– g)(2 m – 0), which gives v 0B = [(4 m)g] . 0B 2 Thus y = y + v t + !at ; B

0B

0

– H = 0 + [(4 m)g]

1/2

2

(6.30 s) + !(– g)(6.30 s) .

2

Combining the equations from A and B gives g = 1.27 m/s and H = 10.9 m. 61. We use a coordinate system with the origin at the ground and up positive. 2

2

For the fall: v1 = v0 + 2a(y1 – y0); 2

2

v1 = 0 + 2(– 9.8 m/s )(0 – 25 m), which gives v1 = – 22 2 2 m/s. For the rebound: v2 = v1 + 2a(y2 – y0); 2 2 0 = (22 m/s) + 2(– 9.8 m/s )(y2 – 0); which gives y2 = 25 2 2 m. For a bounce of 20 m: v4 = v3 + 2a(y3 – y0); 2

2

0 = v3 + 2(– 9.8 m/s )(20 m – 0), which gives v3 = 20 m/s. 62. We use a coordinate system with the origin at the ground and up positive. From the symmetry of the up and down motion, we know that the ball takes !(3.6 s) = 1.8 s to reach the highest point from the ground and !(2.2 s) = 1.1 s to reach the highest point after it passes the window. Thus, after being thrown, the ball takes 0.7 s to reach the window. 2 From y = y0 + v0yt + !at we can write: 2

to reach the window: 10 m = 0 + v0(0.7 s) + !(– g)(0.7 s) ,

2

and to return to the ground: 0 = 0 + v0(3.6 s) + !(– g)(3.6 s) .

2

Solving these two equations for the two unknowns gives us v0 = 18 m/s and g = 9.85 m/s . 63. The velocity is found from v

v

dv =

t

– 4 t dt , which gives v – v0 = –

8 3

8

3/2

t , or v = (15 m/s) –

3

3/2

t

, with v in m/s and t in s.

0

0

If we set v = 0, we find the time to come to a stop: t = 3.16 s. The displacement is found from x

x

dx = 0

t

t

vdt = 0

8 3/2 v0 –3 t dt, which gives x – x0 = v0t –

16 5/2 15 t

, with x inm and t in s.

0

16

The displacement when it stops is x – x0 = 15 m/s 3.16 s – 15 3.16s

5/2

= 28.4m.

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Page 2-16

Fishbane, Gasiorowicz, and Thornton

64. (a) a(t) = dv/dt = 10 – 2t, so t

v=

(b)

t

∫0a(t)dt =∫0(10 − 2t)dt =[10t− t2 ]t0 = 10t− t2. t

x(t) =

t

2

∫0v(t)dt =∫0 (10t − t

2

1

3

t

2

1

3

5t

)dt = − 3 t 0 = 5t − 3 t . 2 3 At t = 5 s, x(t) = 5(5) – @(5) = 83 m.

65. For simplicity, temporarily suppress the units in the calculation below. a(t) = dv/dt = d(4.0 + 8.0 e

–0.5t

)/dt = 8.0(– 0.5) e

–0.5t

=–4e

–0.5t

.

Plug this into the expression for v(t): –0.5t –0.5 t v(t) = 4.0 + 8.0 e = 4.0 – 2 (– 4 e )= 4.0 – 2 a(t), 2 or a(t) = 2 – !v, where t is in s, v in m/s, and a in m/s . 66. The displacement is found from x

t

dx =

0

0

vdt =

t 0

v0

at

v0 e dt, which gives x =

At t = 2 s, we have x = (1 m/s)/(0.5 /s)[e

a

e

at

t

0

(0.5 /s)(2 s)

v at = 0 e –1. a

– 1] = 3.4 m.

67. (a)

y (m)

0.1

0

0.1 0

0.2

0.4

0.6

0.8

1

1.2

t (s)

(c) (d)

The instantaneous velocity is found from the slope. The line at t = 0.15 s gives v0.15 = 0.15 m/s . r ˆ 0.94 m/s . From v = A cos(3πt) j : – 0.147 m/s = A cos[3π(0.15 s)], which gives A =

1.0

Velocity (m/s)

(b)

0.5 0.0

t (s) 0.0

–0.5

–1.0

0.2

0.4

0.6

0.8

1

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Page 2-17

Chapter 2: Straight-Line Motion

68. (a) Take up as positive and put the origin at the ground level. Then the acceleration of the falling body is a(t) = dv/dt = – g(t) = – (g – hg’). To eliminate t, write 0

v = dh/dt, or dt = dh/v, so 2 dv dv dv 1 dv v = = = = −g + g'h, 0 dt dh/v dh 2 dh 2 dv = 2(− g0 + g'h)dh. Integrate both sides: v

∫0

2

dv = 2

y (m) v (m/s) 60 50

h

(−g 0 + g'h)dh,

∫ h0

1

2

v = 2 − g0h+

g'h

2

2

h

2 = 2g0 (h0 − h) + g'(h

2

− h0

),

2

v

10 0

h0 2

y

40 30 20

)

. − h0 v = 2g0 (h0 − h) + g'( h 0 1 2 3 4 5 2 2 (b) a(t) = dv/dt = d h/dt = – g(t) = – (g0 – hg’), or Time (s) 2 2 d h/dt – hg’ + g0 = 0. To avoid solving this 2 the falling distance h0 – h ≈ !g0 t . Thus differential equation for h(t), note that g’ is small so 2 dv/dt = – g0 + hg’ ≈ – g0 + h0 g’ – !g0 g’ t , which we integrate over t to obtain 3 v(t) ≈ (– g0+ h0 g’) t – g0 g’ t /6. 2

69. Because the acceleration is a function of time, a = αt , we obtain the speed by integrating: v

f

t

dv = 0

t

2 αt dt,

3

f

0

0

α3

which gives v =

t.

Thus t =

3v f

1/3

α

.

70. (a) The velocity is found from v 0

t

ge

dv =

– bt

dt , which gives

0

v=–

g – bt e b

t

=

0

g 1–e b

– bt

2

9.8m/s – (0.5 /s)t = 0.5 /s 1 – e = 19.6 m/s 1 – e The displacement is found from y 0

t

t

dy =

vdt = 0

0

y =g t + 1e– bt b

t

– (0.5 /s)t

.

– bt g 1 – e dt, which gives b 2

– bt – (0.5 /s)t – (0.5 /s)t = g bt – 1 + e = 9.8 m/s (0.5 /s)t– 1 + e = 39.2 m (0.5 /s)t – 1 + e .

2

b (0.5 /s) b 0 (c) We find the time to fall 50 m from

2

– (0.5 /s)t

; or 2.276 = 0.5t + e 50 m = 39.2 m (0.5 /s)t – 1 + e A numerical solution of this equation gives t = 4.32 s . (d) When b = 0, we have 2 2 2 y = y + v t + !gt ; 50 m = 0 + 0 + !(9.8 m/s )t , which gives 0

– 0.5t

.

0

t = 3.19 s, less than when air resistance is present.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 2-18

Fishbane, Gasiorowicz, and Thornton

71. If 20 transfers take 15 s, the time for one transfer is τ = 0.75 s. To juggle two objects, one will be in the air for τ, which means !τ to reach the highest point and !τ to fall back to the hand. We use a coordinate system with the origin at the highest point and down positive. 2 2 From y = y0 + v0t + !at ; h2 = 0 + 0 + !g(!τ) , we get 2

h 2 = !g[!(0.75 s)] = 0.7 m. To juggle three objects, two must be in the air during a transfer so each one must be in the air for a time of 2τ or fall for a time of τ. Thus 2

2

h 3 = !g(τ) = !g(0.75 s) = 3 m. 72. We use a coordinate system with the origin at the fingers with up positive. The point on the ruler that will be grabbed is L from the fingers. Its motion can be represented by 2

2

y = y0 + v0t + !at ; 0 = L + 0 + !(– g)t , which gives t = (2L/g)

1/2

2 1/2

= [2(5 in)(1 ft/12 in)/(32 ft/s )]

= 0.16 s.

73. The height reached is determined by the initial velocity. We assume the same initial velocity of the 2

2

jumper on the Moon and Earth. With a vertical velocity of 0 at the highest point, from v = v0 + 2ah we get 2

v0 = 2gEh E = 2gMh M, or h M = (gE/gM)h E = (6)(2 m) = 12 m. 74. The speed attained in a fall through a height h can be found from 2

2

2

v1 = v0 + 2a(y – y0); v1 = 0 + 2gh. If we assume you stop by flexing your knees a distance 2 2 ∆: v2 = v1 + 2a(y – y0); 0 = 2gh + 2a∆, which 2

gives a = – gh/∆ = – 9.8(8 ft/∆ ft) m/s . 75. We use a coordinate system with the origin at the throw and up positive. Then v0 = 7 m/s. From v = v0 + at = v0 – gt; – 4 m/s = 7 m/s – g(7 s), we get g = 1.6 2 m/s . To find the highest point, we have 2

2

v = v0 + 2a(y – y0); 2

2

0 = (7 m/s) + 2(– 1.6 m/s )h, which gives h = 15 m. 76. We use a coordinate system with the origin at June’s initial position and the positive direction toward Bill. 2 2 Then aJ = 1.0 m/s and aB = – 0.9 m/s : 2 2 2 x =x + v t + !a t = 0 + 0 + !(1.0 m/s )t . J 0J 0 J 2 2 2 x =x + v t + !a t = 20 m + 0 + !(– 0.9 m/s )t . B 0B 0 B 2 2 When they meet: x = x , or 0.50t = 20 – 0.45t , which gives J

B

t = 4.6 s and xJ = 11 m. 77. Let da/dt = J = constant. Integrate over time: a(t) = a0 + Jt, where a0 is the initial acceleration at t = 0. Integrate over t again: 2

v(t) = v0 + a0t + Jt /2. Integrate over t once again: 2

3

x(t) = x0 + v0t + a0t /2 + Jt /6 .

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 2-19

Chapter 2: Straight-Line Motion

3

78. v(t) = dx/dt = 4t – t, so the displacement from t1 to t2 is t t2 t2 1 2 2 1 2 3 4 4 1 2 4 ∆x ∫ v(t)dt =∫t = t2 − t2 − t1 (4t − t)dt = t − t − t1 . t 2 2 2 t1 1 1 Plug in t1 = 0.5 s and t2 = 1.5 s to obtain ∆x = 4 m. The average speed is vav = ∆x/∆t = 4 m/(1.5 s – 0.5 s) = 4 m/s. 3 2 a(t) = dv/dt = d(4t – t)/dt = 12t – 1, which is negative before t1 = √(1/12) s = 0.29 s and positive 3 thereafter; while v(t) = 4t – t, which is negative before t2 = 0.5 s and positive thereafter. So between 0.5 s and 1.5 s both v and a have the same direction, indicating that the speed increases monotonically during this time interval. So vmin occurs at t = 0.5 s while vmax at t = 1.5 s. Plug these values of t into the 3 expression v(t) = 4t – t to obtain vmin = 0 and vmax = 12 m/s. Note that vav = 4 m/s is indeed within this range. [Since the acceleration is not a constant here, however, we generally cannot write vav = !(vmin + vmax).] =

79. (a) Take up as the y-axis. The velocity of the ball just before it hits the switch after it falls through a distance h is ˆ ˆ r 1/2 ˆ 2 1/2 . v = (2gh) (– j )= [2(9.8 m/s )(10 m)] (– j ) = (−14 m/s )j (b) The ball rebounds to a height h’, so its the velocity as it leaves the switch is 1/2 ˆ r 1/2 ˆ ˆ 2 v = (2gh’) j = [2(9.8 m/s )(9 m)] j = (13 m/s )j . ˆ ˆ r rr 4 2 ˆ (c) a av = (v f – v i)/∆t = [(13.3 m/s) j – (–14 m/s) j ]/0.002 s = (1.4⋅ 10 m/s )j . i

f

80. Our coordinate system has the origin at the archer, up positive and t = 0 when the balloon is dropped. For the balloon: 2 yB = y0B + v0B t + !at = 200 m + 0 + ! (– 9.8 2 2 m/s )t . For the arrow: 2 2 yA = y0A + v0A t + !gt = 0 + (40 m/s)(t – 5 s) + !(– 9.8 m/s )(t – 5 2 s) . The arrow intercepts the balloon when yB = yA: 2 2 2 2 (200 m) – !(9.8 m/s )t = (40 m/s)(t – 5) – !(9.8 m/s )(t – 5 s) . _

2

2

The time of intercept is t = 5.87 s, which means y = 200 m – !(9.8 m/s )(5.87 s) = 31 m. 81. We use a coordinate system with the origin at the top of the snowbank with down positive. To 2

2

find the distance: v = v0 + 2a(y – y0); 2 2 0 = (40 m/s) + 2(– 50)(9.8 m/s )(y – y0), which gives y – y0 = 1.6 m. To find the time: v = v0 + at; 2

0 = 40 m/s + (– 50)(9.8 m/s )t, which gives t = 0.08 s. 82. We use a coordinate system with the origin at the ground and up positive. For all the weights, v0 = 0 and a = – g. Take t = 0 when the string is released, and the lowest weight is at the ground. The time for the jth sinker to hit the drum is found from: 2

2

y = y0j + v0tj + !(– g)tj ; 0 = y0j + 0 + !(– g)tj , which gives tj = (2y0j/g) 1/2

1/2

, with j = 0, 1, 2, …,

n. If we call t1 = (2L0/g) the time for the second lowest weight to hit the ground, for equal time intervals we have tj – tj–1 = t1 , j = 2, 3, …, n. This gives the sequence t2 = 2t1 , t3 = 3t1 , t4 = 4t1 , …. In terms of the1/2 initial positions: 1/2 2 (2y /g) = j (2L /g) , or y = j L. 0j

0

0j

0

Because each y0 is the sum of the corresponding L’s, we have y02 = L0 + L1 = 4L0 , L0 + L1 + L2 = 9L0 , L0 + L1 + L2 + L3 = 16L0 , …. Successively combining these, we get L1 = 3L0 , L2 = 5L0 , L3 = 7L0 , …, or Lj = (2j + 1)L0 , j = 1, 2,…, n . © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 2-20