4. solve...2.4 chapter 2 object is at rest and has not moved. if velocity is positive or negative,...

2.2 Chapter 2

REFLECT Since acceleration is the change in velocity with time,

0 0.x xx

v va

t Velocity can be positive,

negative, or zero. Acceleration can be zero in any of these cases. An object with zero velocity and zero acceleration

is at rest.

4. SOLVE Yes. Think of throwing an apple straight up in the air. Establish a coordinate system in which “up” is

positive. On its way up, the apple’s velocity is positive. While it is falling back down, the velocity is negative.

During the entire time the apple is in the air, the acceleration due to gravity is negative.

REFLECT While the apple is moving upward, it is slowing down because acceleration points downward. After it

has reached its highest point, acceleration still points downward, and the apple “speeds up” in the negative

direction.

5. SOLVE Yes. If you throw the apple of Question 4 straight upward, it slows down to a velocity of zero at its

highest point. The acceleration due to gravity is still negative at this highest point.

REFLECT Think about a drag race with the positive x-direction toward the finish line. At the instant the race

starts, the vehicles have zero velocity. At the same instant, the acceleration has a large positive value. As a result,

the velocities of the vehicles increase rapidly as they move down the track.

6. SOLVE The speed at the end of each time interval is

distance traveled.v

t

The change in speed, 0v v , is

2 1 2 1distance traveled distance traveled distance traveled distance traveledv

t t t

The change in speed with time is acceleration, so

2 1

2 1

2

distance traveled distance traveleddistance traveled distance traveled

.tat t

Since the time intervals are all equal, their squares are all equal. We are told that the differences in the distances

traveled are also equal. Therefore acceleration is constant.

REFLECT Let us go one step further and add the distances in this series of observations. After each succeeding

time interval, we have traveled the distance obtained by adding all the distances traveled up through the current

time interval, as shown in the table below.

Number of time interval Total distance

traveled

0 0 0

1 0 1 1

2 0 1 3 4

3 0 1 3 5 9

4 0 1 3 5 7 16

5 0 1 3 5 7 9 25

With an initial speed 0 = 0,v these values match the results of the kinematic equation for distance with constant

acceleration:

21

2distance traveled a t

where 22m sa and t increases by one second for each time interval.

https://testbanknetwork.com/download/solution-manual-for-essential-college-physics-1st-edition-by-andrew-rex-richard-wolfson/

link full download:

Motion in One Dimension 2.3

7. SOLVE

REFLECT Look at a point on a velocity-versus-time graph. This point gives the slope of the line on the

corresponding position-versus-time graph at that same time. The slope of the velocity-versus-time graph gives the

location of a point on the acceleration-versus-time graph at that same time.

8. SOLVE Suppose your car has a maximum speed of 100mi h, or about 45m s. At an acceleration of 23.0m s , it

would take 2

45 m s15 s

3.0 m st to reach that maximum speed.

REFLECT This answer assumes that this maximum acceleration remains constant. In practice, acceleration

depends on the gear you are in and the speed of the engine (we’ll study more about what is known as “torque” in

Chapter 8. After the transmission shifts into higher gears, the acceleration decreases. As you speed up, air

resistance also increases. This also contributes to decreasing the acceleration.

9. SOLVE The velocities of each of the two cars are given as constant, with no mention of change in velocity. The

acceleration of each car must be zero.

REFLECT If the problem had stated that the two velocities were given only for a specific point in time, then the

answer might have been ambiguous. An alternative is that the faster car has zero acceleration while the slower car

has a positive acceleration. The faster car passes the slower car, but in a short time, the car that was slower will

accelerate to a speed faster than 25 m s and will pass the first car. This is what happens when you pass a stationary

police car while you are driving at constant speed over the speed limit!

10. SOLVE At a given position, velocity is different, speed is the same, and acceleration is the same.

REFLECT At any particular position or height, velocity has the same magnitude, but opposite sign. Going up, the

sign on velocity is positive. Coming back down, the sign is negative. Speed does not change because ,yv v so

the sign of yv does not matter. Acceleration is constant during free fall, with 29.80 m s .ya g

11. SOLVE If an object’s average velocity is zero, then its displacement must be zero.

xx v t

For 0,xv 0 0 mx t

If the object’s acceleration is zero, then the displacement might be zero, positive, or negative, depending on the

value and sign of the velocity.

REFLECT Even if the object’s instantaneous velocity has not been zero throughout the time interval, a velocity of

zero implies that it has returned to its starting point, so 0.x If acceleration is zero and velocity is zero, then the

2.4 Chapter 2

object is at rest and has not moved. If velocity is positive or negative, then displacement constantly increases in the

same direction as velocity.

12. SOLVE With a sufficient number of steps, you will eventually reach your destination. If we graph displacement

versus time, we see that even though we divide the distance traveled by two in each successive step, we also divide

the time it takes by two. The speed therefore remains constant. Therefore, if you cover half the distance to your

goal in 12

,t you will also cover the remaining half in an additional time of 12

,t no matter the number of parts

into which it is divided.

REFLECT In fact, the distance is divided into an infinite number of steps. Mathematicians have found this to be a

“summable” infinite series, written as

1 1 1 1 1 11

2 4 8 16 32 2n

So the sum of this infinite number of steps toward the goal is simply the distance to the goal. Since the paradox

gives no limit on the number of steps and velocity remains constant, you can reach your goal in exactly twice the

time it takes you to go half the distance toward the goal.

MULTIPLE-CHOICE PROBLEMS

13. ORGANIZE AND PLAN We are to find average speed. We use the definition of average speed, .v x t The

responses are all in m s, so we must convert distance to meters and time to seconds.

Known: 385,000 km;x 2.5 day.t

SOLVE Using the definition of speed,

x

xv

t

385,000 km 1000 m km1800 m s

2.5 day 24 h day 3600 s h xv

The answer is response (C).

REFLECT The escape velocity from Earth is about11,000 m s. A spacecraft traveling toward the moon initially

achieves escape velocity, and then slows down due to Earth’s gravity. If the intention is to make a “soft” landing

on the moon with 0 m s,xv it makes sense that the average speed must be less than the escape velocity.

14. ORGANIZE AND PLAN We are given time and speed and are asked to find distance. We use the definition of

speed, v x t and rearrange it to solve for distance.

Known: 32 m s;v 35 s.t

SOLVE The definition of speed is

xv

t

Rearranging, we get

x v t

Substituting known values,

32 m s 35 s 1100 mx

The answer is response (D).

REFLECT Even though the problem states that the cheetah’s top speed is 32 m s we consider it to be the average

speed for the duration of this problem. A distance of 1100 m is about 1200 yards, the length of 12 football fields.

This is a reasonable distance for a cheetah to chase its prey.

15. ORGANIZE AND PLAN In this problem, the runner has already run part of the race with known quantities. We

have to figure out what’s going to happen during the rest of the race. We’ll use subscript “1” for the first part of the

race, subscript “2” for the remainder of the race, and no subscript if the variable applies to the entire race. We need

to find the speed 2v the runner must maintain until the finish line. We know that it’s a 1500 m race, with 1200 m


already run. Since the runner must finish the race in under 4 minutes, our strategy is to find the time already

elapsed and then the allowed time remaining. From that, we can calculate the necessary speed.

Known: 1500 m;x 1 1200 m;x 1 6.14 m s;v 4 min.t

SOLVE For the part of the race already completed, we use

11

1

xv

t

Solving for 1t , we get

1

11

1200 m195.44 s

6.14 m s

xt

v

2 4 min 60 s min 195.44 s 44.56 st

2 1500 m 1200 m 300 mx

22

2

300 m6.73 m s

44.56 s

xv

t

The answer is response (a).

REFLECT This is nearly a 1-mile race. It is normal for runners to sprint, or run faster, at the end of a long race.

The required average speed to finish under 4 minutes is only a little faster than the average speed in the first part of

the race, which is reasonable.

16. ORGANIZE AND PLAN In this problem, the runner travels forward and backward along the x-axis. We are asked to

calculate the runner’s average velocity. We can’t just average the two velocities given; this would violate math

rules for the definition of velocity. Our strategy is to calculate the total displacement and the total time. From these

we can calculate the average velocity for the entire run. We’ll use subscript “1” for running forward, subscript “2”

when jogging backward, and no subscript for a variable that applies to the entire run. Since we are calculating

average velocity rather than speed, we carefully note the signs of the values.

Known: 1 9.2 m s;xv 2 3.6 m s;xv 1 100 m;x 2 50 m.x

SOLVE First, the total displacement

1 2 100 m 50 m 50 mx x x

The runner ends up only 50 m from the starting point.

Next, the total time

1

11

100 m10.87 s

9.2 m sx

xt

v

22

2

50 m13.89 s

3.6 m sx

xt

v

1 2 10.87 s 13.89 s 24.76 st t t

Finally, the average velocity

50 m2.0 m s

24.76 sx

xv

t


REFLECT Velocity is based upon displacement, not distance. The farther backward (toward the origin) the runner

jogs, the smaller the displacement and the closer the average velocity will approach zero.

17. ORGANIZE AND PLAN This problem is about the properties of displacement in one-dimensional motion and

comparison of displacement with distance.

SOLVE The definition of displacement in one-dimensional motion is the difference between final and initial

positions in a coordinate system:

0x x x

2.6 Chapter 2

A one-dimensional coordinate system extends without bound in both the positive and negative directions from the

origin. Therefore both x and 0x can have values that are positive, negative, or zero, independently of each other.

Since this is the case, the difference between the two values can also be positive, negative, or zero.

The answer is response (b).

REFLECT Distance is the sum of the lengths of all the straight-line segments of a trip, without regard to sign.

Distance is always positive. The displacement value may be equal to the distance value, but does not have to be.

Displacement is never greater than distance.

18. ORGANIZE AND PLAN In this problem we are to find average speed from two different speeds and distances run

at those speeds. We don’t just average the speeds. We must divide the total distance by the total time, so we have

to find each of these first. We’ll use the formula for average speed .v x t

Known: 1 4.0 m s;v 2 6.0 m s;v 1 2 60 m.x x

SOLVE The total distance is just the sum of the two distances.

1 2 60 m 60 m 120 mx x x

Now we find the times for each distance run, using the formula for speed.

1

11

60 m15 s

4.0 m s

xt

v

2

22

60 m10 s

6.0 m s

xt

v

1 2 15 s 10 s 25 st t t

120 m4.8 m s

25 s

xv

t


REFLECT The answer is an average. The value we found lies between the higher and lower values, as it should.

However, note that the answer is not just the arithmetic mean of the two given velocity values.

19. ORGANIZE AND PLAN We must find average acceleration from initial velocity, final velocity, and elapsed time.

This means we use the definition of average acceleration, .x xa v t

Known: 0 1250 m s;xv 1870 m s;xv 35 s.t

SOLVE We use the definition of average acceleration

0 21870 m s 1250 m s17.7 m s

35 s

x x xx

v v va

t t

The answer is response (c).

REFLECT The acceleration of the spacecraft may seem modest, about a 50% increase in velocity. But the velocity

values themselves are not important. Only their difference matters, .xv In this case the acceleration is about 180%

of that due to gravity. If this spacecraft has human occupants, acceleration of this magnitude would certainly be

noticeable. By comparison, a sports car accelerating from zero to 60 miles per hour in 4 seconds only accelerates at

about 68% of that due to gravity.

20. ORGANIZE AND PLAN This problem requires us to find displacement from initial velocity, final velocity, and

acceleration.

Known: 2

1.4 m s ;xa 10 m s;xv 0 0 m s.xv

SOLVE Since we don’t know time, we will make use of the kinematic equation

2 20 2x x xv v a x

Rearranging to find x ,

2 22 20

2

10 m s 0 m s35.7 m

2 2 1.4 m s

x x

x

v vx

a



REFLECT Note that squaring the velocities results in a positive numerator. The sign of the acceleration is

negative, producing a negative displacement, as we would expect from an object starting from rest.

21. ORGANIZE AND PLAN This problem requires us to find acceleration from initial velocity, final velocity, and

displacement.

Known: 0 21.4 m s;xv 0 m sxv 3.75 cm.x

SOLVE Since we don’t know time, we will make use of the kinematic equation

2 20 2x x xv v a x

Since x is given in centimeters, we need to convert it to meters.

Rearranging to find ,xa

2 22 2

200 m s 21.4 m s

6100 m s2 2 3.75 cm 1 m 100 cm

x xx

v va

x


REFLECT The arrow has a relatively high speed and stops in a very short distance. This is like an automobile

traveling 50 mi/h stopping in about 1.5 inches. This requires a very large negative acceleration. The sign on the

acceleration is negative because the initial velocity of the arrow is positive, and it slows down as it travels into the

target.

22. ORGANIZE AND PLAN This is a free-fall problem, with the object in constant acceleration. We want to know how

long it takes to strike the ground, in seconds. We will have to use one of the kinematic equations that contain time

as a variable. Since we know that initial velocity is zero, but we don’t know final velocity, we will use 1 2

0 2y yy v t a t and solve for t.

Known: 442 m;y 2

9.81 m s ;ya g 0 0 m s.yv

SOLVE We start with the formula we chose.

1 20 2y yy v t a t

Since 0 0,yv

1 22 yy a t

22

y

yt

a

2

2 441 m29.5 s

9.80 m sy

yt

a


REFLECT The Sears Tower is one of the world’s tallest buildings. It takes an object less than 10 seconds to fall to

the ground, ignoring air resistance. This emphasizes the important aspect of constant acceleration. The longer the

object falls, the faster it goes, and the greater the distance it covers each second. We see that the height of the

building appears under the radical. Doubling the height of the building will not double the length of time the object

takes to fall from the top. Rather, the time increases by a factor of 2.

23. ORGANIZE AND PLAN This is a free-fall problem. Notice that only an algebraic solution is needed. We need to

find the effect on final velocity if we double the distance an object falls. We won’t use a subscript for the original

height, but we’ll use the subscript “2” for the doubled height. Since time is not included in the problem, we’ll use

the kinematic equation 2 20 2 .y yyv v a y

Known: ;y h 0 m s;yv ;yv v .ya g

SOLVE Substituting the variables given in the problem, we get

2 20 2 yv v a h

2.8 Chapter 2

Since 0 0yv for both the original height and the doubled height,

2 2v ah

2v ah

If we now double y to the value 2 ,h then the new velocity is

2 2 2 2 2 2yv a h ah v


REFLECT Displacement in the y-direction is a function of the square of velocity. It makes sense that velocity is a

function of the square root of height during free-fall.

24. ORGANIZE AND PLAN This is a constant-acceleration problem. Since time is not part of the problem, we can use

the kinematic formula 2 20 2 .x x xv v a x

Known: 0 m s;xv 0 12.8 m s;xv 16.0 m.x

SOLVE Starting with our chosen formula and solving for acceleration,

2 20 2x x xv v a x

2 20

2

x xx

v va

x


20 m s 12.8 m s5.1 m s

2 16.0 mxa


REFLECT We can think of this problem as the opposite of a free-fall problem (and in the horizontal direction).

Since the car initially has positive velocity and is slowing down, the acceleration must be negative.

25. ORGANIZE AND PLAN We are asked to find the shape of the graph of velocity versus time for a moving object

under constant acceleration. Since the object is moving, we know that velocity can be positive or negative, but it

cannot be zero. No calculation is needed.

Known: 0 m s;v 2

0 m s .xa

SOLVE We know that velocity is a function of time and can be expressed as

0x x xv v a t

We rearrange this formula to put it in the slope-intercept form of the equation of a straight line.

0x x xv a t v

We see that xa is the slope and 0xv is the vertical intercept (the intercept on the velocity axis). We are told

that 0.xa The slope of the line must be either positive, sloping upward, or negative, sloping downward. Either of

these possibilities gives a diagonal line on the graph.


REFLECT An acceleration of zero would give us a horizontal line (slope = zero). In order for the graph to have

the shape of a parabola, velocity would have to be a function of 2t . Rather, it is a function of the first power of t , a

linear function.

26. ORGANIZE AND PLAN This problem contains two objects! We must find some common value between them. In

this case, both the time at which they pass and their height y as they pass one another are common. We solve for

the height at which each ball passes the other and set the two equations equal to each other. From this, we find a

numeric value for .t Substituting this value back into the equation for height for either ball 1 or ball 2 gives us

the value for height.

Known: 1 10m s;yv 2

9.80 m s ;ya g 2 10 m s;yv 2 10 m.y

SOLVE The two balls not only pass each other at the same height, the must pass each other at the same time. Our

key is to find that time. For the first ball,

1 21 1 2y yy y v t a t


For the second ball,

1 22 2 2y yy y v t a t

Since the balls meet at the same height y, we can set these two equations equal:

1 12 21 1 2 22 2y y y yy v t a t y v t a t

Canceling like terms and substituting known values, we get

10 m s 10 m+ 10 m st t

10 m

0.5 s20 m s

t

We know that the height is the same for both balls, so we choose the equation for the position of the first ball.

Substituting known values including the time,

221

210 m s 0.5 s+ 9.80 m s 0.5 s 3.8 my


REFLECT The two balls have to meet somewhere between ground level and 10 m above the ground, which they

do. Checking our work with the equation for the position of the second ball,

221

10 m 10 m/s 0.5 s 9.80 m/s 0.5 s2

y

3.8my

PROBLEMS

27. ORGANIZE AND PLAN In this problem we must show the difference between the distance and the displacement

for a round trip between two points. We’ll use “1” as the subscript for the first part of the trip and “2” as the

subscript for the second part of the trip.

Known: 1 200 m.d

SOLVE The distance between two points is always positive regardless of the direction traveled. For a round trip to

the video store, the distance from your friend’s house is 1 200 m.d The distance from the video store back to your

friend’s house is also 2 200 m.d So the total distance for the round trip is

1 2 200 m 200 m 400 md d d

But for displacement, we must take into account the sign of the direction of travel for each part of the trip.

Traveling in the positive direction, from your friend’s house to the video store,

1 200 mx

Returning from the video store in the negative direction,

2 200 mx

The total displacement for the round trip is

1 2 200 m 200 m 0 mx x x

REFLECT A round trip, with the ending position the same as the starting position, always has a positive distance

and a zero displacement.

28. ORGANIZE AND PLAN This problem uses the definition of displacement. We choose a coordinate system with the

positive direction to the east. Since Lincoln is our starting point, we’ll choose the position of Lincoln to be 0x . Our

final position will be Grand Island,160 km to the west. Its position is .x

Known: 0 0 km;x 160 km.x

SOLVE We travel in the negative direction from Lincoln to Grand Island. Using the definition of displacement,

0 160 km 0 km 160 kmx x x

2.10 Chapter 2

REFLECT When you give someone directions to a destination, you must tell them which direction to travel if you

want them to arrive where they intend to go. When traveling in the negative direction, the value of displacement

will have a negative sign.

29. ORGANIZE AND PLAN In this problem, we have to consider what effect a fractional round trip has on distance and

displacement. We set up a coordinate system with the positive direction to the east. We start at Grand Island, so we

can declare this position to be 0.x The position of Lincoln will be .x

SOLVE (a) Here we have three round trips from Grand Island to Lincoln. The displacement 1x traveling from

Grand Island to Lincoln is

1 0 160 km 0 km 160 kmx x x

Traveling from Lincoln back to Grand Island, the displacement is

2 160 kmx

The total displacement for one round trip is

160 km 160 km 0 kmx

Therefore, the displacement for three round trips is

3 3 0 km 0 kmx

The distance traveled from Grand Island to Lincoln is160 km. Since distance is always positive, the distance

traveled from Lincoln back to Grand Island is also160 km. The distance traveled in three round trips is

3 160 km 160 km 960 kmd

(b) Here we have 12

3 round trips. This means three round trips plus one last trip from Grand Island to Lincoln.

Since the displacement from (a) for exactly three round trips was 0 km, the displacement here is the same as one-

half of a round trip, or the displacement from Grand Island to Lincoln, so

0 160 km 0 km 160 kmx x x

However, the distance for 12

3 round trips from Grand Island to Lincoln is

3 160 km 160 km 160 km 7 160 km 1120 kmd

(c) This part asks about displacement and distance for 34

3 round trips. From (b), the displacement for 3 round trips

is zero. During the remaining 34

round trip, we travel from Grand Island to Lincoln in the positive direction, and

then halfway back to Grand Island in the negative direction. The displacement for this trip is

12

160 km 160 km 80 kmx

The distance for each leg of the trip is always positive, so

12

7 160 km 160 km 1200 kmd

REFLECT The displacement for a round trip is always zero. The distance for each leg of a trip is positive,

regardless of direction. Think about how often a car’s owner must fill the fuel tank just from driving round trips to

school!

30. ORGANIZE AND PLAN We must find the average speed in m s of a runner under three different conditions of time

and distance. For each of these, we’ll use the definition of average speed, distance traveled

vt

. However, in each

scenario, we’ll have to convert time or distance, or both, to the proper units.

Known: (a) distance traveled 41 km; 2 h 25 min;t (b) distance traveled 1500 m; 3 min 50 s;t

(c) distance traveled 100. m; 10.4 st

SOLVE

(a) Start by converting 2 h 25 min to seconds. We think of this as 2 h 25 min and convert each term separately:

2 h 3600 s h 25 min 60 s min 8700 st


Then we convert 41 km to m,

distance traveled 41 km 1000 m km 41,000 km

Now we use the definition of average speed.

distance traveled 41,000 m4.7 m s

t 8700 sv

(b) The distance is already given in meters. We convert the time to seconds

3 min 50 s 3 min 60 s min 50 s 230 st

Then

distance traveled 1500 m6.5 m s

t 230 sv

(c) Here the distance is already in meters and the time is already in seconds.

distance traveled 100.m9.62 m s

t 10.4 sv

REFLECT The lengths of these races systematically decrease from very long (a marathon!) to moderate (just

under a mile) and finally to a short dash. We can expect the average speeds to increase from a bit more than a fast

walk ( 4.7 m s ) to a fast run ( 9.62 m s ).

31. ORGANIZE AND PLAN To calculate elapsed time, we need to know average distance between the Earth and the

sun, and the speed of light. We use 83.00 10 m s for the speed of light and 111.50 10 m for the orbital radius of

Earth.

Known: 11distance traveled 1.50 10 m; 83.00 10 m s.v

SOLVE Using the definition of average speed,

distance traveledv

t

and rearranging for ,t

11

8

distance traveled 1.50 10 m500 s

3.00 10 m st

v

REFLECT

One occasionally hears that if the sun suddenly stopped shining, it would take the inhabitants of Earth about 8

minutes to realize this. Checking this value,

1 min500 s 8.33 min

60 sor about 8 minutes and 20 seconds.

32. ORGANIZE AND PLAN This problem requires us to rearrange the definition of average speed, solving for time.

Then we are given two instances in which we must find time from speed and distance.

Known: (A) distance traveled 18.4 m; 44 m s;v (B) distance traveled 18.4 m; 32 m s.v

SOLVE Rearranging the definition of average speed we obtain

distance traveledt

v

(a) 18.4 m

0.42 s44 m s

t

(b) 18.8 m

0.58 s32 m s

t

REFLECT Major league batters have about one-half second to see, track, and hit a pitch. The faster the pitch, the

less time the batter has.

33. ORGANIZE AND PLAN In this problem there are two parts to the motion, at different speeds. We do not simply

average the speeds, even though the distances traveled are both 100.m. We must go back to the definition of

2.12 Chapter 2

average speed, distance traveled

.vt

We’ll use subscript (1) for the 4.0 m s run and subscript (2) for the

5.0 m s run.

Known: 1 4.0 m s;v 1distance traveled 100. m; 2 5.0 m s;v 2distance traveled 100. m.

SOLVE First we find the distance traveled

1 2distance traveled distance traveled distance traveled

Then we find total time

1 21 2

1 2

distance traveled distance traveledt t t

v v

Finally we substitute known values into the definition of average speed

distance traveled 100.m 100.m4.4 m s

25 s 20 sv

t

REFLECT To understand why we can’t just average the speeds, we see that this is a rate problem. We have to add

the reciprocals of the individual speeds to get the reciprocal of the average speed:

20.s 25 s 45 s 1

200.m 200 m 200 m 200 m

t

v

200 m4.4 m s

45 sv

34. ORGANIZE AND PLAN In this problem we are asked to compare average speeds calculated under different

conditions. We use the definition of average speed, distance traveled

vt

. In part (a) time is constant. In part

(b) distance is constant. We will use subscript (1) for the first run and subscript (2) for the second run in each set of

conditions.

Known: (A) 1 10.m s;v 1 100.s;t 2 20.m s;v 2 100.s;t (B) 1 10.m s;v distance traveled 1000.m;

2 20.m s;v distance traveled 1000.m.

SOLVE (a) We must find the distances run at each speed.

1 1 1distance traveled 10.m s 100.s 1000 mv t

2 2 2distance traveled 20.m s 100.s 2000 mv t

Then we use the definition of average speed:

1 2

1 2

distance traveled distance traveled 1000.m 1000.m15 m s

100.s 100.sv

t t

(b) We must find the time it takes to run each leg of the total distance.

11

1

distance traveled 1000.m100.s

v 10.m st

22

2

distance traveled 1000.m50.s

v 20.m st

Using the definition of average speed,

1 2

1 2

distance traveled distance traveled distance traveled 2000.m13 m s

150.sv

t t t

(c) The speeds in (a) and (b) are not identical. In (a) you ran longer at 20.m s than you did in (b). You ran for the

same time at 10.m s in both cases.

REFLECT The only circumstance in which you can find average speed by taking the arithmetic means of two

different speeds is if the times run at each speed are equal. You cannot do so if the distances run are equal.

35. ORGANIZE AND PLAN We are given the distances traveled and the speeds for two legs of a flight, and the layover

time between legs. We calculate the times for each leg using the definition of velocity. Then we add the layover


time to find t. From the total time and total distance, we find average speed. We’ll use subscript (1) for the first

leg of the flight and subscript (2) for the second leg.

Known: 1distance traveled 1100 km; 2distance traveled 550 km; 800.km/h;v layover 80.min.t

SOLVE

(a) First we find the time for the first leg of the flight. Notice that distance is in kilometers and time is in hours.

11

1

distance traveled 1100 km1.375 h

800.km ht

v

22

2

distance traveled 550 km0.688 h

800.km ht

v

Then we convert the layover time to hours:

layover 80.min 1 h 60 min 1.333 ht

1 2 layover 1.375 h 0.688 h 1.333 h 3.4 ht t t t

(b) Average speed is total distance divided by total time, so

1 2

1 2

distance traveled distance traveled 1100 km 550 km 490 km/h

1.375 h 0.688 h 1.333 hlayover

vt t t

+ +

+ + + +

REFLECT This is one kind of problem in physics where it is not necessary to convert all units to SI base or

derived units. Since the problem does not ask for specific units, we are free to express speed in km/h and time in

hours. These units allow us to use values of reasonable magnitude for travel by air.

36. ORGANIZE AND PLAN In a race, the winner is the contestant that starts at position 0x and first reaches the finish

line .x Finishing first means reaching x with a lower time t than the other contestant. Here we will compare the

elapsed times of the two boats. The boat with the lower t is the winner. We will use subscripts (1) and (2) for

boats 1 and 2, respectively, and a second subscript stating the first or second variable of the boat during the race.

Known: distance traveled 2.000 km; Boat 1: 11 4.0 m s;v 12 3.1 m s;v 11distance traveled 1500 m;

Boat 2: 21 3.6 m s;v 22 3.9 m s;v 21distance traveled 1200 m.

SOLVE Since speed is given in m s, we will need to convert the length of the race to meters.

2.000 km 1000 m km 2000.m

Now, for boat 1, the first leg requires time

1111

11

distance traveled 1500 m375 s

4.0 m st

v

The second leg of the race requires

1212

12

distance traveled 2000 m 1500 m161 s

3.1 m st

v

The total time for boat 1 is then

1 11 12 375 s 161 s 536 st t t

Doing the same calculations for boat 2,

2121

21

distance traveled 1200 m333 s

3.6 m st

v

21

2121

distance traveled 2000 m 1200 m205 s

3.9 m st

v

The total time for boat 1 is

2 21 22 333 s 205 s 538 st t t

Boat 1 reaches the finish line first and wins the race.

REFLECT Another way of determining the winner is to calculate the higher average speed of each boat and to

compare these values. The boat with the higher average speed wins. However, average speed is a function of total

time, so this alternate method requires two additional math steps.

2.14 Chapter 2

37. ORGANIZE AND PLAN We are to find both speed and velocity. First we must find both distance and displacement.

To find total distance, we use the definition of speed to find the distance traveled on each leg of the trip. To find

displacement, we take into account the signs of the distances. For this purpose, our coordinate system will establish

east as positive and west as negative. We’ll use subscript (1) for the first leg and subscript (2) for the second leg.

Known: 1 210 km h east;v 1 3.0 h;t 2 170 km h west, or 170 km h;v 2 2.0 h.t

SOLVE First we find the displacement for each leg

1 630 kmx

The second leg of the trip is in the negative direction:

2 170 km h 2.0 h 340 kmx

1 2 630 km 340 km 290 kmx x x

Dividing displacement by time to obtain velocity,

290 km58 km h

5.0 hx

xv

t

Then we find the distance for each leg:

1 1 1distance traveled 210 km h 3.0 h 630 kmv t

Remember that distance is always positive, so we have to use the absolute value of velocity

2 2 2distance traveled 170 km h 2.0 h 340 kmv t

The total distance is

distance traveled 630 km 340 km 970 km

So the average speed is

distance traveled 970 km194 km h

5.0 hv

t

REFLECT Since the plane reverses its path and flies back toward its starting point, the distance is greater than the

displacement and the speed is greater than the velocity.

38. ORGANIZE AND PLAN In this problem, we help a runner plan strategy for a 10K (10-km) race. We know the total

length of the race, but not the lengths of the individual legs. We have enough information to find average speed.

There is only one combination of distances at each of the two given speeds that will result in this calculated

average speed. We can express the distance of the first leg as 1distance traveled . The second leg will then be

110,000 m distance traveled .We will use subscript (1) for the first leg and subscript (2) for the second leg.

Known: distance traveled 10.0 km; 1 4.10 m s;v 2 7.80 m s;v 40.0 min.t

SOLVE First we convert the time to seconds:

40.0 min 60 s min 2400 s

Then we convert the distance of the race to meters:

10.0 km 1000 m km 10,000 m

Finding expressions for the elapsed times of the two legs,

11

1

distance traveledt

v

and

1

22

10,000 m distance traveledt

v

For the entire race,

1 1

1 21 2

distance traveled 10,000 m distance traveledt t t

v v


Now we clear the denominators by multiplying by 1 2v v , group similar terms, and solve for 1distance traveled :

2 1 1 1 1 2distance traveled 10,000 distance traveledv v v v t

2 1 1 1 1 1 2distance traveled distance traveled 10,000 mv v v v v t

1 2 1 1 2 1distance traveled 10,000 mv v v v t v

1 2 11

2 1

10,000 mdistance traveled

v v t v

v v

1

4.10 m s 7.80 m s 2400 s 4.10 m s 10,000 mdistance traveled

7.80 m s 4.10 m s

1distance traveled 9663 m

This is the distance the runner must cover before starting the sprint. Now we subtract this value from the length of the race:

2distance traveled 10,000 m 9663 m 337 m

The runner must start her sprint 337 m before the end of the race.

REFLECT Starting the sprint at 337 m before the finish line will cause the runner to finish the race in exactly

40.0 min. If she starts the sprint sooner, she will finish the race in less time.

39. ORGANIZE AND PLAN This problem emphasizes that velocity takes into account all the elapsed time, not just the

time an object is in motion. Here we will use the definition of velocity. The dogsled goes “straight” so we are free

to establish our own coordinate system, with the “straight” direction of travel being in the positive x-direction.

We’ll use the subscript (1) for the time the dogsled is in motion, and no subscript for the variables pertaining to the

entire 24-hour period.

Known: 24 h;t 1 10 h;t 1 9.5 m s.xv

SOLVE First, we convert the two known times to seconds.

24 h 3600 s h 86,400 st

1 10 h 3600 s h 36,000 st

Then we find the displacement 1x during the 10-hour period when the dogsled is moving.

1 1 1 9.5 m s 36,000 s 342,000 mx v t

This gives us velocity.

1 342,000 m4.0 m s

86,400 sx

xv

t

REFLECT A velocity of 9.5 m s is about the highest velocity a human can achieve for short periods of time. No

wonder that humans in snowy regions of the Earth use dogsleds for transportation.

40. ORGANIZE AND PLAN This problem contrasts speed and velocity. When we calculate speed, we use the absolute

distance traveled on each leg of a trip. When we want velocity, all we care about is how far we end up away from

the starting point. This means we have to use the sign of each direction traveled. We’ll use the subscripts (1), (2),

and (3) for the three legs of the trip.

Known: 1 100.km h;xv 1 30.0 min;t 2 60.0 km h;xv 2 10.0 min;t 3 80.0 km h;xv

3 20.0 min.t

SOLVE We must first convert the elapsed times to hours to be consistent with the velocity units.

1 30.0 min 1 h 60 min 0.500 ht

2 10.min 1 h 60 min 0.167 ht

3 20.0 min 1 h 60 min 0.333 ht

1 2 3 0.500 h 0.167 h 0.333 h 1.000 ht t t t

First we find the average velocity, taking into account the negative sign on 3xv

3 3 3 80 km h 0.333 h 26.6 kmxx v t

1 1 2 2distance traveled and distance traveledx x

2.16 Chapter 2

Adding the three separate displacements,

50.0 km 10.0 km 26.6 km 33.4 kmx

and the average velocity is

33.4 km33 km h

1.000 hx

xv

t

Then we find the total distance traveled:

1 1 1distance traveled 100.km h 0.500 h 50.0 kmxv t

2 2 2distance traveled 60.0 km h 0.167 h 10.0 kmxv t

Since we are calculating distance, not velocity, we use the absolute value of 3xv

3 3 3distance traveled 80.0 km h 0.333 h 26.6 kmxv t

1 2 3distance traveled distance traveled distance traveled distance traveled

distance traveled 50.0 km 10.0 km 26.6 km 86.6 km

So the average speed is

distance traveled 86.6 km87 km h

1.000 hv

t

REFLECT Displacement is never greater than distance, and is always less than distance if a change in direction

takes place during the motion. Likewise, average velocity is never greater than average speed.

41. ORGANIZE AND PLAN In this problem, we have to find the error in an observation. Error

is observed value true value . The observed value is the speedometer reading, 60.0mi h. We must calculate the

true speed from the true distance between highway mileposts and the true elapsed time (measured by your clock).

Known: observed 60.0 mi h;v truedistance traveled 5.00 mi; (a): true 4 min 45 s;t (b): true 65.0 mi h.v

SOLVE (a) First we convert time to hours:

true 4.75 min 1 h 60 min 0.792 ht

Then we calculate our true speed

true

true

distance traveled 5.00 mi63.2 mi h

0.0792 hv

t

observed actualerror 60.0 mi h 63.2 mi h 3.2 mi hv v

(b) If our true speed is 65.0mi h, then

true

true

distance traveled 1.00 mi0.154 h 55.4 s

65.0 mi ht

v

REFLECT Error is one of many statistical functions that we use as tools to compare data from a limited number of

observations to data from a large population.

42. ORGANIZE AND PLAN This problem emphasizes that we must calculate values for each leg of a trip. In the first

case (a) there is no wind, so the bird’s speed in the same whether flying east or west. In the second case (b) with a

tailwind outbound and a headwind inbound, the two speeds are very different. We must calculate the time of each

leg. We must not simply average the two speeds. We’ll use subscript (1) for the eastbound leg and subscript (2) for

the westbound (returning) leg.

Known: 1 10.0 km;x 2 10.0 km;x (a): 1 10.0 m s;v 2 10.0 m s;v (b) 1 15.0 m s;v 2 5.0 m s.v

SOLVE First we convert the displacements to meters:

1 10.0 km 1000 m 1km 10,000 mx

2 10.0 km 1000 m 1km 10,000 mx


(a) Flying east,

1

11

10,000 m1000 s

10.0 m s

xt

v

Flying west toward home,

22

2

10,000 m1000 s

10.0 m s

xt

v

1000 s 1000 s 2000 st

(b) Flying east,

1

11

10,000 m667 s

15.0 m s

xt

v

22

2

10,000 m2000 s

5.0 m s

xt

v

667 s 2000 s 2667 st

(c) The two times are different. We can’t just average two velocities, because time is inversely proportional to

velocity. We would have to average the reciprocals of the velocities to find the reciprocal of the average velocity in

this case. This is the same strategy we would use in a rate problem.

REFLECT The elapsed time to fly to a location and return to your origin depends on the wind. If you have a

constant wind during a trip, the elapsed time will be greater than for no wind. The higher the wind speed, the

greater the elapsed time. Aircraft pilots must be very careful of this, because the aircraft consumes the same

amount of fuel per hour, regardless of groundspeed.

43. ORGANIZE AND PLAN In this problem, two moving objects start at different positions. Here we have to use the full form of

a kinematic equation for position, not just displacement. We’ll establish a coordinate system with both animals traveling in

the positive x-direction. The cheetah starts at the origin and the zebra starts at a position of 35 m. Since both animals end up

at the same spot and at the same time, we can set the equations for each animal’s position equal to each other and solve for

.t We’ll use subscript (1) for the cheetah and subscript (2) for the zebra.

Known: 10 0 m;x 20 35 m;x 1 30.m s;xv 2 14 m s.xv

SOLVE This is a constant velocity problem. For the cheetah,

10 1xx x v t

For the zebra,

20 2xx x v t

Setting these equal,

10 1 20 2x xx v t x v t

Solving for ,t

10 20 2 1x xx x v v t

10 20

2 1

0 m 35 m2.2 s

14 m s 30 m sx x

x xt

v v

REFLECT The answer is just the zebra’s initial lead divided by the difference in velocities. Think of the animals

on a treadmill that is moving at the zebra’s velocity of 14 m s. To a stationary observer standing on the ground

next to the treadmill, the zebra appears to be standing still while the cheetah approaches at

30.m s 14 m s 16 m s . The time it takes the cheetah to reach the zebra under these conditions is given by

solving the definition of velocity .x v t This gives the exact answer we obtained above. We’ll discuss this

notion of relative motion in a later chapter.

2.18 Chapter 2

44. ORGANIZE AND PLAN The parachutist’s average velocity yv while her chute is open is her

displacement y divided by the time her chute is open. We are given the displacement while the chute is open. We

have to find the time by subtracting 10.0 seconds from the elapsed time for the entire trip. We’ll calculate this from

the average velocity for the entire trip. We’ll use the subscripts “ff” for free fall and “chute” for the time her

parachute is open.

Known: 440.m;FFy 10.0 s;FFt 3.45 m s;yv chute 1350 m.y

SOLVE First, the time for the entire trip: We see that the total displacement is the sum of the two displacements

given.

440.m 1350 m518.8 s

3.45 m sy

ytv

Now, the time for which the chute is open:

chute ff 518.8 s 10.0 s 508.8 st t t

and finally the velocity while the chute is open:

chutechute

chute

1350 m2.65 m s

508.8 sy

yv

t

REFLECT The velocity while the chute is open is between the average free-fall velocity and the overall velocity,

as we would expect. To check how reasonable our answer is, we can use the kinematic equation 2 20 2 .y yv v g y

Solving for ,y we find that a final velocity of 2.65 m s is how fast we would strike the ground after jumping

from a height of about 0.36 m, or about 14 inches. This seems to be a reasonable velocity.

45. ORGANIZE AND PLAN We are to find an experimental speed of light and compare it with today’s accepted value

of 83.00 10 m s. We must convert the experimental distance from kilometers to meters and the experimental

time from minutes to seconds.

Known: distance traveled 299,000,000 km;= 22 min.t

SOLVE

8 3

82.99 10 km 10 m kmdistance traveled

2.3 10 m s22 min 60 s min

vt

REFLECT Today we find Römer’s value to be significantly in error, low by 70,000,000 m s , or 23% . We must

remember that only shortly before his work, scientists were still considering the speed of light to be infinite. Only

about 65 years earlier, Galileo Galilei had developed the modern telescope. Before this, Jupiter’s moons had not

been seen at all. In 1675, timekeeping devices were still not accurate. Sixty years later, in 1735, John Harrison

started his work on the first accurate marine chronometer, which was not completed until a voyage to Jamaica in

1761. Römer’s value was quite a scientific feat for his time.

46. ORGANIZE AND PLAN Average velocity is displacement divided by the time interval. Each time interval is 2.0 s.

We must read initial and final positions from the graph shown in Figure P2.46 in the text to find displacement.

SOLVE We summarize the data read from the graph in a table. Then we use the definition of average

velocity xv x t to complete the last column of the table.


Time interval, s x, m 0, mx x

xv

0.0–2.0 4.17 10.0 –5.83 –2.92

2.0–4.0 5.00 4.17 0.83 0.42

4.0–6.0 12.5 5.00 7.50 3.75

6.0–8.0 25.7 12.5 14.2 7.08

REFLECT The average velocity is the slope of a tangent to the curve of the graph at the midpoint of each time

interval, for small intervals.

47. ORGANIZE AND PLAN Here we are to construct a graph of velocity versus time for the 8-second time interval.

From Problem 46, we already have the average velocity for each time interval. Velocity is the slope of a line

tangent to curve of the position-versus-time graph when t is small.

SOLVE The table below summarizes the data from Problem 46 that we need to construct our graph.

Time interval, s Midpoint values, s xv

0.0-2.0 1.0 2.8

2.0-4.0 3.0 0.2

4.0-6.0 5.0 3.8

6.0-8.0 7.0 6.2

Using the values from the table below, we plot the following:

REFLECT Velocity increases steadily with time, from an initial negative (downward) value to positive values.

The original path of the object (from Problem 46) is a parabola, concave upward. It shows an initial downward

velocity, decreasing to zero, and then increasing as t increases.

48. ORGANIZE AND PLAN Average acceleration is found using 0

0

.x xx

v va

t t

We’ll read the velocity values from

the vertical axis of the graph shown in Figure P2.48 in the text. Final and initial values for time come from the

horizontal axis.

Known:

2.20 Chapter 2

SOLVE A systematic way to set up this problem is to create a table. Each line of the table represents one time

period. We calculate average acceleration from the values in each line. Reading time and velocity values from the

graph, we obtain:

0, st , st 0, m sxv , m sxv 2, m sxa

0 5 0 12 2.4

5 10 12 18 1.2

10 15 18 13 –1.0

15 20 13 –3 –3.2

REFLECT This graph consists of three straight line segments connected by smooth curves. The slope of each line

segment is the acceleration. However, the second and third time intervals include parts of the curves, so the

average acceleration values for those intervals do not correspond to any straight line segment.

49. ORGANIZE AND PLAN Here we have to draw a graph of instantaneous acceleration. We’ll need to look at the

graph in Problem 48. This graph has three regions of constant slope. We’ll find the slope for each region using

.xx

va

t

Known: From 0 st to 7.5 st velocity changes from 0 m s to17.5 m s. From 8 st to 12 st velocity is

constant at17.5 m s. From 12.5 st to 20 st velocity decreases from 17.5 m s to 2.5 m s.

SOLVE We’ll indicate the three line segments using the subscripts 1, 2, and 3. The acceleration values for the

three segments are

21

17.5 m s 0 m s2.33 m s

7.5 s

xx

va

t

22

18 m s 18 m s0 m s

12 s 8 s

xx

va

t

23

2.50 m s 17.5 m s2.67 m s

20 s 12.5 s

xx

va

t

REFLECT Acceleration is the slope of the graph of velocity versus time. The graph of instantaneous acceleration

consists of three horizontal line segments, each indicating a constant acceleration during that time interval.

50. ORGANIZE AND PLAN Here we are to draw a motion diagram for the car in Problem 48. We’ll refer to the figure

in that problem for our information. A motion diagram shows the position of an object after equal intervals of time.

Known: Velocity increases steadily from 0 st to 6 s.t Then velocity becomes constant from 8 st to 12 s.t

From 16 st to 20 st velocity steadily decreases to a negative value.

SOLVE


REFLECT In the first part of the graph from Problem 48, the car starts at zero velocity and speeds up at a constant

rate. We know this because the slope is constant and positive. In the middle of the graph, acceleration is zero

because the slope is zero. During the last part of the graph, the car slows down at a constant rate, because the slope

is constant and negative. At the very end of the graph, we see that the car reaches zero velocity and then reverses

direction. We know this because the velocity falls below zero.

51. ORGANIZE AND PLAN Acceleration is the value of the slope of a graph of velocity versus time. We are to

consider the graph in Problem 48. The greatest acceleration occurs when the slope of the graph is most positive.

The least acceleration occurs when the slope is most negative. Acceleration is zero where the graph is a horizontal

line. We use the equation 0x xx

v va

t

to calculate acceleration.

Known: From 0 st to 7.5 st velocity changes from 0 m s to12.5 m s. From 7.5 st to 12.5 st velocity is

constant at17.5 m s. From 12.5 st to 20 st velocity decreases from17.5 m s to 2.5 m s.

SOLVE

(a) Acceleration is greatest between 0 s and 7.5 s.

(b) Acceleration is least between12.5 s and 20 s.

(c) Acceleration is zero between 7.5 s and12.5 s.

(d) For the greatest acceleration,

20 12.5 m s 0 m s2.33 m s .

7.5 s

x xx

v va

t

For the least acceleration,

20 2.5 m s 17.5 m s2.67 m s .

20.0 s 12.5 s

x xx

v va

t

REFLECT This motion is like starting from rest in your car, accelerating to some constant speed, realizing that

you should have turned at the corner, then slowing down and backing up toward the corner. Actually turning the

corner is not included in this problem.

52. ORGANIZE AND PLAN The race lasts10.0 s, so the first half of the race ends at 5.0 s.t We use data read from

the graph in Figure 2.13(b) to find the average acceleration, 0 .x xx

v va

t

Known: 0 0 s;t 5.0 s;t 11.0 m s.xv

SOLVE During the time interval 0t t , velocity increases from 0 0 m sxv to 11.0 m sxv .

20 0

0

11.0 m s 0 m s2.2 m s

5.00 s 0 s

x x x xx

v v v va

t t t

REFLECT Average acceleration is just that — an average. It does not tell us how acceleration has varied during a

time interval.

53. ORGANIZE AND PLAN The stock car’s velocity is related to time by the equation 21.4 1.1 .xv t t We are asked

to find xv after 4.0 s. This means we have to evaluate the given equation at 4.0 s. For the units to cancel properly

the value 1.4 must have units of 3m s and the value 1.1 must have units of 2m s .

Known: 0 0 s;t 4.0 s.t

SOLVE Substituting 4.0 s for ,t we get

22 3 21.4 1.1 1.4 m s 4.0 s 1.1 m s 4.0 s 26.8 m sxv t t

REFLECT We know the car starts from rest at time 0 st since substituting this value into the given equation

gives

22 3 21.4 1.1 1.4 m s 0 s 1.1 m s 0 s 0 m sxv t t

During the time interval of this problem, velocity increases as a quadratic function of time. This situation can’t last.

The engine’s ability to accelerate the car will decrease as engine speed increases past a certain point. Air resistance

also reduces acceleration as the car moves faster.

2.22 Chapter 2

54. ORGANIZE AND PLAN We must graph the velocity of the stock car versus time. We create a table containing

ordered pairs of time and velocity values. We only know that the function 21.4 1.1xv t t is valid

between 0 st and 4.0 s.t

Known: 0 0 s;t 4.0 s.t

SOLVE Using the function 21.4 1.1 ,xv t t we find these values for t and :xv

,st ,m sxv

0 0

1 2.5

2 7.8

3 15.9

4 26.8

Plotting time on the horizontal axis and velocity on the vertical axis, we obtain a graph like this

REFLECT This graph is a segment of a parabola, as we expect from seeing the 2t term in the equation. Velocity

increases more and more rapidly as elapsed time increases.

55. ORGANIZE AND PLAN The instantaneous velocity is the y-value on a graph of velocity versus time. This is the

graph we constructed in Problem 54.

Known: 2.0 st

SOLVE In the present problem, we raise a perpendicular from 2.0 st on the horizontal axis. At the point where

this perpendicular intersects the curve, we draw a horizontal line segment left to the vertical axis. The point on the

vertical axis where the horizontal line intersects it represents the value of the instantaneous velocity.

We see that the horizontal line intersects the vertical axis at the point 0,7.8 m s , which is the instantaneous

velocity.

REFLECT This value is the velocity at the point in time 2.0 s.t We can see from the graph that for any other

time on the graph, the velocity will have a different value.


56. ORGANIZE AND PLAN We are to draw a motion diagram based on Figure 2.15(a) in the text, which is also shown

below. In this figure, the car speeds up during the first equal time interval. Then it moves with constant speed

during the second equal time interval. Finally the car slows down during the third equal time interval.

Known:

SOLVE

REFLECT We see that the positions of the car are closer together at lower velocities than at higher velocities.

Some safety professionals suggest that a driver should maintain a 2-second time interval between his or her car and

the car ahead. This corresponds to a smaller spacing on the roadway at lower speeds, while allowing for a constant

reaction time to apply the brakes.

57. ORGANIZE AND PLAN We are to draw a motion diagram based on Figure 2.15(b) in the text, which is also shown

below. The car speeds up in the negative direction during the first equal time interval. Then it moves with constant

velocity during the second equal time interval. Finally the car slows down and stops during the final equal time

interval.

Known:

SOLVE

REFLECT We cannot tell from the information given whether the vehicle is pointed in the positive or negative

direction. It doesn’t matter, because we model the car as a point, as shown in Section 2.1. We only know that the

car is moving in the negative direction, regardless of whether it is in a forward gear or reverse gear.

2.24 Chapter 2

58. ORGANIZE AND PLAN We are to find the average acceleration between points on the figure.

This figure is a graph of velocity versus time. The slope of a secant line between any two points on this graph is the

average acceleration between these points. We are asked to find average acceleration between the positive diastolic

peak and the negative systolic peak, and then again to the next diastolic peak. We’ll use the kinematic

equation 0

0

.x xx

v va

t t

We’ll need the velocity and time values from the graph.

Known: From the graph, the diastolic peak occurs at the ordered pair 0.45 s, 0.60 m s . The systolic peak occurs

at 0.97 s, 0.55 m s . The next diastolic peak occurs at 1.27 s, 0.60 m s .

SOLVE In the first case, we use the diastolic values as the initial values and the systolic values as the final values.

20

0

0.55 m s 0.60 m s2.2 m s

0.97 s 0.45 s

x xx

v va

t t

In the second case, the systolic value is the initial value:

20

0

0.60 m s 0.55 m s3.8 m s

1.27 s 0.97 s

x xx

v va

t t

REFLECT We can use the same formula for average acceleration even though the acceleration varies

considerably.

59. ORGANIZE AND PLAN We know our initial velocity 0 50.km hxv and the distance to the stoplight, 40.m.x

We see that we need to convert the initial velocity to m s to agree with the units of .x We need to find both

acceleration and how long it takes to stop. We can use 2 20 2x x xv v a x to find the acceleration. Then,

knowing ,xa we use 0x xx

v va

t

to find the stopping time, .t

Known: 40.m;x 0 50 km h.xv

SOLVE First, convert the initial velocity to m s

1000 m 1 h50.km/h 13.9 m/s

1 km 3600 s

(a) Then rearranging

2 20 2x x xv v a x

We get

2 220 0 m s 13.9 m s

2.4 m s2 2 40.m

x xx

v va

x

(b) To find the time it takes to stop, we rearrange the definition of acceleration

0x xx

v va

t

02

0 13.9 m s5.8 s

2.4 m s

x x

x

v vt

a

The acceleration is about the same as that of a car accelerating from 0 to 60mi h in 11 seconds. This seems to be a

reasonable acceleration.

REFLECT According to the sign of our initially velocity, we are traveling in the positive x-direction. We are

slowing down, so the sign of acceleration must be negative, as we calculated.


60. ORGANIZE AND PLAN To calculate acceleration, we need the change in velocity and the elapsed time. Then we

can use the definition of acceleration, 0 .x xx

v va

t

Known: (a) 0 0 m s;xv 90 km h;xv 3.0 s.t (b) 0 0 m s;xv 10 m s;xv 2.0 s.t

SOLVE (a) First we convert the cheetah’s speed to m/s.

1000 m 3600 s90 km h 25 m s

1 km 1 h

Then,

20 25 m s 0 m s8.3 m s

3.0 s

x xx

v va

t

(b) Since the human’s speed is already in m s ,

20 10 m s 0 m s5.0 m s

2.0 s

x xx

v va

t

REFLECT It may surprise us that a human’s acceleration can be as much as 60% of a cheetah’s acceleration.

However, the cheetah can maintain its acceleration for 50% more time than the human, giving the cheetah a final

velocity of 2.5 times that of the human.

61. ORGANIZE AND PLAN This is a comparison problem. We are given the distance to the hole. Now we have to

calculate how far the ball goes, and compare that value to the distance to the hole. We assume for the moment that

the ball comes to rest and use 2 20 2x x xv v a x with a final velocity of zero. If we find that the ball makes it to the

hole, then we can recalculate using the same formula, but instead using 4.8 mx and solving for final

velocity xv at the position of the hole.

Known: 0 2.52 m s;xv 2

0.65 m s ;xa distance to the hole 4.8 m.

SOLVE (a) To find whether the ball makes it to the hole,

2 20 2x x xv v a x

Rearranging,

2 22 20

2

0 m s 2.52 m s4.9 m

2 2 0.65 m s

x x

x

v vx

a

So the ball does reach the hole!

(b) Now, at the hole,

2 20

2x x xv v a x

222

02 2 0.65 m s 4.8 m 2.52 m s

x x xv a x v

0.33 m sxv as the ball passes the hole.

REFLECT Since the problem tells us that the golfer putts straight toward the hole, we have confidence that the

ball goes into the hole at this modest speed. If the ball’s speed were too great, it could strike the far edge of the cup

and bounce out. When we study Chapters 3 and 6, we’ll learn about projectile motion and collisions, which will

help us understand why the ball might not end up in the cup!

62. ORGANIZE AND PLAN We’re given acceleration and time. We have to find the sled’s final speed. We can use

0 .x x xv v a t Then we can use 2 20 2x x xv v a x to find the distance the sled has traveled.

Known: 0 0 m s;xv 2

21.5 m s ;xa 8.75 s.t

SOLVE (a) First calculate the final velocity:

2

0 0 m s 21.5 m s 8.75 sx x xv v a t

188 m sxv

2.26 Chapter 2

(b) Then we can find the distance traveled.

2 20 2x x xv v a x

Rearranging,

2 22 20

2

188 m s 0 m s822 m

2 2 21.5 m s

x x

x

v vx

a

REFLECT A velocity of188 m s is about 420 mi h. The sled travels 822 m just to reach its top speed. After the

end of this problem, the sled will require even more distance to slow down and stop. We would want a very long

and straight path for this sled run. When we study rotational motion in Chapter 8, we’ll learn more about what

would happen to the sled if the path were not straight.

63. ORGANIZE AND PLAN We have a car that speeds up from some initial speed, then slows down and stops in

two distinct time intervals. We need to find the car’s maximum speed, and the total time and distance it

travels. We’ll use the subscripts 1 and 2 to represent what happens during the first and second time intervals.

Between the time intervals, acceleration changes from 1xa to 2xa . We can use the formulas 0 xv v a t

and 21

0 2distance traveled .xv t a t We have to be careful because the car slows down to less than its

original speed.

Known: 1 6.2s;t 2

1 1.9 m s ;xa 2

2 1.2 m s ;xa 0 13.5 m s;v 2 0 m s.v

SOLVE (a) The car’s maximum speed occurs at the end of the first time interval, that is, after the maximum time

with positive acceleration:

2

1 0 13.5 m s 1.9 m s 6.2 s 25.3 m sxv v a t

(b) We use the answer to (a) to find the length of the second time interval:

2 12

2

0 m s 25.3 m s21.1 s

1.2 m sx

v vt

a

total 1 2 6.2 s 21.1 s 27.3 st t t

(c) Since we now know both time intervals, we can calculate the distance traveled for both intervals:

2 221 1

1 0 1 1 12 2distance traveled 13.5 m s 6.2 s 1.9 m s 6.2 s 120.1 mxv t a t

2 221 1

2 1 2 2 22 2distance traveled 25.3 m s 21.1 s 1.2 m s 21.1 s 266.7 mxv t a t

1 2distance traveled distance traveled distance traveled 120.1 m 266.7 m 387 m

REFLECT To solve graphically, we could plot distance traveled versus time, and find distance as the area under

the line. We’d divide the graph into a trapezoid and a triangle and add the areas of each. The numeric equivalent of

this is to multiply the average speed for each time interval by the length of each interval, and add the two values.

64. ORGANIZE AND PLAN To calculate stopping time from an initial speed and acceleration, we can use the

definition of acceleration 0 .x

v va

t

To find stopping distance, we use 1

02distance traveled .v v t

Known: 2

1 2 3.50 m s ;x xa a 10 50.km h;v 220 1.0 10 km h;v 1 2 0 km h.v v

SOLVE First, convert the initial velocities to m s.

10

1 m s50.km h 13.9 m s

3.6 km hv

220

1 m s1.0 10 km h 27.8 m s

3.6 km hv

(a) The stopping time for the first car is

1 101 2

1

0 m s 13.9 m s3.97 s 4.0 s

3.50 m sx

v vt

a


For the second car,

2 202 2

2

0 m s 27.8 m s7.94 s 7.9 s

3.50 m sx

v vt

a

(b) For the first car, the stopping distance is

1 11 10 12 2

distance traveled 13.9 m s 0 m s 3.97 s 27.6 m 28 mv v t

1 12 20 22 2

distance traveled 27.8 m s 0 m s 7.94 s 110.4 m 110 mv v t

(c) The ratio of stopping times is

2

1

7.94 s2

3.97 s

t

t

The ratio of stopping distances is

2

1

distance traveled 110.4 m4

distance traveled 27.6 m

REFLECT The answers make sense since stopping time has only one factor of t in the equation, but stopping

distance uses two factors of .t Therefore, for a doubling of ,t we would expect stopping distance to increase by

a factor of 4, which we see here.

65. ORGANIZE AND PLAN We’ll model the bullet as a point at the front of the bullet. Since we know displacement

but not time, we can use 2 20 2 xv v a x for both parts of this problem.

Known: 0 310 m s;xv 5.0 cm;x (a) 0 m s;xv (b) 50.m s.xv

SOLVE First, we convert the displacement to meters.

1 m5.0 cm 0.050 m

100 cmx

(a) For the bullet stopping in the target,

2 22 220 5

0 m s 310 m s9.6 10 m s

2 2 0.050 m

x xx

v va

x

(b) For the bullet leaving the target at 50.m s ,

2 22 220 5

50 m s 310 m s9.4 10 m s

2 2 0.050 m

x xx

v va

x

REFLECT These are very large values! Since the acceleration is proportional to the difference of the squares of

the velocities, the velocity of the bullet leaving the target would have to be a lot greater to make a significant

difference in the acceleration. We have simplified this problem by specifying constant acceleration and modeling

the bullet as a point. In fact, the bullet begins its negative acceleration at the moment the front of the bullet enters

the target. Its acceleration does not stop until either the bullet stops as in (a) or until the bullet exits the target

completely, as in (b).

66. ORGANIZE AND PLAN Since we don’t know time, we can use 2 20 2x x xv v a x to find the length of the runway,

.x Once this value is known, we can use 21

0 2x xx v t a t to find the time for takeoff.

Known: 0 0 m s;xv 250 km h;xv 2

3.0 m s .xa

SOLVE First, convert the takeoff speed to m s.

1 m s250 km h 69.4 m s

3.6 km hxv

Then we can calculate the runway length:

2 20 2x x xv v a x

2.28 Chapter 2

2 22 20 2

2

69.4 m s 0 m s803 m 8.0 10 m

2 2 3.0 m s

x x

x

v vx

a

Now we find the time the plane requires to lift off.

21

0 2x xx v t a t

Since 0 0xv ,

2

2 803 m223 s

3.0 m sx

xt

a

REFLECT Modern commercial aircraft lift off in a range of about 20 to 35 seconds after beginning their takeoff

runs, depending on the power of the engines and how heavily the aircraft is loaded. Our answer appears to be

reasonable.

67. ORGANIZE AND PLAN The police cruiser and the speeder must end up at the same position at the same time. The

speeder will travel 1.2 km. The cruiser will have to travel the 1.2 km plus100 m, or 1.3 km. We’ll use subscript

1 for the speeder and subscript 2 for the cruiser. Since the speeder travels at constant speed, we can directly

calculate his time to the state line, ,t which will be the same for the cruiser. Once we know the time, we can use

21

0 2x xx v t a t to find the cruiser’s acceleration.

Known: 1 75 mi h;xv 1 1.2km;x 02 0 m s;xv 2 1.3 km.x

SOLVE First, convert 1.2 km and 1.3 km to meters, and 75mi h to m s.

1

1000 m1.2 km 1200 m

1 kmx

Likewise, 2

1.3 km 1300 mx

1

5280 ft 0.3048 m 1 h75 mi h 33.53 m s

mi ft 3600 sxv

Then find the time it takes the speeder to get to the state line:

1 1xx v t

1

1

1200 m35.85 s

33.53 m sx

xt

v

Finally, find the acceleration of the cruiser:

2

12 0 2x xx v t a t

Since2

0 m s,xv

22

2 2

2 1300 m22.03 m s

35.85 sx

xa

t

REFLECT This is a relatively modest acceleration, and the police cruiser should reasonably be able to overtake

the speeder at the state line.

68. ORGANIZE AND PLAN Since we know distance and acceleration, we can use 2 20 2x x xv v a x to find velocity.

Once we have velocity, we can use the definition of acceleration to find time.

Known: 0 0m s;xv 2145.0 10 m s ;xa 15 cm.x

SOLVE First, convert x to meters.

1 m15 cm 0.15 m

100 cmx

(a) Find the electrons’ final velocity 2 2

0 2x x xv v a x


Since 0 0xv ,

214 72 2 5.0 10 m s 0.15m 1.2 10 m sx xv a x

(b) Now find the time it takes an electron to accelerate to that velocity:

0x xx

v va

t

70 8

214

1.2 10 m s 0m s2.4 10 s

5.0 10 m s

x x

x

v vt

a

REFLECT Although the acceleration is very large, the electrons’ speed is still only about 4% of the speed of light.

69. ORGANIZE AND PLAN This is a constant acceleration problem. We know the values of three variables, so we can

find a solution: Initial velocity, final velocity, and displacement. We have the equation 2 20 2x x xv v a x to directly

find acceleration.

Known: 0 m s;xv 0 86.1 m s;xv 31.00 10 m.x

SOLVE 2 2

0 2x x xv v a x

Since 0,xv

22

20

3

86.1 m s3.71 m s

2 2 1.00 10 m

xx

va

x

REFLECT Commercial aircraft approach and land at a comparatively high rate of speed due to the size and shape

the their wings and also so they can gain altitude and “go around” quickly if something unexpected happens on the

runway ahead of them. The wheel brakes would wear out very quickly stopping a plane at this speed. Pilots use a

combination of resistance on parts of the wing and reverse thrust by the engines to slow the aircraft down on the

runway before applying the brakes.

70. ORGANIZE AND PLAN In this problem we have to find out if we can stop in a given distance, and also if we can

travel a certain distance in a calculated amount of time. These are both comparison problems in which we compare

a calculated quantity with a given quantity. Part (a) involves both constant velocity and constant speed using

xx v t and 2 20 2 .x x xv v a x From this, we have to calculate stopping distance. Part (b) is a constant velocity

problem for which we have to calculate time.

Known: light 3.4 s;t 1 0.60 s;t 1 35 m;x road 9.5 m;x 0 50.km h 13.9 m s;xv 2

3.0 m s .xa

SOLVE (a) First we calculate the displacement before we touch the brake pedal:

1 1 13.9 m s 0.60 s 8.34 mxx v t

Then we calculate the displacement after we apply the brakes

2 20 2x x xv v a x

2 22 2

02 2

0 m s 13.9 m s32.2 m

2 2 3.0 m s

x xv vx

a

1 2 8.34 m 32.2 m 41 mx x x

We are not able to stop before the intersection under these circumstances.

Now, supposing we maintain our velocity at13.9 m s, we calculate how long it takes to get across the intersection.

1 road 35 m 9.5 m 44.5 mx x x

44.5 m3.2 s

13.9 m sx

xtv

We can get completely through the intersection before the light turns red 0.2 s later.

REFLECT Here we model the car as a point. In the real world, the car has a dimension of length. At13.9 m s,

the 0.2 s difference between the time it takes you to just get across the intersection and when the light turns red

corresponds to 2.8 m or about 9 ft of car. This means that any part of the car more than 2.8 m behind the point we

modeled will still be in the intersection when the light turns red.

2.30 Chapter 2

71. ORGANIZE AND PLAN This problem requires conversion from English units to SI units. It is a constant

acceleration problem. We know initial and final velocity, and elapsed time, so we can use the definition of

acceleration to find acceleration. We must also find its displacement. We can do that independently using the

average velocity formula 102

.x xx v v t

Known: 0 0 m s;xv 60.mi h;xv 2.8 s.t

SOLVE (a) First, convert the final velocity to SI units:

5280 ft 0.3048 m 1 h60.mi h 26.8 m s

1 mi 1 ft 3600 sxv

Then,

20 26.8 m s 0 m s9.6 m s

2.8 s

x xx

v va

t

(b) Finding the displacement,

1 102 2

0 m s 26.8 m s 2.8 s 38 mx xx v v t

REFLECT The acceleration is very nearly that due to gravity, so the effect is somewhat like falling sideways.

There are custom-built “performance” automobiles that can accelerate from rest to 60 mi h in 2.8 s or less!

72. ORGANIZE AND PLAN In each of the two scenarios, we are to find initial velocity where final velocity is zero.

The two scenarios use the same formula, 2 20 2 .x x xv v a x

Known:2

980.m s ;xa 0 m s;

xv airbag: 20.cm;x seatbelt: 5.0 cmx

SOLVE First, convert displacement to meters. For the airbag,

1 m20. cm 0.20 m

100 cmxv

Likewise for the seatbelt,

5.0 cm 0.050 m.xv

Now we find initial velocity for the airbag scenario.

2 20

2x x xv v a x

Since 0

0 m s,xv

22 2 980. m s 0.20 m 19.8 m s

x xv a x

For the seatbelt scenario,

22 2 980. m s 0.050 m s 9.90 m s

x xv a x

REFLECT For the airbag scenario, the maximum speed before brain injury would occur is about 44 mi h. The

seatbelt only allows a maximum speed of about 22 mi h. Seatbelts and airbags protect our brains only at modest

speeds for the kind of collision described here.

73. ORGANIZE AND PLAN We know three of the five possible variables in this constant acceleration problem,

allowing us to solve it. We solve for final speed using 2 20

2 .y

v v a y We establish a coordinate system in which

“down” is negative.

Known: 00 m s;

yv

29.80 m s ;

ya g 9.5 my

SOLVE Using the chosen equation,

2 20

2y

v v a y

Since 00 m s,

yv

22 2 9.80 m s 9.5 m

yv a y

We know that the flowerpot is traveling downward, but the problem only asks for speed, so:

14 m sv


REFLECT Notice that the displacement and the acceleration are both negative. This gives us a positive number

under the radical. In free-fall problems, we must remember that the sign of the displacement is negative if the

object falls from a higher to a lower position.

74. ORGANIZE AND PLAN Once the book leaves the student’s hand, it is in free fall. In this problem we could either

use the quadratic formula or find the final velocity first, followed by using the definition of acceleration. Since the

final velocity is required as one of the answers, we’ll choose that path.

Known: 1.5 m;y 0 3.9 m s;yv 2

9.80 m s .ya g

SOLVE (a) Finding final velocity,

2 20 2y y yv v a y

2 22

0 2 3.9 m s 2 9.80 m s 1.5 my y yv v a y

Since the book was released above the floor and strikes the floor, we know it must be traveling in the negative

direction, so we choose the negative root.

6.68 m s 6.7 m syv

Now, knowing ,yv

0y yy

v va

t

0

2

6.68 m s 3.9 m s1.08 s 1.1 s

9.80 m s

y y

y

v vt

a

(b) From part (a), 6.7 m syv

REFLECT Solving for t in part (a) using the quadratic formula gives the same answer. It’s just more tedious

unless you have the quadratic formula programmed into your calculator! See Problem 75 for an example of using

the quadratic formula in this kind of problem.

75. ORGANIZE AND PLAN We know the initial velocity, acceleration and displacement. We’ll use the quadratic

formula to find time, .t We start with 21

0 2.y yy v t a t

Known: 0 0 m s;yv 2

9.80 m s ;ya g 25.6 m.y

SOLVE Beginning with the kinematic equation

21

0 2y yy v t a t

We rearrange it to the general form of a quadratic equation,

21

020y ya t v t y

We look for the coefficients to use in the quadratic formula,

2 4

2

b b acx

a

Where ,x t 1

2,y

a a 0,

yb v and c y

The quadratic formula solved for t is then

21

0 0 2

12

4

2

y y y

y

v v a yt

a

2.32 Chapter 2

Which simplifies to

20 0 2y y y

y

v v a yt

a

Choosing the negative root so that time is positive,

2 2

2

0 0 m s 2 9.80 m s 25.6 m

2.29 s9.80 m s

t

To find final velocity, we substitute the value for t into

29.80 m s 2.29 s 22.4 m s

yv a t

REFLECT A time of 2.29 s to drop a distance of 84 feet seems reasonable. The answer is easily shown to be

correct using 21

2 yy a t knowing that 0 0 m s.yv

76. ORGANIZE AND PLAN This is a free-fall problem. We make a reasonable assumption that the first baseman will

catch the ball at about the same height above the ground that the batter originally struck it. This assumption of

0y gives us the three known values that allow us to solve this problem. We’ll use the equation

21

0 2.y yy v t a t In this case, we’ll set the right side equal to zero and factor out t . Taking 0t , we can

solve the other factor for .t

Known: 0 19.5 m s;yv 2

9.80 m s ;ya g 0 m.y

SOLVE Using the chosen formula,

21

0 2y yy v t a t

10 2

0 y yt v a t

One of the two factors must equal zero, but when 0,t we are at the moment when the batter strikes the ball, not

when it is caught. So we choose

10 2

0y yv a t

0

2

2 2 19.5 m s3.98 s

9.80 m s

y

y

vt

a

REFLECT A time of about four seconds seems to be reasonable to catch a pop ball. The strategy we illustrated

here can be used with any free-fall fall problem in which the object’s motion begins and ends at the same

y-position. This will also be useful in Chapter 3 when we study projectile motion.

77. ORGANIZE AND PLAN We are to graph the position of the rock and its velocity versus time. We need to know

two pieces of information. First, in order to properly scale the horizontal (time) axis, how long will it take the rock

to hit the ground? Second, to perhaps simplify our task, what are the shapes of the two graphs? We know initial

velocity and acceleration. For any given value of ,t we can find vertical displacement using

21

0 2.y yy v t a t We see that this is a quadratic equation that will have the shape of a parabola concave

downward. We’ll construct a table of values to plot y-position versus time. For velocity, the equation we’ll use

is 0 .y y yv v a t We see that velocity is a linear function of time, and we only need two points to graph this line.


9.80 m s .ya g

SOLVE First, let’s find out how long it takes for the rock to return to the ground. Ignoring air resistance, we see

that final velocity is just the opposite of initial velocity, and

0

2

16.5 m s 16.5 m s3.37 s

9.80 m s

y y

y

v vt

a

So a range of 0–4 seconds on the horizontal will produce an acceptable graph for either position or velocity.

First, we complete a table of values for y and t using

21

0 2y yy v t a t


,t s ,y m

0 0

0.5 7.03

1.0 11.60

1.5 13.73

2.0 13.40

2.5 10.63

3.0 5.40

3.37 0

Plotting these points and drawing a smooth curve through them, we obtain the graph shown in the figure below.

From our calculation of ,t we see that the initial and final velocities are of the same order of magnitude as the

positions. It would be possible to plot both position and velocity on the same graph, but for clarity we will make

two separate graphs this time. From that same calculation, we have two points on this linear graph, the ordered

pairs 0 s,16.5 m s and 3.37 s, 16.5 m s . Plotting these points and connecting them with a straight line, we

obtain the graph shown in the figure below. We check the linearity with a third point, 1.50 s,1.80 m s .

REFLECT The graph of position versus time is clearly in accordance with our common experience.

The object goes up, stops momentarily, and descends. The graph of velocity versus time is also reasonable. The

rock starts out going up quite fast, slows to a stop at the highest point in its path, and then speeds up as it comes

back down.

78. ORGANIZE AND PLAN In this problem, we use the acceleration due to gravity on the moon and on the Earth to

compare time in free fall. Since we are given displacement, we’ll use 21

0 2.y yy v t a t We don’t have to

2.34 Chapter 2

solve a quadratic equation because the object starts from rest and 0 0 m syv . We’ll use the subscripts “moon”

and “Earth” for the values that apply only to the moon or only to the Earth.

Known: 1.0 m;y 2

,moon 1.6 m s ;ya 2

,Earth 9.80 m s .ya

SOLVE For the moon,

21

0 2y yy v t a t

Since 0 0,yv

moon 2

2 1.0 m21.1 s

1.6 m sy

yt

a

For the Earth, all that changes is the value of ya , so

Earth 2

2 1.0 m0.45 s

9.80 m st

REFLECT Since acceleration on the moon is smaller than on the Earth, we expect the time in free fall to be

greater on the moon. More precisely, we see that ya is in the denominator under the radical, so we expect the two

values of t to differ by a factor of ,moon ,Earthy y

a a which is what we see in the answer.

79. ORGANIZE AND PLAN Here we have to find the value of g on a different planet by experimentation! Since

we know time and displacement, we can use 21

0 2.y yy v t a t We’ll substitute Marsg for ya and solve

for Mars.g

Known: 5.01 s;t 45.2 m.y

SOLVE

21

0 2y yy v t a t

Since Mars,ya g

21

0 Mars2yy v t g t

The value of 0 0yv because the rock starts from rest:

2

Mars2 y g t

2

Mars 2 2

2 45.2 m23.60 m s

5.0 s

yg

t

REFLECT Since Marsg is smaller than Earth,g we expect the rock to take longer to fall than it would on Earth.

The units come out to be 2/m s so we are confident of our answer.

80. ORGANIZE AND PLAN The tennis-ball gun launches balls into free fall. We can use 21

0 2y yy v t a t to

solve. We’ll use subscripts 1 and 2 for the first and second balls. The key is that the balls have to be in the same

place at the same time at the end of the problem. For (a) we set the equations for height equal to each other and

solve for .t For (b) we use the value of t from (a) and solve for y using 21

0 2.y yy v t a t For (c) we

rearrange the definition of acceleration0y y

y

v va

t

to solve for the velocity of each ball.


9.80 m s ;ya g 1 2 2.0 s.t t

SOLVE (a) We’ll write equations of motion for each ball:

21

1 0 1 12y yy v t a t

21

2 0 2 22y yy v t a t

Because of the interval between balls launched by the tennis-ball gun,


1 2 2.0 st t

Also, 1 2y y at some time .t For ball 1, that’s 1;t for ball 2, it’s 2.t We set the two equations equal to

each other and substitute 2 2.0st for 1 :t

2 21 1

0 1 1 0 2 22 2y y y yv t a t v t a t

2 21 1

0 2 0 2 2 22 22.0 s 2.0 sy y y yv t v t a t a t

Canceling like terms,

1 20 22

2.0 s 4.0 s t 4.0 sy yv a

Solving for 2,t

1 20 2

2

2 2.0 s 4.0 s

4.0 s

y y

y

v at

a

21 22

2 2

2 18.8 m s 2.0 s 9.80 m s 4.0 s

9.80 m s 4.0 st

2 0.8877st

The time after launch of the first ball when the two balls meet is

1 2 2.0 s 2.89 s 2.9 st t

(b) We substitute the value for 1t into the equation of motion for ball 1. We use the unrounded value from (a).

21

1 0 1 12y yy v t a t

221

1 218.8 m s 2.89 s 9.80 m s 2.89 13 my s

(c) We remember that the times for the two balls are different. For ball 1,

2

1 10 1 18.5 m s 9.80 m s 2.89 s 9.8 sy y yv v a t

For ball 2,

2

2 20 2 18.5 m s 9.80 m s 0.888 s 9.8 sy y yv v a t

REFLECT This is no surprise! The speed of an object in free fall is the same at a given height in its path above its

launch point regardless of whether it’s going up or coming down.

81. ORGANIZE AND PLAN In this problem we have to use the player’s maximum height above the floor to calculate

time in the air to reach that point. Then we can show that the time interval to fall back to the floor is the same.

Adding these gives us total time in the air. Knowing ,t we graph y versus time. We calculate 10 points and

complete a table of ordered points ,t y .

Known: 1.1 m;y 0 m s;yv 2

9.80 m sya g .

SOLVE (a) For the athlete traveling from the playing surface to the highest point in the path,

2 20 2y y yv v a y

2 202y y yv a y v

2

0 2 2 9.80 m s 1.1 m 4.64 m sy yv a y

On the way back down, just before the player touches the floor

2

0 2 9.80 m s 1.1 m 4.64 m syv

2.36 Chapter 2

Considering the entire path of the player and using the definition of acceleration,

0

2

4.64 m s 4.64 m s0.95 s

9.80 m s

y y

y

v vt

a

(b) Now that we know t for the entire path of the player, we construct a table of values:

,st ,my

0 0

0.1 0.42

0.2 0.73

0.3 0.95

0.4 1.02

0.5 1.07

0.6 1.10

0.7 0.85

0.8 0.58

0.9 0.21

0.95 0

Graphing these points, we get the graph we need.

(c) Between 0.3 s and 0.7 s, the player is at least 0.85 m off the ground. This means that the player is at least 77%

of maximum height for 43% of the time. This is the reason the player appears to “hang” in the air near the top of

the jump.

REFLECT If an object starts and ends at the same y-position, the graph of position versus time is symmetrical

about the vertical line 1total2

.t t Mathematically, we can use 2 20 2y y yv v a y to show that 0y yv v if

0.y

82. ORGANIZE AND PLAN This is a variation on a stopping-distance problem. We’ll use conventional Cartesian

coordinates where “up” is positive on the y-axis. In this case, we have to find acceleration using 2 20 2 .y y yv v a y

First we use free fall to find your velocity when you first touch the ground. Then we find your acceleration as your

velocity decreases to zero.

Known: In free fall: 2.5 m;y 0 0 m s;yv 2

9.80 m s .ya g While accelerating to a stop:

55 cm;y 0 m s.yv

First, when you are in free fall,

2 20

2y y yv v a y

2 22

0 2 0 m s 2 9.80 m s 2.5 my y yv v a y

Choosing the negative root because you are falling in the negative direction,

7.0 m syv


Second, while you are crouching to stop,

1 m55 cm 0.55 m

100 cmy

Then

2 20 2y y yv v a y

220 27.0 m s

45 m s2 2 0.55 m

y

y

va

y

REFLECT The units come out in 2

m s . Your velocity is negative and the acceleration is positive, slowing you

down. This is the result we would expect. This is a significant acceleration, but you are jumping from a height of

over eight feet!

83. ORGANIZE AND PLAN Each of the two parts here is a constant acceleration problem. The first is a powered ascent

for which we’ll use the subscript 1. The second is a period of free fall. We’ll use the subscript 2 for the period of

free fall until maximum height, and the subscript 3 for the descent to the ground. In the first part we’ll use

0 yv v a t to find final speed when the engine cuts off. We’ll also find the height from 2 2

02 .

yv v a y During

free fall, we’ll reset 0 m sv and use the same formula to find the change in height between the times the

engine cuts off and the rocket reaches maximum height. We’ll find t for both the powered ascent and free-fall

descent. Adding these gives the total time to reach the ground.

Known: For powered ascent: 0

0 m;y 2

112.6 m s ;

ya

111.0 s;t

00 m s.v For free fall:

2

29.80 m s ;

ya g

20 m sv

SOLVE

(a) For the powered ascent,

1 0 1 10 m s 12.6 m s 11.0 s 139 m s

yv v a t

(b) For the free fall to maximum height, first we find the change in height after the engine cuts off,

2 22 1 2

2y y yv v a y

221

2 22

139 m s985.8 m

2 2 9.80 m sy

vy

a

Now we need the height the rocket reached during powered ascent, 1.y

2 221 1

1 0 12 20 m s 11.0 s 12.6 m s 11.0 s 762.3 m

yy v t a t

0 1 20 m 762.3 m 985.8 m 1750 my y y y

(c) Now considering the entire free fall, we want to find the velocity when the rocket strikes the Earth.

2 23 1 2

2y

v v a y

We’re seeking speed, so

2 2

3 1 2 12 139 m s 2 9.80 m s 762.3 m 185 m s

y yv v a y

(d) In order to calculate total time of flight, we need the time the rocket was in free fall.

3 1free-fall 2

2

185 m s139 m s33.1 s

9.80 m sy

v vt

a

total 1 free-fall11.0 s 33.1 s 44.1 st t t

REFLECT In (c) we could have just as easily considered only the descent from maximum height. The velocity as

the rocket strikes the Earth is negative since it is traveling in the negative y-direction.

84. ORGANIZE AND PLAN (a) We know that the window is 5.6 m above the ground. We must find the ball’s velocity

when the observer first sees it in the window. Using that value, we can calculate the velocity with which it is

2.38 Chapter 2

thrown. (b) Once we know the velocity with which the ball is thrown, we use one of the equations of motion to

find the total time it is in the air.

Known: For the ball’s path beginning at the window, 3.2 s,t 5.6 m.y

SOLVE (a) First, the ball’s velocity in the window:

1 20 2yy v t gt

Since the ball rises above the window and returns to the same y-position, we set 0y , rearrange and divide by one

factor of .t

10 2yv gt and

10 2

9.81 m s 3.2 s 15.7 m syv

Now, since we don’t know how long it took the ball to get to the window after it was thrown, we use

2 2

02

y yv v gy

2 2

02

y yv gy v and

22 2

0 2 15.7 m s 2 9.81 m s 5.6 m 18.9 m sy yv v gy

Notice that we use the positive root of the expression for initial velocity, since the ball has a positive velocity when

it is thrown. Once we know the initial velocity of the ball, we start with the same formula as in (a), but this time we

want to find .t

0

2

2 2 18.9 m s3.85 s

9.81 m s

yvt

g

REFLECT It doesn’t take much more time for the ball to travel its entire path than it does to appear and then

reappear in the window on its way up and then down. This implies that the ball passes the window shortly after it

was thrown. To further support this, the ball loses only a small part of its initial velocity from the time it was

thrown to the time it first appeared in the window.

85. ORGANIZE AND PLAN We aren’t told what the height is, so we’ll call it y, taking the horizontal surface to be zero

on the coordinate system. The vertical distance the ball travels is then y . We’ll use trigonometry to find the

distances the ball rolls down the 30° incline and up the 45° incline, calling them 1d and 2.d Since the ball rolling

uphill is analogous to it rolling downhill, the acceleration 2a is related to the same quantities, angle and ,y

that the initial acceleration is. We’ll find an expression for the velocity of the ball at the bottom of the 30° incline

and find an expression for the acceleration required to reduce that velocity to zero at the top of the 45° incline. To

do this, we’ll use 2 20 2v v ad where d is the distance the ball rolls.

Known: 1 30 ; 2 45 ; 2

1 3.50 m s ;a 2 0 0 m s.v v

SOLVE Rolling downhill,

2 21 0 12v v a d

2 21 0 1

1

2sin

yv v a

Rolling uphill,

2 22 1 2

2

2sin

yv v a

and

2 21 2 2

2

2sin

yv v a

Setting the two expressions for 21v equal,

2 20 1 2 2

1 2

2 2sin sin

x xv a v a


Since 2 0 0 m s,v v

1 2

1 2sin sin

a y a y

Canceling and rearranging for 2,a

2 222 1

1

sin sin453.50 m s 4.95 m s

sin sin30a a

REFLECT The magnitude of the acceleration is greater rolling up the steeper slope. It is also negative, because the

ball is slowing down. In this problem, neither the initial nor final motion is parallel to either the x- or y-axis. Since

the angles aren’t complementary, there’s no advantage to tilting the axes as in Figure 2.2. If it appears that there is

not enough information given to find the answer to a problem, the best strategy is to set the problem up and solve it

algebraically before trying to substitute known values. The unknown variables will cancel, such as 0,v 2,v

and y did in this problem. You will be left with a simplified and more easily solvable problem.

86. ORGANIZE AND PLAN This is a free-fall problem involving a wrench, but we only know ya and .y We need one

more variable to solve this problem. We can calculate the initial speed of the wrench from the motion of the

helicopter. The helicopter is in constant acceleration from rest on the ground. We know the helicopter’s initial

speed, displacement and acceleration and can calculate its speed at 20 m.y Since the wrench is in the

helicopter, the initial speed of the wrench as it falls out is the same as the final velocity of the helicopter. Now we

can use 2 20

2y

v v a y to find the final speed. Then we can use the definition of acceleration to find time.

Known: 0

0 m s;hv

20.40 m s ;

yha

29.80 m s ;

ywa g 20. m;

hy 20. m

wy

SOLVE First, find the upward speed of the helicopter (and hence the wrench) at 20.0 m.

2 20

2yh yh yhv v a y

2 22

02 0 m s 2 0.40 m s 20. m 4.0 m s

h h yhv v a y

Next, use the velocity of the helicopter to find the final velocity of the wrench as it strikes the ground.

04.0 m s

w hv v

2 2

02

w w ywv v a y

The wrench is falling downward, but the problem only asks for speed, so

2 2

4.0 m s 2 9.80 m s 20. m 20.2 m s 20. m swv

Now, using the definition of acceleration,

0

2

20.2 m s 4.0 m s2.5 s

9.80 m s

w ww

yw

v vt

a

(b) For the wrench’s free fall to the ground, from (a)

20. m swv

REFLECT The wrench’s initial velocity is positive and its final velocity is negative, as we would expect. The time

to fall 20.0 m is reasonably between 2 and 3 seconds. It is a math coincidence that the magnitude of ywv and

wy are the same. This has no special meaning and is due to the choice of values in the problem.

87. ORGANIZE AND PLAN Now that we know the initial velocity and the time to impact of the wrench, we can graph

position versus time and velocity versus time.

Known: 0 20.m;wy 0 4.0 m s;ywv 2

9.80 m s .ywa g

SOLVE To find ordered pairs of position and time, we use 21

0 0 2w w yw ywy y v t a t between 0t and

2.5 s.t To find ordered pairs of time and velocity, we use 0 .yw yw ywv v a t

2.40 Chapter 2

We construct a table with (arbitrarily) 12 ordered pairs:

,st ,mwy ,m sywv

0 20.0 4.0

0.2 20.6 2.0

0.4 20.8 0.1

0.6 20.6 1.9

0.8 20.1 3.9

1.0 19.1 5.8

1.2 17.7 7.8

1.4 16.0 9.7

1.6 13.9 11.7

1.8 11.3 13.6

2.0 8.4 15.6

2.2 5.1 17.6

2.5 0 20.2

Now we can plot position versus time, shown in the first figure.

The graph of velocity versus time is shown in the second figure.

REFLECT It makes sense that the wrench continues upward for a short time and then its velocity becomes

negative, responding to the acceleration of gravity. We notice from the table and graphs that the wrench seems to

“hang” in the air near its highest point, like the volleyball player in Problem 81.

88. ORGANIZE AND PLAN This is a constant acceleration problem, since the flea is in free fall after it leaves the

ground. We know its displacement, final velocity and acceleration, so we can solve the problem. We’ll use


2 20 2y y yv v a y to find initial velocity. Then we’ll use the definition of acceleration

0y yy

v va

t

to find the time

to reach its final height.

Known: For the flea ascending though the air: 0 m s;yv 2

9.80 m s ;ya g 30.cm.y For the flea

during the act of jumping: 0 0 m s;yv 0.90 mm.y

SOLVE (a) First we convert the height to meters.

1 m30.cm 0.30 m

100 cmy

Then we find the takeoff speed, 0yv 2 2

0 2y y yv v a y

Since 0,yv

2

0 2 2 9.80 m s 0.30 m 2.42 m sy yv a y

(b) Then we find the time the flea takes to reach 0.30 m

0

2

0 m s 2.42 m s0.25 s

9.80 m s

y y

y

v vt

a

(c) When the flea is in the process of jumping into the air,

41 m0.90 mm 9.0 10 m

1000 my

2 20 2y y yv v a y

Since 0 0 m syv ,

2 2y yv a y and

222

4

2.42 m s3300 m s

2 2 9.0 10 m

yy

va

y

This acceleration is about 330 times g .

REFLECT The velocity of the flea in the air is positive and ya is negative, so the flea slows down as we expect.

The acceleration of the flea is very large as it launches itself into the air. There is a great difference between a

human brain and the nervous system of a flea (see Problem 72).

89. ORGANIZE AND PLAN We’ll find the final speed of a cat falling through a distance of 6.4 m. We know

displacement, initial speed, and acceleration, so we’ll use 2 20

2 .y

v v a y Then we’ll use the same formula with

the final velocity we just calculated to find the acceleration of the cat as it crouches to a stop.

Known: When the cat is in free fall: 0

0 m s;v 6.4 m;y 2

9.80 m s .ya g As the cat crouches:

14 cm.y

SOLVE (a) To find final velocity in free fall,

2 20

2y y yv v a y

Since 0

0 m s,v

22 2 9.80 m s 6.4 m 11 m s

yv a y

(b) First convert y to meters:

1 m14 cm 0.14 m

100 cmy

Then as the cat crouches,

2 20

2y

v v a y

2.42 Chapter 2

Using unrounded values,

2 22 2

00 m s 11.2 m s

448 m s2 2 0.14 m

y

v va

y

REFLECT Cats can endure greater acceleration than humans (see Problem 72).

90. ORGANIZE AND PLAN The motion in this problem is part constant acceleration and part constant velocity. We’ll

divide the problem into these two parts to solve. We’ll use the subscripts 1 and 2 for the two parts. During constant

acceleration we use 0x x xv v a t to find final velocity. To find distance, we’ll use 21

0 2.x xx v t a t

During the constant velocity portion, we’ll use xx v t to find displacement.

Known: Constant acceleration: 0 m s;xov 2

2.0 m s ;xa 5.0 s.t Constant velocity: 5.0 s.t

SOLVE (a) Your final velocity is reached at the end of the period of constant acceleration:

2

0 0 m s 2.0 m s 5.0 s 10.m sx x xv v a t

(b) The displacement during constant acceleration is

2 2 221 1

1 0 2 20 m s 2.0 m s 5.0 s 25 mx xx v t a t

During constant velocity,

2 10.m s 5.0 s 50.mxx v t

total 1 2 25 m 50.m 75 mx x x

(c) We complete a table from which we can select ordered pairs.

,st ,mx ,m sxv

0 0 0

1 1 2

2 4 4

3 9 6

4 16 8

5 25 10

6 35 10

7 45 10

8 55 10

9 65 10

10 75 10

We graph ordered pairs ,t x as shown in the figure below.


Then we graph ordered pairs , xt v as shown in the figure below.

REFLECT Your displacement increases faster and faster until you reach 5.0 s.t Then it increases linearly

with time. Velocity increases linearly with time until you reach 5.0 s.t Then it “levels off” to 10.m s.xv

91. ORGANIZE AND PLAN Here we know initial and final velocity and displacement. We need to find acceleration.

The problem is independent of time, so we use 2 20 2 .x xxv v a x Once we have found xa we find time using the

definition of acceleration,

0 .x xx

v va

t In the second part of the problem, we have a constant-velocity period

before we accelerate. For this period we use 0x x xv v a t to find the displacement, 1.x Finally we find the

acceleration needed to stop in the remaining distance, 2.x

Known: (a) and (b) 0 13.4 m sxv ; 0 m s;xv 15.0 m.x (c) Constant velocity: 0.60 s;t 0 m s.xv

SOLVE (a) Rearranging for ,xa

2 20 2x x xv v a x

2 22 2

200 m s 13.4 m s

5.99 m s2 2 15.0 m

x xx

v va

x

(b) Rearranging for ,t

0x xx

v va

t

Using the unrounded value for xa

02

0 m s 13.4 m s2.24 s

5.98 m s

x x

x

v vt

a

(c) First we find how far we went in the 0.6 s reaction time:

1 13.4 m s 0.6 s 8.04 mxx v t

This leaves the distance in which we have to stop as

2 115.0 m 15.0 m 8.04 m 6.96 mx x

Using the same formula as in (a),

2 22 2

200 m s 13.4 m s

13 m s2 2 6.96 m

x xx

v va

x

REFLECT With a positive velocity, we expect negative acceleration in order to slow down. The acceleration in

(c) is a rather rapid acceleration, about one-third greater than that of gravity.

92. ORGANIZE AND PLAN A ball is projected up an incline. We are to describe the motion of the ball. We’ll use

102 x xx v v t to find how far up the incline the ball reaches before rolling back down. Finally, we’ll draw

two graphs showing acceleration versus time and position versus time for the rolling ball.

2.44 Chapter 2

Known: (a) 4.0 s;t 0 1.0 m s;xv 1.0 m s;xv (b) 2.0 s;t 0 1.0 m s;xv 0 m s;xv

SOLVE (a) The ball will start up the incline at 1.0 m s, accelerate to a stop, and then roll back down,

accelerating to 1.0 m s. It would look like “free fall at an angle.”

(b) We find the distance up the ramp at the point where 0 m sxv and 2.0 st from the figure above.

1 102 2

0 m s 1.0 m s 2.0 s 1.0 mx xx v v t

(c) We see from the figure above that xa is constant. We calculate

20 1.0 m s 1.0 m s0.5 m s

4.0 s

x xx

v va

t

and plot the line 2

0.50 m s ,xa which appears as a horizontal line on our graph, shown in the figure below.

Now we’ll need to construct a table in order to find ordered pairs ,t x to plot position versus time. We’ll get

our values of x from 102

.x xx v v t We’ll calculate a value for every 0.50 s increment since these values

are directly readable from the figure in part (a).

,st ,mx

0.0 0.0

0.5 0.4

1.0 0.8

1.5 0.9

2.0 1.0

2.5 0.9

3.0 0.8

3.5 0.4

4.0 0.0


Then we plot the ordered pairs ,t x . Our graph is shown in the figure below.

REFLECT This problem is just like a free-fall problem except the value for acceleration is less than .g In Chapter

3, we’ll learn how to use and evaluate x- and y-components of a value that has a direction, such as “up the incline.”

93. ORGANIZE AND PLAN We know the values of three variables in this constant acceleration problem. In order to

graph velocity and position versus time, we need to find acceleration. We’ll use the definition of acceleration,

0x xx

v va

t

. Once we know xa we can construct a table of values to graph, using 2 2

0 2x x xv v a x to find

position.

Known: 0 25 m s;xv 0 m s;xv 10.0 s.t

SOLVE (a) Using the definition of acceleration,

20 0 m s 25 m s2.5 m s

10.0 s

x xx

v va

t

Rearranging the definition of acceleration to find velocity,

0x x xv v a t

Calculating xv at 2.0-s intervals,

, st vx.m/s

0 25

2.0 20

4.0 15

6.0 10

8.0 5.0

10.0 0

Plotting ordered pairs of , xt v we get the graph shown in the figure below.

2.46 Chapter 2

(b) Finding values of x at the end of each 2.0-s time interval using 21

0 2,x xx v t a t

, st x

0 0

2.0 45

4.0 80

6.0 105

8.0 120

10. 125

Plotting ordered pairs of ,t x we get the graph shown in the figure below.

(c) A motion diagram shows position at equal time intervals. We can use the data from the table in (a) to do this:

REFLECT The slope of the graph in (b) represents the acceleration. The slope is downward and to the right,

corresponding to a negative acceleration. Looking at the graph in (c), we see that forward motion continues ever

slower, until the vehicle stops.

94. ORGANIZE AND PLAN In order to graph the motion of the ball, we need an equation that describes its motion. We

need to use the value of acceleration in order to calculate velocity and position at each point in time, so we’ll

calculate xa first. We know time, displacement, and initial velocity, so we can solve for displacement using

21

0 2.x xx v t a t For velocity, we’ll use .x xo xv v a t We take the x-axis to be parallel to the incline.

Known: 0 2.40 m sxv ; 6.0 st 0 m.x

SOLVE (a) First, we declare the initial position to be zero

0 0 mx

21

0 2x xx v t a t

Since 0,x

10 2

0 x xt v a t

Setting the second factor equal to zero,

10 2

0 x xv a t

202 2.40 m s2

0.80 m s6.0 s

xx

va

t

Knowing ,xa we can complete a table showing values of xv and x as functions of time. We’ll use time intervals of 0.50 s.


,st xv ,mx

0 2.40 0

0.50 2.00 1.10

1.00 1.60 2.00

1.50 1.20 2.70

2.00 0.80 3.20

2.50 0.40 3.60

3.00 0.00 3.60

3.50 –0.40 3.50

4.00 –0.80 3.20

4.50 –1.20 2.70

5.00 –1.60 2.00

5.50 –2.00 1.10

6.00 –2.40 0

Graphing ordered pairs of velocity versus time , xt v we get the graph shown in the figure below.

(b) Then graphing ordered pairs of position versus time ,t x we get the graph shown in the figure below.

(c) Finding average velocity,

0 m0 m s

6.0 sx

xv

t

Then we find the total distance traveled. The distance to the highest point on the incline is 3.60 m,

so 1distance traveled 3.60 m. Likewise, the distance from the highest point back down to the point of beginning

is 2distance traveled 3.60 m. Total distance traveled is

total 1 2distance traveled distance traveled distance traveled 3.60 m 3.60 m 7.20 m

2.48 Chapter 2

REFLECT The graph of velocity versus time is a straight line, as we expect from the linear relationship

0 .x x xv v a t It has a negative slope, reflecting the negative value of .xa The graph of position versus time is a

parabola, as we expect from the term 2

x contained in the formula for position.

95. ORGANIZE AND PLAN In this problem we’re asked to find the time(s) when the ball is 5.20 m above its launch

point. Those parentheses are important! The only case in which there would be only one time is if the ball is at its

maximum height at that time. We can test this by using the quadratic formula to see if we get two positive roots,

which are solutions for time. After that, we can use 2 20 2x x xv v a x to find the velocities at each of these points in

time.


9.80 m s ;ya g 5.20 m.y

SOLVE (a) We start with

21

0 2x xx v t a t

Rearranging to the general form of a quadratic,

21

020y ya t v t y

Substituting the coefficients into the quadratic formula,

21

0 0 2

12

4

2

y y y

y

v v a yt

a

Simplifying,

2

0 0 2y y y

y

v v a yt

a


2 2

2

12.1 m s 12.1 m s 9.80 m s 5.20 m

9.80 m st

The roots of this equation are

12.1 6.67 m s 0.554 s

9.80t and

12.1 6.67 m s 1.92 s

9.80t

(b) To find the final velocity, we use

2 20 2y y yv v a y

22

0 2 12.1 m s 2 9.80 m s 5.20 m 6.67 m sy y yv v a y

REFLECT We notice that the expression for final velocity also appears in the numerator of the quadratic formula.

We can deduce that on the way back down, the ball will strike the ground at 1.915 s 0.554 s 2.47 s . The

0.554 s is the same time interval it took for the ball to reach a height of 5.20 m on the way up.

96. ORGANIZE AND PLAN This problem is similar to Problem 80 except we have no evidence that the upward and

downward velocities have the same magnitude. It would be a coincidence if they did, so all we know is that the

position and time are the same for both balls. We’ll solve for position in terms of time for each ball. Then we’ll set

the two equations equal and solve for time. Once we know time, we can solve for the position of one ball, which

will be the same for the other ball. We’ll use subscripts 1 and 2 to identify the variables that apply to each ball.

Known: Ball 1: 0 11.0 m s;yv 2

9.80 m s ;ya g 1 1.50 m.y Ball 2: 0 11.0 m s;yv

2

9.80 m s ;ya g 2 12.6 m.y


SOLVE First, the position of ball 1

21

01 01 2y yy y v t a t

For ball 2,

21

02 02 2y yy y v t a t

Combining these two equations,

2 21 1

01 01 02 022 2y y y yy v t a t y v t a t

Simplifying,

01 02 02 01y yy y v t v t and

01 02

02 01

1.50 m 12.6 m0.505 s

11.0 m s 11.0 m sy y

y ytv v

Substituting the value of t into the equation of position for ball 1,

221

21.50 m 11.0 m s 0.505 s 9.80 m s 0.505 s 5.81 my

The two balls meet at a height of 5.81 m at 0.505 s after they are thrown.

REFLECT

Examining the solution, we see that this is just like assuming that one ball is standing still and both ball 2 and the

surroundings are moving.

97. ORGANIZE AND PLAN Here we have to describe the motions of the friends in a frame of reference that is moving

with respect to the “stationary” fixed ends of the “moving sidewalk.” We’ll use subscripts 1 and 2 for the friends,

and 3 for the sidewalk.

Known: 3 1.0 m sxv with respect to the fixed ends; 1 4.00 m sxv with respect to the sidewalk; 2 4.0 m sxv

with respect to the sidewalk.

SOLVE From the frame of reference of the sidewalk, each of the two friends, traveling at the same speed, will

travel 25 m, meeting in the middle of the sidewalk. For each person,

xx v t

25 m6.25 s

4.0 m s

xt

t

In the same time period, the entire frame of reference of the sidewalk has moved

3 3 1.0 m s 6.25 s 6.25 mxx v t

So in the frame of reference of the fixed ends of the sidewalk, the friends meet at 25 m 6.25 m 31 m from one

of the fixed ends.

An alternative solution is to consider the relative velocities of the two friends with respect to the fixed ends. The

first person will be traveling 5.0 m s while the second person will be traveling 3.0 m s.

1 5.0 m sx t

2 3.0 m sx t

1 1 50.m

5.0 m s 3.0 m s

x xt

1

5.0 m s 50.m31.25 m 31 m

8.0 m sx

So the answer is 31 m from one end.

REFLECT This is an example of relationships between the motions of objects compared in two frames of

references that are moving at constant velocity with respect to each other.

2.50 Chapter 2

98. ORGANIZE AND PLAN In this problem, we have to compare the position of the gazelle with the position of the

cheetah at the moment the cheetah would tire, after which the cheetah would accelerate to a stop. If the cheetah has

not caught the gazelle during this time interval, then the cheetah will not catch the gazelle at all. The gazelle has a

constant velocity and a head start of 25 m, so we use 0 .x x xv v a t The cheetah accelerates to a constant

velocity, so for the cheetah, we can use a combination of 21

0 2x xx v t a t and the equation we used for the

gazelle. We’ll use the subscripts c and g to represent the cheetah and the gazelle, and subscripts a and v for periods

of constant acceleration and velocity.

Known: 0 25 m;gx 20.m s;xgv 0 m;cx 0 0 m s;x cv 60.mi h;xcv during cheetah’s constant

acceleration: 3.0 s;at during cheetah’s constant velocity: 10.s.vt

SOLVE (a) First we convert the speed of the cheetah to SI units:

5280 ft 0.3048 m 1 h60.mi h 27 m s

1 mi 1 ft 3600 sxcv

Then we calculate the distance traveled by the cheetah during constant acceleration:

2 20 2x x xv v a x

Since 0 0 m s,x cv

22

2

26.8 m s40.m

2 2 8.93 m s

xca

y

vx

a

During constant velocity, the cheetah runs

26.8 m s 10.s 268 mcv xcx v t

The cheetah can run a total of

40.2 m 268 m 310 mc ca cvx x x

(b) The total time of this event is

total 3.0 s 10.0 s 13 sa vt t t

At the end of this time interval, the position of the gazelle is

0 total 25 m 20.m s 13 s 285 mg g xgx x v t

Since the gazelle can only travels 285 m in the time it takes the cheetah to travel 310 m, yes, the cheetah catches

the gazelle before tiring.

REFLECT We see that the motion of each animal is stated separately as if the other did not exist. If we were to

calculate the time at which the cheetah catches the gazelle, we would see that it’s likely the cheetah doesn’t

actually run the entire10. s, but stops for a meal instead.

99. ORGANIZE AND PLAN The runners meet at the same time, .t We can consider this a frame-of-reference

problem. Suppose that Runner A is standing still and Runner B’s initial velocity is 0 m s. They are still separated

by 85 m, but it has become easier to calculate the time interval in which Runner B has traveled the 85 m.

Known: 0 0 m s;xv 2

0.10 m s ;xa 85 m;x for constant acceleration, 10.s.t

SOLVE First, we calculate how far Runner B travels during constant acceleration.

2 221 1

0 2 20 m s 10.s 0.10 m s 10.s 5.0 mx xx v t a t

After the time interval of 10.s , Runner B is traveling

0 0 m s 0.10 m s 10.s 1.0 sx x xv v a t

Now Runner B has

85 m 5.0 m 80.mx

left to run. At 1.0m s,xv

80.m80.s

1.0 m sx

xtv


Runner B’s total time to overtake Runner A is then

total 10.s 80.s 90.sa vt t t

REFLECT We would get exactly the same result if we considered the velocities of the two runners with respect to

the track. Notice that we have not taken into account the actual distance traveled with respect to the track. This

would become important if the question asked for the winner of the race!

100. ORGANIZE AND PLAN Since the rocks are in free fall, we can use 2 20 2 yv v a y to find the final speed of each

rock. Then we can find the definition of acceleration to find the time. We’ll use subscripts 1 and 2 to refer to the

two rocks.

Known: 01 10.m syv ; 02 10.m s;yv 12 my ; 2

9.80 m s .ya g

SOLVE

2 20

2y y yv v a y

For the first rock,

2 21 01

2y

v v a y

2 22

1 012 10.m s 2 9.80 m s 12 m 18.3 m s

yv v a y

We chose the negative root because the rock is falling downward.

The elapsed time is

11 012

y v v t

1

1 01

2 12 m22.89 s

18.3 m s 10.m s

yt

v v

For the second rock,

2 2

110.m s 2 9.80 m s 12 m 18.3 m sv

But

2

2 12 m0.85 s

18.3 m s 10.m st

REFLECT Objects in free fall launched from the same point at the same speed (in the vertical direction) strike the

ground with the same final velocity! However, they strike the ground at different times. We’ll look at this in

another way when we study kinetic and potential energy in Chapter 5.

101. ORGANIZE AND PLAN Here the two trains must end up in the same place at the same time. We will use the

kinematic equations xx v t to find displacement at constant velocity and 21

0 2x xx v t a t to find

displacement at constant acceleration. We’ll set these two equations equal to each other and solve for .t Then we

can find where the two trains meet again by substituting t into one of the equations for position. We’ll indicate

variables for the two trains by using subscripts 1 and 2.

Known: 1 11 m s;xv 02 0 m s;xv 2

2 1.5 m s .xa

SOLVE For the first train,

1 1xx v t

1

1x

xtv

For the second train,

21

01 2x xx v t a t

Since 01 0 m sxv ,

21

2 xx a t

2.52 Chapter 2

Setting the two equations equal,

21

1 2x yv t a t

112

0 x xt a t v

0st when the second train initially passes the first. Therefore, when the two trains meet again,

112 x xa t v

1

2

2 11 m s214.7 s 15 s

1.5 m s

x

x

vt

a

Now we substitute t into one of the original equations:

1 11 m s 14.7 s 160 mxx v t

REFLECT At an acceleration of 2

1.5m s , the second train would have a velocity equal to that of the first train in

a little over 7 s, even though the second train would still be behind the first. Since the second train is still

accelerating, it is reasonable to think that it will overtake the first train in about 15 s.

102. ORGANIZE AND PLAN In this problem, we establish an origin where the front ends of the trains meet. We write

equations for the position of each train, set the equations equal, and solve for .t In (a) this is straightforward,

using the formula for displacement with constant velocity. In (b), however, we use the quadratic formula to solve

directly for .t We’ll use the subscripts 1 and 2 with variables that apply to train 1 and train 2.

Known: (a) 0 0 m;x 1 22.5 m s;xv 2 22.5 m s;xv 1 110.m;x 2 110.mx (b) 0 0 m s;x

01 22.5 m s;xv 2

1 1.0 m s ;xa 2 02 22.5 m s;x xv v 1 110.m;x 2 110.m.x

SOLVE First we establish an origin. We declare that the locomotives of each train pass each other at the origin at

0 s.t Then the position of locomotive 1 is

1 0 xx x v t

Since locomotive 2 is traveling the opposite direction, its position is

2 0 xx x v t

Now we write expressions for the locations of the ends of the trains:

1 0end 110.mxx v t and

2 0end 110.mxx v t

Since the ends are in the same position at the same time, we can set these two equations equal to each other:

0 0110.m 110.mx xx v t x v t

Simplifying and collecting like terms,

2 220.m 0xv t and

110.mxv t

110.m

4.89 s22.5 m s

t

Since 110.m

x

tv

is just the time it takes one whole train to pass the origin, both ends will pass the origin at this

elapsed time, but traveling in opposite directions.

(b) Now, if one train accelerates, the position of the first locomotive is

21

1 0 01 2end x xx v t a t

The equation for the position of locomotive 2 at constant velocity remains as it was in (a),

2 0end 110.mxx v t


Setting the equations for the positions of the two locomotives equal, we get

21

0 01 02110.mx x xx v t a t x v t

Simplifying and putting in the general form of a quadratic,

21

022 220 m 0x xa t v t

Using the quadratic formula,

2 22 10 0 2

212

2 22.5 m s 45 m s 2 1.0 m s 220.m2 2 4 220.m

2 1.0 m s

x x x

x

v v at

a

The roots of this equation are

4.65 st and 94.6 st

In this problem, only the positive root has any physical meaning, so

4.65 st

REFLECT The time for the ends of the trains to meet when one train is accelerating is less than when both trains

are at constant velocity, which is what we would expect. If we extend the conditions of (b) backward in time, we

find that the other root of the quadratic refers to a point in time past when the first train passed the second train

while traveling in the same direction as the second train at a high velocity, slowing down, reversing direction, and

then meeting the second train again as this problem states.

103. ORGANIZE AND PLAN Since we are only given two known variables in this free-fall problem, we must consider

the first half of the motion of the ball, where it reaches maximum height and a final velocity of 0 m s.yv We’ll

use 2 20 2y y yv v a y to find initial velocity. Then we can use the definition of acceleration to find the time to reach

maximum height. From symmetry considerations, we can double the time to find when the ball is caught.

Known: 17.2 m 1.6 m 15.6 m;y 0 m s;yv 2

9.80 m s .ya g

SOLVE First we find initial velocity:

2 20 2y y yv v a y

2

0 2 2 9.80 m s 15.6 m 17.49 m sy yv a y

Now, to find time to maximum height,

0

2

0m s 17.49 m s1.784 s

9.80 m s

y y

y

v vt

a

Since the ball was caught at the same height at which it was thrown upward, we know from symmetry that the ball

takes the same time interval to come back down, and

total 2 1.784 s 3.57 st

With 4.8 s on the clock, this leaves 1.2 s for the opponent to take a shot.

REFLECT Realistically, if the opposing player were facing the basket and could make the catch and release a shot

in one smooth motion, there might be a chance of a basket. Otherwise the game is over.

4. solve...2.4 chapter 2 object is at rest and has not moved. if velocity is positive or negative,...

Documents