Solutions manual fishbane physics for scientists and engineers 3rd

The velocity of the chalk is zero at that point, but the acceleration remains g, downward.. As it speeds up, however, it picks up more and more air resistance so its acceleration gradua

C HAPTER 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 g x , the relation between fall distance d and fall time t is, assuming the falling object starts from rest, d = !g x t2 Thus for fixed d, t = (2d/g x)1/2 The variation from

planet to planet for t, that is, with g x , is then (g x) – l/2 The larger g x, the smaller the fall time The speed

of the object at the end of the fall is v = g x t = (2dg x)1/2 The speed increases with g x like (g x)1/2

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 Treat the jump of the astronaut as a projectile motion Then the height he can reach is h = v02/2g, which is inversely proportional to g Since the astronaut can jump 1.2 m on the surface of Earth, assuming that his

initial jumping speed does not change, then he would be able to jump as much as (0.8 m)[(9.8 m/s2)/(1.6 m/s2) ] = 5 m on the surface of the Moon Note that we neglected the height of the

astronaut To get a more precise result, we need to find how much the center of mass of the astronaut can rise on the surface of Earth, and multiply that number by [(9.8 m/s2)/(1.6 m/s2) ] = 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!

13 Suppose that the object in question travels x0 from to x Then v is proportional to (x – x0)1/2, so v2 is

proportional to (x – x0) Compare this with the equation v2 = v02 + 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

downward Then v2 = 2gx As the object has fallen twice as far x doubles to 2x, at which time v2 = 2g(2x)

= 4gx So v2 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

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 average acceleration a by applying the formula h = ! a ( t)2, or a = 2h/(∆t)2

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

(a) object at rest (b) object moving slowly (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

© 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently

(b) From the plot, we see that the two automobiles meet twice.

The position of the second auto as a function of time is 20

At t2 = 37 min, x2 = 15 km = x1; the two autos meet at 0 20 40 60

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.

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

Fishbane, Gasiorowicz, and Thornton

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

For the second half: v av = ∆y /∆t = (20 mi) j /[(40 min)(1 h/60 min/h)] = (30 mi/h)j

(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

v

av = ∆y /∆t = (4.25 mi) j /7.55 min = (0.56 mi/min ) j

© 2005 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently

10 Given x = x0 + v0t – !gt2, we find the velocity from v = dx/dt = v0 – gt Because the average velocity is defined as vav =

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

11 Because the displacement is the area under the v-t curve, we can approximate the position as

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

v

av = ∆y /∆t= (25 m){cos[π(3 s)/(6 s)] – cos[π(2 s)/(6 s)]} j /(1s) = (−13 m/s )j

v

av = ∆y /∆t = (25 m){cos[π(4 s)/(6 s)] – cos[π(3 s)/(6 s)]} j /(1 s) = (−13 m/s )j

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

Fishbane, Gasiorowicz, and Thornton

13 If we add the distances for the different trips, we 80

get a total distance of

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

Fishbane, Gasiorowicz, and Thornton

20 For uniformly accelerated motion we use v = v + a∆t and ∆x = v ∆t + !a∆t2.

∆x = (60 km/h)(1.5 min)(1 h/60 min) = 1.5 km. (k m/ h) Car B

II We need the acceleration, which we can get from

∆x = vav∆t = [(60 km/h)/(60 min/h)](1.6 min) = 1.6 km. The total distance for B is 3.2 km.

21 We will take the origin as the location at t = 0 s, so we have x = x0 + v0t + !at2 = !at2

22 We choose the origin at the first gate For constant acceleration we have x = x0 + v0t + !at2

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

The magnitude of the acceleration is maximum at x = ±A and zero at the origin

26 From x = At2 + Be at we can get the speed: v = dx/dt = 2At + Bae at and the acceleration: a = dv/dt = 2A +

B a2e at

Putting the initial conditions into x, we get – 1.5 m = A(0)2 + Be a(0) , or B = – 1.5 m

Putting the initial conditions into v, we get 0.25 m/s = 2A(0) + Bae a(0);

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

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

O

Time

28 For constant acceleration we have x = x0 + v0t + !at2 and v2 = v02 + 2a(x – x0)

From the v-equation: (128 mi/h)2 = 0 + 2a[# mi/ – 0], which gives a = 3.28 ⋅ 104 mi/h2 From

the x-equation: # mi = 0 + 0 + ! (3.28 ⋅ 104 mi/h2)t2, which gives t = 0.00391 h = 14.1 s

29 We convert the speeds to ft/s: (25 mi/h)(5.28 ⋅ 103 ft/mi)/(3600 s/h) = 36.7 ft/s; 50 mi/h = 73.3 ft/s For constant acceleration:

v = v 2 + 2a(x – x ); (73.3 ft/s)2= (36.7 ft/s)2 + 2a(1000 ft – 0), which gives

a = 2.02 ft/s2 (0.61 m/s2)

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:

v1 = 0 + (0.50 m/s2)(1.0 s) = 0.50 m/s;

v2 = 0 + (0.50 m/s2)(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:

x = !(0 + 4.2 ⋅ 103 mi/h)(125 s)(1 h/3600 s), which gives x = 73 mi

33 (a) For constant acceleration: v = v0 + at = 8.0 m/s + (– 0.50 m/s2)t

= 8.0 – 0.50t, easterly with t in s, v in m/s

(b) For the position: x – x0 = v0t + !at2 = (8.0 m/s)(5.0 s) + !(– 0.50 m/s2)(5.0)2 = 34 m to the east

34 (a) For constant acceleration: v = v0 + at = 10 m/s + (– 0.2 m/s2)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

air is v0 = (2gh)1/2, where h = 6 m The weight then falls through a distance ∆x in the mud, undergoing

an acceleration a, reaching final speed v = 0 From v2 = v02 + 2a∆x = 2gh + 2a∆x = 0