CHAPTER

4 .Acceleration.................

~ Demon DropThe Demon Drop ride lasts only 1.5 seconds. How fast areyou going when you are slowed to a halt? How far do youfall?

Amusement park rides thrill customers by changing thevelocities of the riders quickly and in unexpected direc-

tions. The ride may be a gentle, circling carousel or a rap-idly-swinging tilt-o-whirl. It may be the rapid ascents anddescents of a roller coaster. For pure heart-stopping thrills,however, nothing beats the ride on the Demon Drop. Helddown with shoulder bars, you stop for a moment at the top.The supports under the car are suddenly released, and youdrop with close to the acceleration of gravity.

Chapter Outline4.1 WHAT IS

ACCELERATION?· Average Acceleration· Average and Instantaneous

Acceleration· Velocity of an Object With

Constant Acceleration

4.2 DISPLACEMENT DURINGCONSTANTACCELERATION

· Displacement When Velocityand Time Are Known

· Displacement WhenAcceleration and Time AreKnown

· Displacement When Velocityand Acceleration Are Known

· Acceleration Due to Gravity

VConcept CheckThe following terms or conceptsfrom earlier chapters areimportant for a goodunderstanding of this chapter. Ifyou are not familiar with them,you should review them beforestudying this chapter.· graphing, slopes, Chapter 2· manipulating algebraic

equations, Chapter 2· position and velocity,

Chapter 3• graphs of motion at constant

velocity, Chapter 3

63

Objectives· define average and instantaneous

acceleration.• be able to calculate average

acceleration, given two velocitiesand the time interval between them.

· be able to calculate average andinstantaneous acceleration from avelocity-time graph.

· be able to calculate final velocity inthe case of uniform acceleration.

Vector Conventions

• Velocity vectors are red.

• Displacement vectors are green.

~ Acceleration vectors are pur-ple.

FIGURE 4-1. The air speedindicators above the runway showthe velocities at the times shownbelow the runway.

64 Acceleration

4.1 ...................................WHAT IS ACCELERATION?

The faster that the velocity of a ride changes, the greater the thrill.How fast does velocity change? The rate at which velocity changes

is such a useful concept that it is given a special name, acceleration. Inthis chapter, as in Chapter 3, we will consider motion in only one di-mension. Changes in velocity can either be positive or negative; direc-tion will be indicated by plus or minus.

Average AccelerationAn airplane is at rest at the start of the runway, Figure 4-1; it has

zero velocity. When cleared for takeoff by the tower, the pilot opensthe plane's throttle. After 10 seconds, the airspeed indicator shows+ 30 m/s. After 20 seconds, it shows + 60 m/s. When 30 seconds haveelapsed, the airspeed is + 90 mls and the plane lifts off. In each 10-second interval the airplane's speed increased by 30 m/s. Thus, in eachsecond, the velocity increased 3 m/s. The airplane has accelerateddown the runway.

Let the change in velocity be Llv, and the time interval during whichLlv

the velocity changes be Llt. Consider the ratio -. When is it large? TheLlt

ratio is large when there is a large velocity change in a small time inter-val. The ratio is called the average acceleration between the two times.

Suppose an object has velocity V1 at time t1 and velocity V2 at timet2· The change in velocity is Llv = V2 - v.. The change takes placeduring the time interval Llt = t2 - t1• Thus, the average acceleration,the change in velocity divided by the interval of time, is given by

Llv

Llt

Velocity is measured in meters per second, mis, so acceleration ismeasured in (m/s)/s, or m/s/s. This unit is read "meters per second persecond." More often the unit for acceleration is written as m/s2 and read"meters per second squared." In either case, acceleration tells howmany meters per second the velocity changes in each second.

Just as position and velocity are vectors, numbers with both magni-tude and direction, so is acceleration. In straight-line motion, accelera-tion can be either positive or negative. When the velocity increases, itschange is a positive number and the acceleration is positive; when ve-locity decreases, the change is negative, and so is the acceleration.

Example ProblemCalculating AverageAcceleration

The velocity of a car increases from 2.0 m/s at 1.0 5 to 16 m/s at4.5 s. What is the car's average acceleration?

Given: first velocity, Unknown: acceleration, aVI = 2.0 m/s B.. _ ..1vsecond velocity, asic equation: a = ..1tV2 = 16 m/stime interval,tl = 1.0 s; t2 = 4.5 s

Solution: velocity change, ..1v = V2 - VI = 16 m/s - 2.0 m/s= 14 m/stime interval, ..1t = t2 - tl = 4.5 s - 1.0 s = 3.5 s

. ..1V 14 m/sacceleration, a = - = --- = 4.0 m/s2..1t 3.5 s

Does an object with a negative acceleration always slow down?Consider the following Example Problem.

Example ProblemAcceleration of a Car Backing Up

A car goes faster and faster backwards down a long driveway. Wedefine forward velocity as positive, so backward velocity is negative.The car's velocity changes from - 2.0 m/s to - 9.0 m/s in a 2.0-stime interval. Find its acceleration.

Unknown: acceleration, aB

. • ..1VaSlc equation: a = -..1t

Given: first velocity,VI = -2.0 rn/s

second velocity,V2= -9.0 m/slit = 2.0 s

Solution: velocity change, ..1v = - 9.0 m/s - (- 2 m/s)

= - 7.0 m/s..1V -7.0 m/s

average acceleration, a = At =LI 2.0 s

= - 3.5 rn/s"

BIOLOGYCONNECTION

The cheetah runs at a maxi-mum speed of 60-63 mph. Itsrate of acceleration is awe-some-from a starting speed ofo to 45 mph in two seconds.

The units of acceleration are usuallym s/s or rn.s".

FIGURE 4-2. The car backing downthe driveway is increasing itsvelocity but in the negative direction.

4.1 What is Acceleration? 65

POCKET

UNIFORM OR NOT?Set up a U channel on a

book so it acts as an inclinedramp (or make a channel fromtwo meter sticks taped togetheralong the edges). Release asteel ball so that it rolls downtheramp. Using a stopwatch, mea-sure the time it takes to roll0.40 m. Write a sentence de-scribing the motion of the ball.Predict how much time it wouldtake to roll 0.80 m. Explainyourprediction.

Acceleration is positive if the velocitychange is positive.

Acceleration is the slope of a velocity-time graph.

FIGURE 4-3. A velocity-time graphfor uniformly accelerated motion.

66 Acceleration

Thus, the car's speed increases, but its acceleration is negative. Whenthe car reaches the end of the driveway, the driver puts on the brakesand comes to a stop. Then, the final velocity will be less negative thanthe initial velocity. The acceleration will be positive, even though thecar is moving slower.

Practice Problems1. An Indy-500 race car's velocity increasesfrom +4.0 m/s to + 36 m/s

over a 4.0-s period. What is its average acceleration?2. The same race car slows from + 36 rn/s to + 15 rn/s over 3.0 s. What

is its average acceleration over this time interval?3. A car is coasting backwards down a hill at - 3.0 m/s when the driver

gets the engine started. After 2.5 s, the car is moving uphill at avelocity of +4.5 m/s. What is the car's average acceleration?

~ 4. A bus is moving at 25 rn/s. The driver steps on the brakes, and thebus stops in 3.0 s.a. What is the average acceleration of the bus while braking?b. Suppose the bus took twice as long to stop. How would the ac-

celeration compare to the acceleration you found above?

Average and Instantaneous AccelerationIn Chapter 3, velocity-time graphs were used to find the object's dis-

placement. Now we will use them in another way. Figure 4-3 lists thevelocities of a jet rolling down a runway for takeoff, and a graph of thedata. As time increases, so does the velocity of the plane. The plane isaccelerating. The change in velocity is the change in the ordinate, orrise, of the curve. The time interval is the change in the abscissa, orrun. Average acceleration is the change in velocity divided by the timeinterval, so the average acceleration is given by the slope of the curve.

rise Llvslope = - = - = a

run Llt

Velocity versus Time

Time(s) Velocity (m/s) v

0 0 ~+100

+20 E. +80~ +602 +40 '00 +403 +60 ~ +204 +80

A5 +100 0 2 3 4 5 6 7

Time(s)

In this case, the curve on a velocity-time graph is a straight line, andso the change in velocity is the same in each time interval. The slope isconstant, as is the acceleration of the object. For the airplane in Figure4-3, the acceleration is 20 rn/s/s. Its velocity increases 20 rn/s for everysecond.

In many cases, however, acceleration changes in time. At a giventime the instantaneous acceleration is the slope of the tangent to thecurve at that time. Figure 4-4 shows the change in velocity with timeof an automobile. In this case, the slope, and thus the acceleration, isgreatest just as the car accelerates from rest. This car's acceleration issmallest when its velocity is largest.

Example ProblemFinding the Average Acceleration of a Runner

Figure 4-5 is a velocity-time graph of a sprinter during a 100-mdash. Find the sprinter's average acceleration between a. 0.0 sand1.0 s. b. 2.0 sand 4.0 s.

Given: velocity-time graph Unknown: average acceleration, a. . Llv rise

Basic equation: a = - = -Llt run

12

~-Sz- 6"(50

~

o 10

Solution: From the graph, we get Table 4-1.

Thus,Llv risea.a = - = - =Llt run

b. a = Llv = rise = _0_.8_m_/s= 0.4 rn/s"Llt run 2.0 s

7.8 m/s1.0 s

7.8 m/s'

The instantaneous acceleration is given by the slope of a tangentto the curve on a velocity-time graph. You will find that a tangent tothe curve as shown on Figure 4-5 at 1.0 s has a slope of 3.9 rn/s'.

Velocity versus Time

W 25.!'! 20-S~ 15·0o~ 10

5

o 5 10 15 20 25 30

Time(s)

FIGURE 4-4. The changes in velocitywith time of a car accelerating as thedriver shifts from first to fourth gear.

FIGURE 4-5. Use with the ExampleProblem.

Table 4-1

Time Velocity(s) (m/s)

0.0 0.0

1.0 7.8

2.0 10.0

4.0 10.8

4.1 What is Acceleration? 67

Physicsandtechnology

Air bags can be installed onboth the driver and passengersides of a car. A crash at 10 to15 mph with its sudden accel-eration, triggers sensorsthat ac-tivate an igniter. This decom-poses sodium azide pellets,producing nitrogen gas thatrushes into the nylon bag, forc-ing it out of its storage compart-ment as it inflates. The wholeprocess takes place in less than40 milliseconds. The air bagspreads the time over which theoccupant's speed is brought tozero. Injuries to the head andchest are reduced or elimi-nated. The occupant forces thegas out of the cushion throughholes in the bag. The bag de-flates within two seconds, al-lowing the driver to maintaincontrol of the car.. Crash tests show that air bagsdramatically reduce injury in

frontal collisions. However,protection is poor in side im-pacts and in rollovers. Can youthink of any other disadvan-tage?CUSHIONING THE BLOW

The automobile has givenpeople freedom, but over

the years, its mass and speedhave resulted in many deathsand serious injuries in colli-sions. To improve safety, man-ufacturers have installed de-vices either to avoid accidentsor to protect occupants of a carin an accident. The latest pas-sive restraint devices includeshoulder straps that move intoplace automatically and airbags. These devices allow thecar occupants to slow downgradually with the car insteadof being thrown into the carbody or into the air where theirspeed goes to zero very rapidly.

Practice Problems5. Describe, in words, the velocity of the toy train shown in Figure 4-

6 between 0 and 20 s.6. Figure 4-6 shows a velocity-time graph of a toy train.

a. During which time interval or intervals is the speed constant?b. During which interval or intervals is the train's acceleration posi-

tive?c. During which interval or intervals is its acceleration less than

zero?d. During which time interval is the acceleration most negative?

7. For Figure 4-6, find the average acceleration during the given timeintervals.a. 0 to 5 s c. 15 to 20 sb. 0 to 10 s d. 0 to 40 s

~ 8. a. Draw a velocity-time graph for an object whose velocity is con-stantly decreasing from 10 rn/s at t = 0.0 s to -10 m/s at t

= 2.0 s. Assume it has constant acceleration.b. What is its average acceleration between 0.0 sand 2.0 s?c. What is its acceleration when its velocity is 0 m/s?

12

~10

.s 8.?:- 6'0

40a5> 2

0 10 20 30 40Time(s)

FIGURE 4-6. Use with PracticeProblems 5, 6, and 7.

68 Acceleration

Velocity of an Object With Constant AccelerationAcceleration that does not change in time is uniform or constant ac-

celeration. In this case, the velocity-time graph of constant accelerationis a straight line. The velocity when the clock reading, or time, is zerois called the initial velocity, Vi' The slope of the line is the acceleration,a, so the equation that describes the curve, and thus the velocity at t, is

I Vf = Vi + at. I

Example ProblemFinal Velocity After Uniform Acceleration

If a car with a velocity of 2.0 m/s at t = 0 accelerates at a rate of+4.0 rn/s" for 2.5 s, what is its velocity at time t = 2.5 s?

Given: Vi = + 2.0 m/sa = + 4.0 m/s2t = 2.5 s

Unknown: final velocity, Vf

Basic equation: vr = Vi + at

Solution: Vf = Vi + at+ 2.0 m/s + (+ 4.0 m/s2)(2.5 s)+ 12 m/s

Practice Problems9. A golf ball rolls up a hill toward a Putt-Putt hole.

a. If it starts with a velocity of + 2.0 rn/s and accelerates at a con-stant rate of - 0.50 m/s", what is its velocity after 2.0 s?

b. If the acceleration occurs for 6.0 s, what is its final velocity?c. Describe, in words, the motion of the golf ball.

10. A bus traveling at + 30 krn/h accelerates at a constant + 3.5 m/s 'for 6.8 s. What is its final velocity in km/h?

11. If a car accelerates from rest at a constant 5.5 rn/s", how long willbe required to reach 28 m/s?

~ 12. A car slows from 22 m/s to 3 m/s with a constant acceleration of- 2.1 m/s". How long does it require?

........................CONCEPT REVIEW1.1 Describe, in words, how you calculate average acceleration be-

tween two times.1.2 a. If an object has zero acceleration, does that mean its velocity is

zero? Give an example.b. If an object has zero velocity at some instant, does that mean its

acceleration is zero? Give an example.1.3 Is krn/h/s a unit of acceleration? Is this unit the same as km/s/h?

Explain the meaning of the two.1.4 Critical Thinking: Figure 4-7 is a strobe photo of a horizontally

moving ball. What information about the photo would you need andwhat measurements would you make to estimate the acceleration?

F. Y. I.One of Galileo's contributions

to the study of motion was torecognize that the correctmethod to use in solving a prob-lem is to remove complicatingfactors, like air resistance, tomake clear the important factor,the effect of Earth's gravity onthe motion of bodies.

If a velocity-time graph is a straight line,the acceleration is uniform or constant.

FIGURE 4-7. Use with ConceptReview 1.4.

4.1 What is Acceleration? 69

7. Place the ball on the ramp at the reference line.Start the car behind the reference line and re-lease the ball as the cart reaches the referenceline.

8. Which object won the race? Was it close?

PHYSICSLAB· : Ball and Car RacePurposeTo compare the motion of constant velocity to themotion of constant acceleration.

Materials· battery-powered car• 1/1 steel ball· 10 ern of tape· 90 ern length of U channel· short length of plastic straw· 100 ern of fishing line· stopwatch

Procedure1. Tape the straw to the bottom of the car. Thread

the fishing line through the straw and securelytape both ends as shown below. (The car willnow be constrained to move in a straight line.)

2. Place a piece of tape on the table to serve as areference line.

3. Turn the car on and start it behind the referenceline. (See sketch.)

4. Measure the average time needed for the car totravel from the reference Iine to the end of theramp (at least 3 trials).

5. Place a book or wood blocks and so on underone end of the U channel and measure the timeit takes for the ball to roll down the ramp whenstarting from the reference line!

6. Adjust the height of the end of the ramp untilthe time it takes the ball to roll is within the0.1 s of the time needed for the car.

Observations and Data1. Describe the motion of the car in words.2. Describe the motion of the ball in words.3. Did it appear that the objects ever had the same

speed during the race? If so, then where?

Analysis1. Set up a graph of position (vertical axis) vs time

(horizontal axis).2. Assume that the car had a steady speed. Show

this motion on the graph.3. Place a marker at the halfway point for the ball.

Measure the time needed for the ball to roll half-way.

4. Assume that the ball was gaining speed (accel-erating) for the entire time. Draw a smoothcurve to show the motion of the ball. (Hint: Thiscurve should have the same starting and endingpoints as the car, but also include the data pointfor the halfway distance.)

5. Did the two graphs ever have equal slopes(equal velocities)? Where?

Applications1. Why does it take so long for a policeman in a

parked car to catch up after you speed by him?

book steel ball

\ "U"channel

/reference line fishing line

70 Acceleration

/plastic straw finish line

4.2 DISPLACEMENT DURINGCONSTANT ACCELERATION

The distance traveled by a moving object is much easier to measurethan its velocity. For that reason, we will develop three equations

that relate the displacement to two of the three other quantities we useto describe motion-time, velocity, and acceleration. Theseequationsare correct only when acceleration is constant. Fortunately, for manytypes of moving bodies, including those falling due to gravity, this con-dition is true when air resistance is ignored.

Displacement When Velocity and Time Are KnownWhen an object is moving with constant velocity, its displacement

can be found by multiplying its velocity by the time interval. To find thedisplacement if the object is uniformly accelerating, the velocity is re-placed by the average velocity,

\! = 112(Vf+ v.).

To show this, consider the velocity-time graph for an object with con-stant acceleration, Figure 4-8. The displacement is the total area underthe curve. The total area is the sum of the area of the rectangle, Vit, andthe area of the triangle, 1/2(Vr - Yilt. The total area is the displacement,d = vit + 1/2(Vr - Yilt, or

r-I d-=-l-h-(V-r-+-V-i -)t-,.I

Example ProblemDisplacement When Velocity and Time Are Known

What is the displacement of a train as it is accelerated uniformlyfrom + 11 m/s to +33 m/s in a 20.0-s interval?Given: initial velocity,

Vi = + 11 m/sfinal velocity,Vr = +33 m/stime interval,t = 20.0 s

Solution: d = 1/2(Vr + Yilt = 1/2(+ 33 m/s + 11 m/s)(20.0 s)= +4.4 x 102 m

Unknown: displacement, dBasic equation: d = 1/2(vr + Yilt

Objectives· be able to calculate the

displacement of an objectundergoing uniform accelerationwhen you know two out of thethree quantities: acceleration, time,velocity.

· be able to solve problems of themotion of objects uniformlyaccelerated by gravity.

· learn to use an organized strategyfor solving motion problems.

4.2 Displacement During Constant Acceleration 71

v

v, r------"71""\

V;

v,-v;=at

'-~o t

FIGURE 4-8. A velocity-time graphof uniformly accelerated motion. Theshaded area representsdisplacement.

72 Acceleration

Practice Problems13. A race car traveling at +44 m/s is uniformly accelerated to a veloc-

ity of + 22 m/s over an 11-s interval. What is its displacement dur-ing this time?

14. A rocket traveling at + 88 m/s is accelerated uniformly to + 132 m/sover a 15-s interval. What is its displacement during this time?

15.A car accelerates at a constant rate from 15 m/s to 25 m/s while ittravels 125 m. How long does this motion take?

~ 16. A bike rider accelerates constantly to a velocity of 7.5 m/s during4.5 s. The bike's displacement is + 19 m. What was the initial ve-locity of the bike?

Displacement When Accelerationand Time Are Known

If the initial velocity, acceleration, and time interval are known, thedisplacement of the object can be found by combining equations al-ready used. The final velocity of a uniformly accelerated object is Vr= Vi + at. The displacement of an object with uniform acceleration isd = 1/2(Vr+ Vi)t.

Substitute the final velocity from the first equation into the secondequation.

d = l/z(vr + Vi)t = 1/2((Vi+ at) + Vi)t = 1/2(2vi + at)t

! d = Vit + IhatZ!

There are two terms in this equation. The first term, Vii, correspondsto the displacement of an object if it were moving with constant veloc-ity, Vi' The second term, 1/2at2, gives the displacement of an object start-ing from rest and moving with uniform acceleration. The sum of thesetwo terms gives the displacement of an object that starts with an initialvelocity and accelerates uniformly. For an object that starts from rest,the equation reduces to d = 1fzatz.

You can see these relationships by looking at a velocity-time graph ofuniformly accelerated motion. As you know, the area under the curveof a velocity-time graph is the displacement of the object.

Look at the shaded area under the curve between time ti = 0 andtime t, Figure 4-8. The shaded area is made up of a rectangle with atriangle on top. The area of the rectangle is its height, the initial veloc-ity, Vi, multiplied by its base, the time t. That is, area = v.t. This areais the displacement the object would have if it had the constant velo-city Vi'

The altitude of the triangle is the change in velocity of the uniformlyaccelerated object. Since Vr = Vi + at, (v, - v;l = at. The base is againthe time t. The area of a triangle is half the base times the altitude, or1/2(at)(t) = 112at2•This area is the displacement the object would have ifit were uniformly accelerated from rest.

The total displacement is the sum of the two areas under the curve,d = v;t + 1/2at2.

Position versus Time

Time(s) Position (m)

0 +01 +102 +403 +904 +1605 +250

280

240

I 200c

1600:-e(f)

0 1200...

80

40

0 2 4 6

Time(s)

What kind of curve is the position-time graph of an object acceleratedfrom rest at a constant rate? Figure 4-9 includes a table of the positionsof the jet plane for the first five seconds of its acceleration. The positionswere found from the area under the curve of the velocity-time graph,Figure 4-3.

The data were then plotted to give the position-time graph shown.The curve of the position-time graph is half a parabola. When onequantity varies directly with the square of the other, a parabolic curveresults.

As you have seen, the slope of a position-time graph is the velocityof the object. When an object's velocity is constant, its position changesthe same amount each second. Therefore, the position-time graph is astraight line with a constant slope. If, however, the object is acceler-ated, during each second its displacement is larger than it was thesecond before. The position-time graph for accelerated motion is aparabola, and the slope does change. Figure 4-10 again shows theposition-time graph for the jet plane. Suppose you want to find the slopeof the curve at point P. Draw a line tangent to the curve at point P. Theslope of the tangent line is the instantaneous velocity at point P. PointP is at the time 3 seconds. A tangent line is drawn, and

rtse + 270 m - 0 mv = slope =

run 6.0 s - 1.5 s270 m

= -- = +60 rn/s.4.5 s

280

240

I 200c

1600

""en0 1200...

80

40

0 2 4 6

Time(s)

FIGURE 4-9. A position-time graphfor uniformly accelerated motion.

FIGURE 4-10. The slope at any pointon a position-time graph indicatesthe velocity of the object.

4.2 Displacement During Constant Acceleration 73

HELP WANTEDAIR TRAFFIC CONTROLLERIf you can think fast and commu-

nicate clearly and decisively whilejuggling dozens of details in yourmind, you might be the person weare seeking. Three years of work ex-perience or 4 years of college andthe ability to pass the Federal CivilService Exam may qualify you to en-ter a training program. Advancementis based on job performance, andcompletion of on-the-job training pro-grams. Knowledge of foreign lan-guages is a plus. For information

agement

FIGURE 4-11.

74 Acceleration

The velocity at the end of 3 seconds for an object that accelerates at+20 m/s2 is +60 rn/s, which agrees with the velocity-time graph forthe jet plane.

Example ProblemCalculating Displacement From Acceleration and Time

A car starting from rest accelerates uniformly at + 6.1 m/s" for7.0 s. How far does the car move?Given: initial velocity,

Vi = 0 m/sacceleration,a = + 6.1 rn/s"final time, t = 7.0 s

Solution: d = vit + 1/2ai= (0 m/s)(7.0 s) + V2(+6.1 m/s2)(7.0 S)2

= 0 m + 150 m = + 150 m

Unknown: displacement, dBasicequation: d = Vit + 1/2ai

Practice Problems17. An airplane startsfrom rest and acceleratesat a constant + 3.00 rn/s"

for 30.0 s before leaving the ground. What is its displacement dur-ing this time?

18. Starting from rest, a race car moves 110 m in the first 5.0 s of uni-form acceleration. What is the car's acceleration?

19. A driver brings a car traveling at + 22 m/s to a full stop in 2.0 s.Assume its acceleration is constant.a. What is the car's acceleration?b. How far does it travel before stopping?

~ 20. A biker passesa lamppost at the crest of a hill at +4.5 m/s. Sheaccelerates down the hill at a constant rate of + 0.40 m/s2 for 12 s.How far does she move down the hill during this time?

Displacement When Velocity and Acceleration AreKnown

We can combine the equations for final velocity and displacement toform an equation relating initial and final velocities, acceleration, anddisplacement in which time does not appear. First, recall the two mo-tion equations,

d = 1/2(vr + Vi)t and v, = Vi + at.

The second equation is now solved for t and substituted in the first,resulting in

a aSolving for v/ yields

I v? = Vi2 + 2ad·1

Example ProblemCalculating Acceleration From Displacement and Velocity

An airplane must reach a velocity of 71 m/s for takeoff. If the run-way is 1.0 km long, what must the constant acceleration be?

Given: initial velocity,Vj = 0 m/sfinal velocity,Vf = 71 m/sdisplacement,d = + l.0 km

= + 1.0 X 103 m

Unknown: acceleration, aBasic equation: v? = Vj2 + 2ad

Solution: v? = Vj2 + 2ad, so av? - Vj2

2d(71 m/s)2 - 0

2 (+ l.0 x 103 m)+2.5 rn/s'

Consider the reverse of this problem. The plane lands going 71 m/s.It slows to a halt with an acceleration of -2.5 m/s'. How muchrunway does it require? We use the same basic equation, solved forthe displacement,

v? - Vj2

2a(0 m/s)2 - (71 m/s)2

2( - 2.5 m/s2)+ 1.0 x 103 m.

Notice that the displacement is the same. The motion in these twocases is said to be symmetrical. The conditions at the beginning andend of the motion are reversed, and the solutions are equivalent.Symmetry often plays an important role in physics problems.

d =

Practice Problems21. An airplane accelerates from a velocity of 21 m/s at the constant

rate of 3.0 m/s2 over + 535 m. What is its final velocity?22. The pilot stops the same plane in 484 m using a constant accelera-

tion of - 8.0 m/s'. How fast was the plane moving before brakingbegan?

23. A person wearing a shoulder harness can survive a car crash if theacceleration is smaller than - 300 m/s:'. Assuming constant accel-eration, how far must the front end of the car collapse if it crasheswhile going 101 km/h?

~24. A car is initially sliding backwards down a hill at - 25 km/h. Thedriver guns the car. By the time the car's velocity is + 35 km/h, itis + 3.2 m from its starting point. Assuming the car was uniformlyaccelerated, find the acceleration.

POCKETLAB

DIRECTION OFACCELERATION

Tape a bubble level onto thetop of a laboratory cart. Adjustthe level until the bubble is cen-tered. Watch the motion of thebubble as you begin to pull thecart forward. Which way doesthe bubble move? The bubbleshows the acceleration of thecart. Which way is the cart ac-celerating as it starts to moveforward? Try to keep the cartmoving so that the bubble re-turns to the center position. Whatdirection is the acceleration ofthe cart now? What happens tothe bubble when the cart coaststo a stop? What does that indi-cate about the acceleration?Predict what would happen tothe bubble if you tie the string tothe back of the cart and repeatthe experiment. Draw a sketch ofthe cart and use arrows to indi-cate the direction of accelerationas the cart starts to move back-wards.

4.2 Displacement During Constant Acceleration 75

When air resistance is ignored, theacceleration of a falling body doesn'tdepend on mass, initial height, andinitial velocity.

FIGURE 4-12. Strobe photograph ofa ball in freefall. Images are at1/120-s intervals.

76 Acceleration

Acceleration Due to GravityGalileo was the first to show that all objects fall to Earth with a con-

stant acceleration. There is no evidence that he actually dropped twocannon bails from the leaning tower of Pisa, but he did conduct manyexperiments with balls rolling down inclined planes. He had previouslyshown that the inclined plane "diluted" gravity, slowing motion so hecould make careful measurements.

We know from experiments that no matter what the mass of the ob-ject is, whether it is dropped from 1 m, 10m, or 100 rn. or whether itis dropped or thrown, as long as air resistance can be ignored, the ac-celeration due to gravity is the same for all objects at the same locationon Earth.

Acceleration due to gravity is given a special symbol, g. Since accel-eration is a vector quantity, g must have both magnitude and direction.We will choose "up" as our positive direction, so a falling object has anegative velocity. The acceleration of a falling object is also negative.On the surface of Earth, a freely falling object has an acceleration, g, of-9.80 m/s'. Actually, g varies slightly, from -9.790 m/s2 in southernFlorida to - 9.81 0 m/s2 in northern Maine. It is also smaller at highaltitudes; g = - 9.789 m/s' on the top of Pike's Peak. We will use atypical value of -9.80 rn/s'. Thus, if a ball is dropped from the top ofa building, it starts with zero velocity and in each second of fall gains adownward velocity of - 9.80 rn/s.

Strobe photos of a dropped ball are shown in Figure 4-12. The timeinterval between photos is 11120 s. The distance between each pair ofimages increases. Since v = lldillt, the speed is increasing. Because wehave taken the upward direction to be positive, the velocity is becomingmore and more negative.

Assuming no air resistance, all problems involving motion of fallingobjects can be solved by using the equations developed in this chapterwith the acceleration, a, replaced by g.

Vf = Vi + gtv? = Vi

2 + 2gd

d= Vit + 1/2gt2

If you throw a ball straight up, it leaves your hand with a positivevelocity of, say, + 20.0 m/s. Positive velocity means it is rising. Theacceleration of the ball is - 9.80 rn/s", so after 1 second its velocity is+ 10.2 m/s. After 2 seconds the velocity is + 0.40 m/s. After 3 seconds,the velocity is - 9.40 m/s. Now the velocity is no longer positive. Thenegative velocity indicates that the ball is no longer rising, it is falling.After 4 seconds, the velocity is -19.20 mis, so it is falling faster andfaster. See the velocity-time graph, Figure 4-13a. Let's examine whathappens around 2 seconds, as shown in Figure 4-13b. The velocitychanges smoothly from positive to negative at about 2.04 s. For just aninstant, at the top of its flight, the ball has zero velocity.

How does the ball's position change? The position-time graph, Figure4-13c, has a large positive slope at t = 0, when the velocity is largeand positive. As time increases, the velocity and the slope decrease.When the slope (and velocity) is zero, the ball is at its maximum height,as shown in the detailed position-time graph, Figure 4-13d. On the

20 5 25 20.41

~ I I I-sc c 0.£0 z- 0 0 0 ~

'(3 '';:::; '';:::; c:ito 'Ui 'Ui0 0 N

~ ~0 0CL CL

o- .5 L--,-~_-,-----,-~

4 2

ba Time(s)2.05

Time(s)

OE--_'--'--_'--_-J2.1 0

c2.1

way down, the ball's velocity and the slope of the position-time graphare negative and become more negative. After it has fallen 2.04 sec-onds, the ball will be once again in your hand. It spends as much timerising as falling. It rises the same distance during the first 2.04 secondsas it falls during the final 2.04 seconds. Note that, although the totaldistance the ball travels is the distance up plus the distance down, thetotal displacement, the change in position, is zero.

Example ProblemFreely Falling Objects

The time the Demon Drop ride at Cedar Point, Ohio is freely fail-ing is 1.5 s. a. What is its velocity at the end of this time? b. How fardoes it fall?Given: acceleration of gravity,

g = - 9.80 m/s2initial velocity,Vi = 0 m/stime interval,t = 1.5 s

Solution:a. Vf = Vi + gt = 0 m/s + (-9.80 rn/s'') (1.5 s)

Vf = -15 m/sb: d = Vit + 1/2gt2 = 0 m/s (1.5s)+ 112(- 9.80 m/s2) (1.5 S)2

= 1/2 (- 9.80 m/s2) (2.25 S2) = -11 mNegative displacement means that the final position is lower than theinitial position.

Unknowns: a. final velocity, Vf

b. displacement, dBasic equations:

a. Vf = Vi + gt

b. d = Vit + 112gt2

Practice Problems25. A brick falls freely from a high scaffold.

a. What is its velocity after 4.0 s?b. How far does the brick fall during the first 4.0 s?

~ 26. Now that you know about acceleration, test your reaction time. Aska friend to hold a ruler just even with the top of your fingers. Thenhave your friend drop the ruler. Taking the number of centimeters

20.39 L-_---'- ---I

2 3 4Time(s) d

2 2.05Time(s)

FIGURE 4-13. Velocity-time andposition-time graphs for a tennis ballthrown straight up.

~ .Demon Drop

FIGURE 4-14. The thrill of theDemon Drop ride comes from beingin freefall.

4.2 Displacement During Constant Acceleration 77

FIGURE 4-15. A tennis ball thrownup vertically.

78 Acceleration

that the ruler falls before you can catch it, calculate your reactiontime. An average of several trials will give more accurate results.The reaction time for most people is more than 0.15 s.

Several equations for solving problems of uniformly accelerated mo-tion have been introduced in this chapter. How do you know whichone(s) to use in solving a problem? The equations are listed in Table 4-2, together with a list of the quantities related by each equation.

Table 4-2

Equations of Motion for Uniform Acceleration

Equation Variables

v, = Vi + at Vi v, a td = 1f2(Vf + Vi)t Vi d Vf td = v;t + 1f2af Vi d a tVf2 = Vi2 + 2ad Vi d Vf a

Note that each equation has at least one of the four variables, d, vr.a, or t missing. All include the initial velocity, Vi, as well as the initialtime (0) and initial position (also zero).

PROBLEM SOLVING STRATEGYWhen solving problems, use an orderly procedure.

1. Read the problem carefully. Try to visualize the actual situation.Make a sketch if necessary.

2. Identify the quantities that are given in the problem.3. identify the quantity that is unknown, the one you have to find.4. Select the equation or equations that will relate the given and un-

known quantities.5. Make sure the equations can be applied to the problem, that is, is

the acceleration constant?6. Rewrite equations as needed to solve for the unknown quantity.7. Substitute the given values including proper units into the equation

and solve. Be sure your answer is in the correct units.8. Make a rough estimate to see if your answer is reasonable.

Example ProblemFlight of a Tennis Ball

A tennis ball is thrown straight up with an initial velocity of + 22.5m/s. it is caught at the same distance above ground from which itwas thrown. a. How high does the ball rise? b. How long does theball remain in the air?Given: Vi = + 22.5 mis,

upwardg = - 9.80 rn/s",downward

Unknowns: a. d b. tBasic equations: Vi = Vi + gt

d = vit + 1/2gt2

V? Vi2 + 2gd

d = 1/2(vr + Vi)t

Solution:a. At the top of its flight, the instantaneous velocity of the ball is zero.

Thus, the final velocity for the upward part of the ball's flight willbe zero. Therefore, we know Vj, g, and need d. Use the equationv? = Vj2 + 2gd, solving it for d.

With v, = 0, 0 '= Vj2 + 2gd or 2gd = - Vj2.

Thus,

d = - Vj2 = - (22.5 m/s)2 = +25.8 m.2g 2(-9.80 rn/s'')

The direction is up (positive), so d = +25.8 m.b. From symmetry, we know that the times required for the rising and

falling parts of the flight will be the same. That is, the time to riseis half the total time. Thus, we know Vf and g but need t. UseVf = Vj + gt. With Vf = 0, we solve for t, obtaining

- Vj - 22.5 m/st = - = = + 2.30 s.

g -9.80m/s2

Thus, the total trip time is 4.60 s.

Example ProblemFinding Displacement When Velocities and Times Are Known

A spaceship far from any star or planet accelerates uniformly from+65.0 m/s to + 162.0 m/s in 10.0 s. How far does it move?Given: Vj = + 65.0 m/s Unknown: displacement, d

Vf = + 162.0 m/s Basicequation: d = 1/2(Vf + Vj)t

t = 10.0 sSolution: d = 1/2(Vf + Vj)t

1/2(+162.0 m/s + 65.0 m/s)(10.0 s)+ 1.14 X 103 m = + 1.14 km

Practice Problems27. If you drop a golf ball, how far does it fall in 1/2 s?28. A spacecraft traveling at a velocity of + 1210 m/s is uniformly ac-

celerated at -150 m/s'. If the acceleration lasts for 8.68 s, what isthe final velocity of the craft? Explain your results in words.

29. A man falls 1.0 m to the floor.a. How long does the fall take?b. How fast is he going when he hits the floor?

30. On wet pavement, a car can be accelerated with a maximum ac-celeration a = 0.20 g before its tires slip.a. Starting from rest, how fast is it moving after 2.0 s?b. How far has it moved after 4.0 s?

IiCAicuUioRCalculators can be used to evaluateequations. Note that each of thefactors in the denominator must bedivided into the numerator.

2

d= -~2g

(22.5)@J8 -506 25808(9.80180 25. 82Q082

d = 25.8 m

FIGURE 4-16. A spacecraftaccelerates vertically at takeoff.

4.2 Displacement During Constant Acceleration 79

POCKETLAB

A BALL RACEAssemble an inclined ramp

from a piece of U channel, or twometer sticks taped togetheralong the edges, and a book.Mark a position 40 cm and80 cm from the top. Start oneball at the top and one ball at the40 cm position. When the ballsare released at the same instant,will the balls get closer togetheror farther apart as they roll downthe ramp? Why? Try it. Explainwhat happened in terms of thevelocities. What does this tellyou about the acceleration ofeach ball? Release one ball fromthe top. Release the other ballfrom the top as soon as the firstball reaches the 40 cm mark. Dothe balls stay the same distanceapart? As they roll down the hill,do they ever have the same ve-locities? Do they have the sameacceleration?

31. A pitcher throws a baseball straight up with an initial speed of27 m/s.a. How long does it take the ball to reach its highest point?b. How high does the ball rise above its release point?

~ 32. A motor of a certain elevator gives it a constant upward accelerationof 46 m/min/s. The elevator starts from rest, accelerates for 2.0 s,then continues with constant speed.a. Explain what this statement of acceleration means.b. What is the final speed after 2 s?c. Calculate speed after 0.5, 1.0, 1.5, 2.0, 3.0, 4.0, and 5.0 s.

Sketch a graph showing speed vs time.d. How far has it risen 1.0, 2.0, 3.0, 4.0, and 5.0 s after start?

Sketch the graph.

........................CONCEPT REVIEW2.1 If you are given a table of velocities of an object at various times,

how could you find out if the acceleration is constant?2.2 If initial and final velocities and constant acceleration are given and

displacement is unknown, what equation would you use?2.3 The gravitational acceleration on Mars is about 113 that on Earth. If

you could throw a ball up with the same velocity on Mars as onEarth,a. how would its maximum height compare to that on Earth?b. how would its flight time compare?

2.4 Critical Thinking; When a ball is thrown vertically upward, it con-tinues up until it reaches a certain point, then it falls down again.At the highest point, its velocity is instantaneously zero. It is notmoving up or down. a. Is it accelerating? b. Give reasons. c. Devisean experiment to find out.

CHAPTER 4 REVIEW····························--------------------------------------------SUMMARY

4.1 What Is Acceleration?

· Acceleration is the ratio of the change in velocityto the time interval over which it occurs.

· Constant acceleration is called uniform acceler-ation.

· The slope of the line on a velocity-time graph isthe acceleration of the object.

· A velocity-time graph for a uniformly acceleratedobject is a straight line.

80 Acceleration

4.2 Displacement During ConstantAcceleration

· Three equations relate the displacement of auniformly accelerated object to velocity, time,and acceleration.

· A position-time graph for a uniformly acceleratedobject is half a parabola. Position varies with thesquare of the time.

· The acceleration of gravity is - 9.80 rn/s" nearEarth.

· Problems of uniformly accelerated motion canbe solved using one or more of the formulas inTable 4-2.

KEY TERMSaverage accelerationinstantaneous

accelerationconstant or uniform

acceleration

initial velocityfinal velocityacceleration due

to gravity

REVIEWING CONCEPTS1. A car is traveling on an interstate highway.

a. Can the car have a negative velocity and apositive acceleration at the same time?

b. Can the car change the direction of its ve-locity when traveling with a constant accel-eration?

2. Can the velocity of an object change when itsacceleration is constant? If you answer yes,give an example. If you answer no, explainwhy not.

3. What does the slope of the tangent to a veloc-ity-time graph measure?

4. If a velocity-time curve is a straight line paral-lel to the t-axis, what can be deduced aboutthe acceleration?

5. If you are given a table of velocities of an ob-ject at various times, how could you find out ifthe acceleration of the object is constant?

6. If initial and final velocities and accelerationare given, and the distance is unknown, whatequation would you use?

7. Write a summary of the equations for dis-placement, velocity, and time of a body expe-riencing uniformly accelerated motion.

8. Explain why an aluminum ball and a steel ballof similar size and shape, dropped from thesame height, reach the ground at the sametime.

9. Give some examples of falling objects forwhich air resistance cannot be ignored.

10. Give some examples of falling objects forwhich air resistance can be ignored.

APPLYING CONCEPTS1. Figure 4-4 shows the velocity-time graph for

an accelerating car. The three "notches" inthe curve occur where the driver shifts gears.a. Describe the changes in velocity and ac-

celeration of the car while it is in first gear.

b. Is the acceleration just before a gearchange larger or smaller than the acceler-ation just after the gear change? Explainyour answer.

2. Study Figure 4-4. During what time interval isthe acceleration largest? smallest?

3. Solve the equation VI = Vi + at for accelera-tion. What is the equation you get?

4. Look at Figure 4-17, which is a position-timegraph of uniform acceleration.a. What type of curve does this graph repre-

sent?b. What does the slope of a tangent line

taken at any point represent?c. How would slopes taken at higher points

on the line differ from those taken at lowerpoints?

Time

FIGURE 4-17. Use with ApplyingConcepts 4.

5. An object shot straight up rises for 7.0 s be-fore it reaches its maximum height. A secondobject falling from rest takes 7.0 s to reach theground. Compare the displacements traveledby the objects during the 7.0-s period.

6. Describe the changes in the velocity of a ballthrown straight up into the air. Then describethe changes in its acceleration.

7. The value of 9 on the moon is 1/6 of its valueon Earth.a. Will a ball dropped from the same height

by an astronaut hit the ground with asmaller or larger speed than on Earth?

b. Will the ball take more or less time to fall?8. A ball is thrown vertically upward with the

same initial velocity on Earth and on planetDweeb, which has three times the gravita-tional acceleration as Earth.a. How does the maximum height reached by

the ball on Dweeb compare with the maxi"\..mum height on Earth?

b. If the ball on Dweeb were thrown withthree times greater initial velocity, howwould that change your answer to a?

Chapter 4 Review 81

9. Rock A is dropped from a cliff; rock B isthrown downward.a. When they reach the bottom, which rock

has a greater velocity?b. Which has a greater acceleration?c. Which arrives first?

PROBLEMS

4.1 What Is Acceleration?

1. Find the uniform acceleration that causes acar's velocity to change from 32 m/s to 96 m/sin an 8.0-s period.

2. Rocket-powered sleds are used to test the re-sponses of humans to acceleration. Startingfrom rest, one sled can reach a speed of444 m/s in 1.80 s and can be brought to astop again in 2.15 s.a. Calculate the acceleration of the sled when

starting and compare it to the accelerationdue to gravity, 9.80 rn/s".

b. Find the acceleration of the sled whenbraking and compare it to the magnitude ofthe acceleration due to gravity.

3. Use Figure 4-18 to find the acceleration ofthe moving objecta. during the first 5 s of travel.b. between the fifth and the tenth second of

travel.c. between the tenth and the fifteenth second

of travel.d. between the twentieth and twenty-fifth sec-

ond of travel.

z!1j

20-Sz-'0 100Q)>

10 15 20 25 30 t

Time (s)

FIGURE 4-18. Use with Problem 3.

o 5

4. To accompany each of the graphs in Figure4-19, drawa. a velocity-time graph.b. an acceleration-time graph.

82 Acceleration

·~LL2~ §~~ ~ ~~ ~ ~

a Time b Time c Time

FIGURE 4-19. Use with Problem 4.

5. A car with a velocity of 22 m/s is accelerateduniformly at the rate of 1.6 m/s'' for 6.8 s.What is its final velocity?

6. The velocity of an automobile changes overan 8.0-s time period as shown in Table 4-3.a. Plot the velocity-time graph of the motion.b. Determine the displacement of the car dur-

ing the first 2.0 s.c. What displacement does the car have dur-

ing the first 4.0 s?d. What displacement does the car have dur-

ing the entire 8.0 s?e. Find the slope of the line between t = 0 s

and t = 4.0 s. What does this slope rep-resent?

f. Find the slope of the line between t = 5.0 sand t = 7.0 s. What does this slope indi-cate?

Table 4-3

Time Velocity Time Velocity(s) (m/s) (s) (m/s)

0.0 0.0 5.0 20.01.0 4.0 6.0 20.02.0 8.0 7.0 20.03.0 12.0 8.0 20.04.0 16.0

~ 7. Figure 4-20 shows the position-time and ve-locity-time graphs of a karate expert using afist to break wooden boards.a. Use the velocity-time graph to describe the

motion of the expert's fist during the first10 ms.

b. Estimate the slope of the velocity-timegraph to determine the acceleration of thefist when it suddenly stops.

c. Express the acceleration as a multiple ofthe gravitational acceleration, 9 = 9.80 rn/s''.

d. Determine the area under the velocity-timecurve to find the displacement of the fist inthe first 6 ms. Compare with the position-time graph.

E 10

2CQ)

EQ)(J

"' 5 10 15 20 25 35Ciif)

i:5-5 Time (ms)

ti b.u;

0

:i[ -5E-.c·0 -100OJ>

-15

FIGURE 4-20. Use with Problem 7.

8. A supersonic jet flying at 145 m/s is acceler-ated uniformly at the rate of 23.1 m/s" for20.0 s.a. What is its final velocity?b. The speed of sound in air is 331 m/s. How

many times the speed of sound is theplane's final speed?

9. Determine the final velocity of a proton thathas an initial velocity of 2.35 x 105 mis, andthen is accelerated uniformly in an electricfield at the rate of - 1.10 x 1012 rn/s" for 1.50x 10-7 s.

4.2 Displacement During ConstantAcceleration

10. Determine the displacement of a plane that isuniformly accelerated from 66 m/s to 88 m/sin 12 s.

11. How far does a plane fly in 15 s while its ve-locity is changing from 145 m/s to 75 m/s at auniform rate of acceleration?

12. A car moves at 12 m/s and coasts up a hillwith a uniform acceleration of - 1.6 rn/s".a. How far has it traveled after 6.0 s?b. How far has it gone after 9.0 s?

13. Four cars start from rest. Car A accelerates at6.0 rn/s"; car B at 5.4 rn/s"; car C at 8.0 rn/s",and car D at 12 m/s",a. In the first column of a table, show the ve-

locity of each car at the end of 2.0 s.b. In the second column, show the displace-

ment of each car during the same 2.0 s.c. What conclusion do you reach about the

velocity attained and the displacement of abody starting from rest at the end of thefirst 2.0 s of acceleration?

14. An astronaut drops a feather from 1.2 mabove the surface of the moon. If the accel-eration of gravity on the moon is 1.62 rn/s"down, how long does it take the feather to hitthe surface?

~ 15. Table 4-4 is a table of the displacements andvelocities of a ball at the end of each secondfor the first 5.0 s of free-fall from rest.a. Use the data in the table to plot a velocity-

time graph .b. Use the data in the table to plot a position-

time graph.c. Find the slope of the curve at the end of

2.0 'and 4.0 s on the position-time graph.What are the approximate slopes? Do thevalues agree with the table of velocity?

d. Use the data in the table to plot a positionversus time-squared graph. What type ofcurve is obtained?

e. Find the slope of the line at any point. Ex-plain the significance of the value.

f. Does this curve agree with the equationd = %gr?

Table 4-4

Velocity(m/s)

0.0-9.8

-19.6-29.4-39.2-49.0

Time(s)

0.01.02.03.04.05.0

Displacement(m)

0.0-4.9

-19.6-44.1-78.4

-122.5

16. A plane travels 5.0 x 102 m while being ac-celerated uniformly from rest at the rate of5.0 rn/s". What final velocity does it attain?

17. A race car can be slowed with a constant ac-celeration of -11 m/s".a. If the car is going 55 mis, how many me-

ters will it take to stop?b. Repeat for a car going 110 m/s.

18. An engineer must design a runway to accom-modate airplanes that must reach a groundvelocity of 61 m/s before they can take off.These planes are capable of being acceler-ated uniformly at the rate of 2.5 m/s".a. How long will it take the planes to reach

takeoff speed?b. What must be the minimum length of the

runway?

Chapter 4 Review 83

19. A rocket traveling at 155 m/s is accelerated ata rate of - 31.0 m/s".a. How long will it take before the instanta-

neous speed is 0 m/s?b. How far will it travel during this time?c. What will be its velocity after 8.00 s?

20. Engineers are developing new types of gunsthat might someday be used to launch satel-lites as if they were bullets. One such gun cangive a small object a velocity of 3.5 km/s mov-ing it through only 2.0 cm.a. What acceleration does the gun give this

object?b. Over what time interval does the accelera-

tion take place?21. An express train, traveling at 36.0 mis, is ac-

cidentally sidetracked onto a local train track.The express engineer spots a local train ex-actly 1.00 x 102 m ahead on the same trackand traveling in the same direction. The engi-neer jams on the brakes and slows the ex-press at a constant rate of - 3.00 rn/s", Thelocal engineer is unaware of the situation. Ifthe speed of the local is 11.0 mis, will the ex-press be able to stop in time or will there bea collision? To solve this problem, take theposition of the express when it first sights thelocal as a point of origin. Next, keeping inmind that the local has exactly a 1.00 x 102 mlead, calculate how far each train is from theorigin at the end of the 12.0 s it would takethe express to stop.a. On the basis of your calculations, would

you conclude that there is or is not a colli-sion?

b. The calculations you made in part a do notallow for the possibility that a collisionmight take place before the end of the12 s required for the express to come to ahalt. To check on this, take the position ofthe express when it first sights the local asthe point of origin and calculate the posi-tion of each train at the end of each sec-ond after sighting. Make a table showingthe distance of each train from the origin atthe end of each second. Plot these posi-tions on the same graph and draw twolines.

c. Use your graph to check your answer topart a.

22. Highway safety engineers build soft barriersso that cars hitting them will slow down at a

84 Acceleration

safe rate. A person wearing a safety belt canwithstand an acceleration of - 300 rn/s". Howthick should barriers be to safely stop a carthat hits a barrier at 110 km/h?

23. A baseball pitcher throws a fastball at a speedof 44 m/s. The acceleration occurs as thepitcher holds the ball in his hand and movesit through an almost straight-line distance of3.5 m. Calculate the acceleration, assuming itis uniform. Compare this acceleration to theacceleration due to gravity, 9.80 rn/s".

24. If a bullet leaves the muzzle of a rifle with aspeed of 600 mis, and the barrel of the rifle is0.9 m long, what is the acceleration of the bul-let while in the barrel?

~ 25. A driver of a car going 90.0 km/h suddenlysees the lights of a barrier 40.0 m ahead. Ittakes the driver 0.75 s before he applies thebrakes, and the average acceleration duringbraking is -10.0 rn/s".a. Determine if the car hits the barrier.b. What is the maximum speed at which the

car could be moving and not hit the barrier40.0 m ahead? Assume the accelerationrate doesn't change.

~ 26. Data in Table 4-5, taken from a driver'shandbook, show the distance a car travelswhen it brakes to a halt from a specific initialvelocity.

Table 4-5

Braking distance(m)1020345070

Initial velocity(m/s)1115202529

a. Plot the braking distance versus the initialvelocity. Describe the shape of the curveyou obtain.

b. Plot the braking distance versus thesquare of the initial velocity. Describe theshape of the curve you obtain.

c. Calculate the slope of your graph from partb. Find the value and units of the quantity1/slope of the curve.

d. Does this curve agree with the equationVj2 = - 2ad? What is the value of a?

27. A car moving with a constant accelerationcovers the distance between two points 60 mapart in 6.0 s. Its velocity as it passes the sec-ond point is 15 m/s.a. What was the speed at the first point?b. What is the constant acceleration?c. How far behind the first point was the car

at rest?~ 28. As a traffic light turns green, a waiting car

starts with a constant acceleration of 6.0rn/s". At the instant the car begins to acceler-ate, a truck with a constant velocity of 21 m/spasses in the next lane. Hint: Set the two dis-tance equations equal to each other.a. How far will the car travel before it over-

takes the truck?b. How fast will the car be traveling when it

overtakes the truck?29. Use the information from the previous prob-

lem.a. Draw velocity-time and position-time

graphs for the car and truck.b. Do the graphs confirm the answer you cal-

culated for the exercise?30. A stone falls freely from rest for 8.0 s.

a. Calculate the stone's velocity after 8.0 s.b. What is the stone's displacement during

this time?31. A student drops a rock from a bridge to the

water 12.0 m below. With what speed doesthe rock strike the water?

FIGURE 4-21.

32. Kyle is flying a helicopter when he drops abag. When the bag has fallen 2.0 s,a. what is the bag's velocity?b. how far has the bag fallen?

~ 33. Kyle is flying the same helicopter and it is ris-ing at 5.0 m/s when he releases the bag. After2.0 s,a. what is the bag's velocity?

b. how far has the bag fallen?c. how far below the helicopter is the bag?

~ 34. Kyle's helicopter now descends at 5.0 m/sas he releases the bag. After 2.0 s,a. what is the bag's velocity?b. how far has the bag fallen?c. how far below the helicopter is the bag?d. what is common to the answers to Prob-

lems 32, 33 and 34?35. A weather balloon is floating at a constant

height above Earth when it releases a pack ofinstruments.a. If the pack hits the ground with a velocity

of - 73.5 mis, how far does the pack fall?b. How long does the pack fall?

36. During a baseball game, a batter hits a highpop-up. If the ball remains in the air for 6.0 s,how high does it rise? Hint: Calculate theheight using the second half of the trajectory.

37. A tennis ball is dropped from 1.20 m abovethe ground. It rebounds to a height of 1.00 m.a. With what velocity does it hit the ground?b. With what velocity does it leave the

ground?c. If the tennis ball were in contact with the

ground for 0.010 s, find its accelerationwhile touching the ground. Compare to g.

USING LAB SKILLS1. After completing the Physics Lab on page 70,

predict what would happen to the distance be-tween two tennis balls when you drop them atthe same time, but start one ball 0.5 m higherthan the other. Why? Test your prediction bydropping two balls down a stair well or from asecond story window while another student ob-serves.

THINKING PHYSIC-LYWhich has the greater acceleration: a car thatincreases its speed from 50 to 60 km/h, or abike that goes from 0 to 10 km/h in the sametime? Explain.

Chapter 4 Review 85