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KINEMATICS

INTRODUCTIONKinematics is the study of an object's motion in terms of its displacement, velocity, and acceleration.Questions such as How far doesthisobject travel?or How fast andinwhat directiondoes it move?or Atwhatrate does its speed change? all properly belong to kinematics. In the next chapter, we will studydynamics, which delves more deeply into why objects move the way they do.

DISPLACEMENTDisplacement is an object's change in position. It's the vector that points from the object's initialposition to its final position, regardless of the path actually taken. Since displacement means changein position, it is generically denoted e.s, where e. denotes changeinand s means §patiallocation. (Theletter p is not used for position because it's reserved for another quantity; momentum.) If it's knownthat the displacement is horizontal, then it can be called e.x; if the displacement is vertical, then it'se.y. The magnitude of this vector is the net distance traveled and, sometimes, the word displacementrefers just to this scalar quantity. Since a distance is being measured, the SI unit for displacement isthe meter: [e. s] = m.

KINEMATICS . 13

: -, .

, -.,

Example 2.1 A rock is thrown straight upward from the edge of a 30 mcliff, rising 10 m then falling all the way down to the base of the cliff.Findthe rock's displacement.

: ~ ! .~.

Solution. Displacement only refers to the object's initial position and final position, not the details ofits journey. Since the rock started on the edge of the cliff and ended up on the ground 30 m below, itsdisplacement is 30 m, downward.

~....,

f::. .

Example 2.2 An infant crawls 5 m east, then 3 m north, then 1 m east.Find the magnitude of the infant's displacement. '

Solution. Although the infant crawled a totaldistance of 5 +3 + 1 = 9 m, this is not the displacement,which is merely the net distance traveled.

.-.=-.finish~

,

i Liy= 3 m,,

star t __ uuhu.h.mum.m.mm.OLix=5m +1m= 6m

Using the Pythagorean Theorem, we can calculate that the magnitude of the displacement is

I

-, :", Example 2.3 In a track-and-field event, an athlete runs exactly oncearound an oval track, a total distance of 500 m. Find the runner's dis­placement for the race.

".

Solution. If the runner returns to the same position from which she left, then her displacement iszero.

start-finish

The total distance covered is 500 m, but the net distance-the displacement-is O.

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SPEED AND VELOCITYWhen we're in a moving car, the speedometer tells us how fast we' re going; it gives us our speed. Butwhat does it mean to have a speed of say, 10m/s ?It means that we're covering a distance of 10 metersevery second . By definition, average speed is the ratio of the total distance traveled to the timerequired to cover that distance:

dtotal distance

average spee . "time

t he car's speedometer doesn't care in what direction the car is moving (as long as the wheels aremoving forward). You could be driving north, south, east, west, whatever; the speedometer wouldmake no distinction. 55 miles per hour, north. and 55 miles per hour, east register the same on thespeedometer: 55 miles per hour. Speed is a scalar.

Howev er, we will also need to includ e direction in our descriptions of motion. We just learnedabout displacement, which takes both distance (net distan ce) and direction into account. The singleconcept that embodies both speed and direction is called velocity, and the definition of averagevelocity is:

displacementaverage velocity

time

_ ~s'1= -

M

(The bar over the v means average.) Because ~s is a vector, v is also a vector, and because M is apositive scalar, the direction of v is the same as the direction of ~s. The magnitude of the velocityvector is called the object's speed, and is expressed in units of meters per second (m/s).

Notice the distinction between speed and velocity. In everyday language, they'r e often usedinterchangeably. However, in physics, speedand velocity are technical terms whose definitions are notthe same. Velocity is speed plus direction?

Example 2.4 If the infant in Example 2.2completes his journey in 20sec, .find the magnitude of his average velocity.

Solu tion. Since the displacement is 6.7 m, the magnitude of his average velocity is

v=Ss] M = (6.7 m)/ (20 s) = 0.34ml s

.' Example 2.5 Assume that the runner in Example 2.3 completes the racein 1 minute and 18 seconds. Find her average speed and the magnitudeof her average velocity.

.~ .

Solution. Average speed is total distance divided byelapsed time. Since the length of the track is 500m, therunner's average speed was (500 m)/(78 s) = 6.4 m/ s. However, since her displacement was zero, heraverage velocity was zero also: v= ~s l ~t =(0 m)/ (78 s) =0 m/s.

I Technical note;Themagnitude of the velocity is thespeed.However (and this is perhaps a bit unfortunateand can beconfusmgf themagnitude of the average velocity is notcalled theaverage speed. Average speedis defined as the totaldistance traveled divided by the elapsed time. On the other hand, the magnitude of the average velocity is the netdistance traveled divided by the elapsed time.

KINEMATICS . 15

Example 2.6 Is it possibleto move-with constantspeed but not constant. velocity? Is it possible to move with constant velocity but not constant" ': speed? :

Solution.Theanswerto the firstquestion isyes. Forexample,if yousetyourcar'scruise control at55 milesper hourbut turn thesteeringwheel. to follow a curvedsection of road, thenthedirection ofyourvelocity .changes (which meansyour velocity is not constant), even though your speed doesn't change.

Theanswerto thesecond question isno.Velocity meansspeedand direction; if thevelocity is constant,then that means bothspeed and direction are constant. If speed wereto change, thenthe velocity vector'smagnitude would change (by definition), which immediately implies that the vector changes.

ACCELERATIONWhenyou step on the gas pedal in your car, the car's speed increases; step on the brake and the car'sspeed decreases. Turn the wheel, and the car's direction of motion changes. In all of these cases, the .velocity changes. To describe this change in velocity, we need a new term: acceleration. In the.sameway that velocitymeasures the rate-of-ehange of an object'sposition, acceleration measures the rate­of-change of an object's velocity. An object's average acceleration is defined as follows:

average acceleration = change in velocitytime

_ !:J.va=-

l it

Theunits ofacceleration are meters per second, per second: fa] = m/s' . Because !J.v is a vector, a is

alsoa vector; and because !J.t is a positive scalar, the directionof a is the sameas the directionof !J.v.Furthermore, if we take an object'soriginal directionofmotion to be positive, then an increase in

speed corresponds to a positive acceleration, while a decrease in speed corresponds to a negativeacceleration (deceleration),

Notice that an objectcan accelerate even if its speed doesn't change. (Again, it's a matter of notallowing the everyday usage of the word accelerate to interfere with its technical, physics usage.) Thisis because acceleration depends on !J.v, and the velocity vector v changes if (l) speed changes, or (2)direction changes, or (3) both speed and direction change. For instance, a car traveling around acircularracetrack is constantly accelerating even if the car's speed is constant, because the directionofthe car's velocity vector is constantly changing.

...... Example 2.7 A car is traveling in a straight line along a highway at aconstant speed of 80miles per hour for10 seconds. Find its acceleration.

Solution.Since the car is travelingat a constant velocity, its acceleration is zero.If there's no changein velocity, then there's no acceleration.

' r:~T_,?j:':r:? ~~.-.r. · :ty.: ~ V

·":::./'-'.L ':' Example 2.8A car is traveling along a straight highway, ala speed of 20~ : . : . :.: ;'.-;; m/s. The driver steps on the gas pedal and, 3 seconds later, the cars{,'." -.>.'''-'i~ speed is 32 m/ s. Find its a'lerage acceleration. .,' . . ' ...''; ~ "i;'~ ·kl;~". 0'''; , . -4~ . ~: . ~.. • " " . • ",.1. • • .,,-~':' i: :.:. ...~.... ""-c-"""""''''''''' .!'Il!n

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Solution. Assuming that the direction of the velocity doesn' t change, it's simply a matter of dividingthe change in velocity, 32 m/s - 20 m/ s = 12 mis, by the time interval during which the changeoccurred: Ii = I'. vl M =(12 m/ s) I (3 s) =4 m/ s' .

. ,Example 2.9 Spotting a police car ahead, the driver of the car in theprevious example slows from 32 mls to 20 ml s in 2 sec. Find the car's .'.average acceleration.

, ".<

Solution. Dividing the change in velocity, 20 m/ s - 32 m/ s = - 12 mi s, by the time interval duringwhich the change occurred, 2 s, gives us it = I'.vI M =(- 12 ml s) I (2 s) =-6 ml S2 The negative signhere means that the direction of the acceleration is oppos ite the direction of the velocity, which .describes slowing down.

UNIFORMLY-ACCELERATED MOTION AND THEBIGFIVEThe simplest type of motion to analyze is motion in which the acceleration is constant(possibly equalto zero). Although true uniform acceleration is rarely achieved in the real world, many commonmotions are governed by approximately constant acceleration and, in these cases, the kinematics ofuniformly accelerated motion provide a pretty good description of wha t's happenin g. Notice that ifthe acceleration is constant, then taking an avera ge yields nothing new, so Ii = a.

Another restriction that will make our analysis easier is to only consider motion that takes placealong a straight line. In these cases, there are only two possible directions of motion . One is positive,and the opposite direction is negative. Most of the quantities we've been dealing with-displace­ment, velocity, and acceleration-are vectors, which means that they include both a magnitude anda direction. With straight-line motion, direction can be specified simply by attaching a + or - sign tothe magnitude of the quantity. Therefore, although we will often abandon the use of bold letters todenote the vector quantities of displacement, velocity, and acceleration, the fact that these quantitiesinclude direction will still be indicated by a positive or negative sign.

Let's review the quantities we've seen so far. The fundamental quantities are displacement (As),velocity (v), and acceleration (a). Acceleration is a change in velocity, from an initial velocity (vio r v~to a final velocity (v, or simply v-with no subscript). And, finally, the motion takes place duringsome elapsed time interval, M. Therefore, we have five kinematics quantities: I'.s, v" v, a, and M.

These five quantities are related by a group of five equations that we call TheBig Five.They workin cases where acceleration is uniform, which are the cases we' re considering.

Variable that's missing

Big Five #1: S s > li M a

Big Five #2: I'. v= aM I'.s

Big Five #3: I'.s =val'.t+.!a(M)' v2

Big Five #4: 1 ( z!!.s= vM- - a !!.t) va2

Big Five #5: v' =V; +2al'.s M

KINEMATICS . 11

Some of these equations will be provided on the non-multiple-choice section of the AP Physicsexam, but they will not be provided in the multiple-choice section. In BigFive #1, the average velocity

is simply the average of the initial velocity and the final velocity: 'if=1.(vo+v). (This is a consequence' .

of the fact that the acceleration is constant.) Also, if we agree to sta?t our clocks at time ti= 0, then

M =tf - ti =t - 0 = i, so we can just write t instead of M in the first four equations. This simplificationin nota tion makes these equations a little easier to write down.

Each of the Big Five equations is missing one of the five kinematic quantities. The way you decidewhich equation to use when solving a problem is to determine which of the kinematic quantities ismissing from the problem-that is, which quantity is neither given nor asked for-and then use theequation that doesn't contain that variable. For example, if the problem never mentions the finalvelocity-v is neither given nor asked for-the equation that will work is the one that's missing v.That's Big Five #3.

Big Five #1 and #2 are simply the definitions of v and ii written in forms that don' t involvefractions. The other Big Five equations can be derived from these two definitions and the equation

'if=~(vo +v), using a bit of algebra.

-::" .,.,.,..~ . ' ...... -"""" .~ . _ -. . ~ -. . .:~ 'Co.,""'~. "'"""",~:·:;~·...'jl};·~,:~:: Ex~mple 2:10 An object with an initi~l velocity of 4m!smo~es alon~ a i:</;~'}; ',..: ':: '. ':; '. . straight axis under constant acceleration. Three seconds later, Its velocity " ', '" . .: " .c.::.;.:e,;'.( . is 14 m/s. How far did it travel during this time? :,. . c. ,::' ,'. :'.~ :"' ~- " . ~:cc '.:; , . '. . ' ~ .

Solution. We're given va' M, and v, and we're asked for Ll.s. So a is missing; it isn't given and it isn'tasked for, and we use Big Five #1;

1 . 1LI.s =iiLl.t =2:(vo+v)t =2:(4 mls + 14 m/ s)(3 s) =27 m

Note: This last equation could also have been wri tten as

11 'LI.s ='ifLl.t =2: (vo+v)t =2:(4 + 14)(3) =27 m

That is, it's OK to leave off the uni ts in the middle of the calculation as long as you remember toinclude them inyour final answer. Leaving units off of your final answer will cost you points on the APexam.

.':~""7::: :.-,~:: ,!,,:~:_, ,; ' ••' W'a- "c{ ~ .....':" ..... • " \ •• ' '''" • • • ."., .~ •• ,,:; . ....~::.~:::t~.: . '

.:.:.r.'.";:...;•."::i: EXample 2.11 A car that's initially traveling at 10 ml s accelerates uni- : :.".:: ·:. :,:.-i:' .':!

:~·.1: '::' :::._.:.; formly for 4 seconds at a rate of 2 mis', in a straight line. How far does : ' ~; .... ::.:'"

f{~ti!. ;?i~.~.~L :~~_:a~ ~~~.eI d~~g this}~:-~ ,: .../'{~:~' !~ :'>:'~': ~1r1~·}::::~S~·"::·:~ :: ~::.~~.~:;'~~."'~~'~::~

18 . CRACKI NG THEAP : PHYSICS EXAM

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Solution. We're given va' t, and a, and we're asked for ·/1s. Sov is missing; it isn' t given and it isn'tasked for, and we use Big Five #3: '

l1s =voM+.!a(/1t)' =(10 m / s)(4 S)+ .! (2 m / s' )(4 s)' =56 m '. 2 2 '

~ ."· tf· .. - . .- " ' ..

»;:':~0:,:\'X Example z.iz Arock is dropped off a cliff that's 80 m high. If it strikes :\',,<:,"Y:\',:,i''':··'\.'::,';',:·,':. the ground with an impact velocity of 40 mis, what acceleration did it .j;\' ."~, }.:~

r~'t\h\'::':X~t:Xp:rien: e,during its ~~cent1 ~: " , . ~ ,. ;"'::;:~<:,:,,,::--.'::'.';.; .::.'~C::)':':" ':/'3~

Solution. If something is dropped, then that means it has no initial velocity: va = O. So, we're given va'S s, and v, and we're asked for a. Since t is missing, we use Big Five #5:

v' = v~ +2al1s => v' =2al1s (since Vo =0)

a = v' J 40 m/ s)' 10m / s'. 2l1s 2(80 m)

Notice that since a has the same sign as S», the acceleration vector points in the same direction as thedisplacement vector , This makes sense here, since the object moves downward and the accelerationit experiences is due to gravity, which also points downward.

KINEMATICS WITH GRAPHSSo far, we have dealt with kinematics problems algebraically, but you should also be able to handlekinematics questions in which information is 'given graphically. The two most popula r graphs inkinematics are position-vs.-time graphs and velocity-vs-time graphs . For example, consider anobject that's moving along an axis in such a way that its position x as a function of time t is given bythe following position-vs-time graph:

10

~ 5S~

'"~ 0 time (s)0

'.0 4 5 6,~

en0 -50..

-10

What does this graph tell us? It says that at lime t = 0, the object was at position x = 0, Then, in thenext two seconds, its position change d from x =0 to x =10 m.

KINE MATICS . 19

W3583

6

end.'" ---. --. --. -----. ------. . ---. .

begin. --- ---------- ---- · ·> •

I-10

I-5

o

C)Io

I5

I10

Then, at time t = 2 s, it reversed dir ection and headed back toward its start ing point, reachingx = 0 at time t = 3 s, and continued, reaching position x = -5 m at time t = 3.5 s. Then the objectremained at this position, x = -5 m, at least through time t '"6 s. Notice how economically the graphembodies all this information!

We can also determine the object' s average velocity (and average speed) during particular timeintervals. For example, its average velocity from time t =0 to time t =2 s is equal to the distancetraveled , 10- 0 =10 m, divided by the elapsed time, 2 s:

v=t.x (10 - 0) m 5 m /st:.t (2-0)s

Notice, however, that the ratio that defines the average velocity, Sx ] jI" t, also defines the slope ofthe x vs, t graph. Therefore, we know the following important fact:

The slope of a position-vs.-time graph gives the velocity.

What was the average velocity from time t =2 s to time t =3.5 s? The slope of the line segmentjoining the point (t, x) =(2 s, 10 m) to the point (t, r) =(3.5 s, -5 m) is

- Sx (-5 - 10) m 10 Iv =- = - m St:.t (3.5-2) s

The fact that v is .negative tells us that the object's displacement was negative during this timeinterval; that is, it moved in the negative x direction. The fact that v is negative agrees with theobservation that the slope of a line that falls to the right is negative. What is the object's averagevelocity from time t = 3.5 s to time t = 6 s? Since the line segment from t = 3.5 s to t = 6 s is horizontal,its slope is zero, which implies that the average.velocity is zero, but we can also figure this out fromlooking at the gra ph, since the object's position did not change during that time.

Finally, let's figure out the object's average velocity and average speed for its entire journey (fromt =0 to t =6 s). The average velocity is

v- t.x _(-5-0) m -0.83 m / s- t:.t - (6- 0) s

20 • CRACK IN GTHE AP : PHYSI CS EXAM

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This is the slope of the imagined line segment that joins the point (I, x) = (0s, 0 m) to the point(I, x) = (6s, - 5 m), Theaverage speed is the total distance traveled by theobjectdivided by theelapsedtime. In this case, notice that the object traveled 10 m in the first 2 s, then 15 m (albeit backwards) inthe next 1.5s; it covered no additional distance from t =3.5 s to t =6 s. Therefore, the total distancetraveled by the object is d =10 +15 =25 m, which took 6 s, so .

d 25maverage speed = - =--= 4.2m Is

. . t'1t 6s

Let's next consider an object moving along a straight axis in such a way that its velocity, v, as afunction of time, t, is given by the following velocity-vs-time graph:

10

~

'" 5<,

S~

'" time (s)i- 0. ~ 4 5 6u0- -5'";>

-10

What does this graph tell us? It says that, at time t =0, the object'svelocitywas v =O. Over the firsttwo seconds, its velocity increased steadily to 10 ml s. At time t = 2 s, the velocity then began todecrease (eventually becoming v = 0, at tiine t = 3 s), The velocity then became negative after t = 3 s,reaching v = -5 m/ s at time t = 3.5 s. From t = 3.5 s on, the velocity remained a steady -5 m/s.

What can we ask about this motion? First, the fact that the velocity changed from t =0 to t =2 stells us that the object accelerated. The acceleration during this time was

_l1v (10-0)m/s 5 12a-- m sI1t (2- 0) s

Notice, however, that the ratio that defines the acceleration, I1vll1l, also defines the slope of thev vs. t graph. Therefore,

The slope of a velocity-vs.-time graph gives the acceleration.

What was the acceleration from time t =2 s to time t =3.5s? The slope of the line segment joiningthe point (I, v) =(2 s, 10 m/s ) to the point (t, v) =(3.5 s, -5 m/s ) is

a=l1v J-5-10)m / s . 10 m / s2t'1t (3.5- 2) s

The fact that a is negative tells us that the object's velocitychange was negative during this timeinterval; that is, the object accelerated in the negative direction. In fact, after time t = 3 s, the velocitybecamemore negative, indicating that the direction of motion was negative at increasing speed.What,

KINEM ATICS . 21

is the object's acceleration from time t = 3.5 s to time t = 6 s? Since the line segment from t = 3.5 s tot = 6 s is horizontal, its slope is zero, which implies that the acceleration is zero, but you can also seethis from looking at the graph; the object's velocity did not change during this time interval.

Another question can be asked when a velocity-vs.-time graph is given:How far did the object travelduring a particular time interval? For example, let's figure out the displacement of the object from timet = 4 s to time t = 6 s. During this time interval, the velocity was a constant -5 mis, so the displacementwas !!.X = vM = (-5 m/s)(2 s) = -10 m.

Geometrically, we've det ermined the area between the graph and the horizontal axis. After all, thearea of a rectangle is base x height and, for the shaded rectangle shown below, the base is M, and theheight is v. So, base x height equals M x v, which is displacement.

10

u;-S<,

E'".s 0

. ~

u0~ -5'";>

-10

1

4 5 6

2

vM

Signed area = displacement

We say signed area because regions below the horizon tal axis are negative quantities (since theobject's velocity is nega tive, its displacement is negative). Therefore, counting areas above thehorizontal axis as positive and areas below the horizontal axis as negative, we can make the followingclaim:

Given a velocity-vs-time graph, the area between the graph and the t axis equals the object'sdisplacement.

What is the object's displacement from time t = 0 to t = 3 s? Using the fact that displacement is thearea bounded by the velocity graph, we figure out the area of the triangle shown below:

10

~

height = 10 rn / s"' 5<,

E 4 5 6'" time (s).s 0

. ~ 1 2u0

- 5 v~

'" base= 3s;>

- 10

Since the area of a triangle is 0 /2) x base x height, we find that Sx = ~(3 s)(10 m/ s) = 15 m.

22 • CRAC KING THE AP : PHY SICS EXAM

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A NOTE ABOUT CALCULUSThe use of graphs for solving kinematics questions .provides a link to some basic definitions incalculus . The slope of a curve is the geometric definition of the 'derivative of a function. Since theslope of a position-vs-time graph gives the velocity, the derivative of a position function gives thevelocity. That is, given an equation of the form x = x(l) for the position x of an object as a function oftime t, the derivative of x(I)-with respect to I-gives the object's velocity; v (t):

v(l) = dxdl

Often, a derivative with respect to time is denoted by a dot above the quanti ty, so an alternativenotation for the equation above is v(l) =x(I ).

Next, since the slope of a velocity-vs.-time graph gives the acceleration, the derivative of a velocityfunction gives the acceleration.That is,given an equation of the form v = v(t) for the velocityv of an objectas a function of time t, the derivative of v(I)- with respect to t-gives the object's acceleration, a(I):

dva(t) = dt

Combining this equation with the previous one, we see that the acceleration is the secondderivative of the position:

Using the dot notation, these last two equations may be written as a(t) = iJ (I) and a(l ) = x(I),respectively. .

The area bounded by a curve and the horizontal axis is the geometric definition of the definiteintegral of a function . Since the area bounded by a velocity-vs.-time graph gives the displacement,the definite integral of a velocity function gives the displacement. That is, given the equation v =v(l )

for the velocity v of an object as a function of time t, the integral of v( l) from time I = I, to time t = I,equals the displacement during this time interval:

rdisplacement = J, v(l)dl. ~

For the velocity-vs.-time graph examined above, the slope of the line segment from(I, v) = (0, 0) to (I, v) = (2, 10) can be described by the equation v(t ) = 51, and the line segment from (t,v) = (2, 10) to (I, v) = (3, 0) can be described by the equation v(t) = -101+30 (with i in seconds and v

in ml s in both equations). Therefore, the acceleration of the object from time I =ato t =2 s is

dv d ( ) 2a=- =- 51 =5 mlsdt di

as we found above.

KIN EMATICS . 23

Also, the displacement of the object from time I =0 to I =3 s is the value of the integral from I =0 toI =3 s, which we can get by adding the integral of v(l) =51 from I =0 to I =2 s to the integral ofv(l) =- 101+ 30 from I =2 s to I =3 s:

)

displacement = J~5I dl+ J; (- 101 +30)dl

= [%12J: + [-51

2+301]:

= %(22.; 02

) +[(-5 32 + 3~ ' 3) _(_5 .22 +30 .2)]

=10 + [45-40]

= 15 m

-. '.

--'.'-..• : '" :;: •• _. ..·1

~.. ~ .

D

A

J-- - ----'''f---- ---- tim e

At which point (A, B, C, D, or E) is the magnitude of the acceleration thegreatest? How would you answer this same question if the graph shownwere a position-vs-time graph? .. '

.- ..

which also agrees with the value computed above.

.;~'f::: ';"i:~. 'Example 2.13 The velocity of an object as a function of time is given by.. .. . :.: ." .... the following graph: ... .. _... .: ; .." : . '':'.' '.

, . .... k1

Solution, The acceleration is the slope of the velocity-vs.-time graph. Although this graph is notcomposed of straight lines, the concept of slope still applies; at each point, the slope of the curve is theslope of the tangent line to the curve. The slope is essentially zero at Points A and D (where the curveis flat), small and positive at B, and small and negative at E. The slope at Point C is large and positive,so this is where the object's acceleration is the greatest.

If the gra ph shown were a position-vs-time graph, then the slope would be the velocity. The slopeof the given graph starts at zero (around Point A), slowly increases to a small positive value at B,continues to slowly increase to a large positive value at C, then, at around Point D, this large positiveslope decreases quickly to zero. Of the points designated on the graph, Point D is the location of thegreatest slope change, which means that this is the point of the greatest velocity change . Therefore,this is the point at which the magnitude of the acceleration is greatest.

24 • CRACKI NGTHEAP: PHY SIC S EXAM

FREE FALLThe simplest real-life example of motion under pretty constant acceleration is the motion of objectsin the earth's gravitational field, near the surface of the earth and ignoring any effects due to the air(mainly air resistance). With these effects ignored, an object can fall freely, that is, it can fall experienc­ing only accelerat ion due to gravity. Near the surface of the earth, the gravita tional acceleration hasa constant magnitude of about 9.8 m/ s'; this quantity is denoted g (for gravitational acceleration).'And, of course, the gravitational acceleration vector, g, points downward.

_Since the acceleration is constant, we can use the Big Five with a replaced by +g or -g. To decidewhich of these two values to use for a, make a decision at the beginning of your calculations whetherto call "down" the positive direction or the negative direction. If you call "down" the positivedirection, then a =+g. If you call "down" the negative direction, then a=-g. It might make thingseasier if you always take the direction of the object's displacement as positive .

In each of the following examples, we'll ignore effects due to the air .

\o: ...:.~-;..., 0,\ Ex~pie 2.14 A rock is drop~ed from a~ 80-mete~ cliff H;w i;g do~s '>~.:t'\..;;.:,,:;:·:i·,··::·.. ···:. ". '.' ittake to reach the ground? .. '.. '., ' ,:., :-- '... ..',: ""..'. '; ,'" .:; \ .

••• - •• ~~ :...' . , ' . ~ -. " > ~- - • • ' • ",-' • : ~

Solution. Since the rock's displacement is down, we call down the positive direction, so a = +g. We'regiven va' !J.s, and a, and asked for t. So v is missing; it isn't given and it isn' t asked for, and we use BigFive #3;

2(+80m)

+9.8 m l S2

)1 2 1 2 ( , )!J.s = vot +-at ~!J.s=-at since Vo=02 2

t= F~

=~~~ = 4.0 s

, ~ .Example 2.15 A baseball is thrown straight upward with an initial speed

"'. of25 m/ s. How high will it go? .': ·i,· . .'." ' ,."~. . ~.

~. ..

Solution. Since here we are concerned only with the baseball' s upward motion, the baseball'sdisplacement is up, so we call up the positive dire ction. Therefore, down is the negative direction, soa= -g. The ball's velocity drops to zero at the instant the ball reaches its highest point, so we're givena, va' and v, and asked for !J. s. Since t is missing, we use Big Five #5:

(+25 m IS)'2(-9 .8 m / s' )

v' = v~ +2a!J.s ~ 2a!J.s = -v~ (because v = 0)2

Ss = _Vo2a

v2= 0 =

- 2(-g) 32 m

'On the AP Exam, you may use g =10 mi s' as a simple approximation to g =9.8 m/ s'.

KIN EM ATICS . 2S

J;:;--:r:r::-';, ",ttl U ... ., " ~

;~gJ;~ ;.:;:;:::;::>Example 2.16 One second after being thrown straight down, an object is , .. ,~;.'F:f ,t); falling with a speed of,20m/s. flow fast will it be falling 2secondslater? ';• •>;!t ..;., -.. ~ A. ' . ' "'., :.,. , ,", -., ..

)

.Solution. Call down the positive direction, so a = +g and Vo= +20 m/ s. We're given v.. a, and t, andasked for v. Since /!,S is missing, we use Big Five #2:

S » =atv - va = at

v = Vo+ at = (+20 m/s) + (+9.8 m/s')(2 s) = 40 m / s

. . '::' Example 2.17 Ifan object is thrown straight upward with an initial speed.,~:. ,~, -.: " of 8 mls and takes 3 seconds to strike the ground, from what height was . ''~'jf.:: .. ':;, the object thrown? '" • :' " . ,,: ,'. ,'.

"'1 .:.; _0: .._.";_-:" •• .... ". ~....! -- ""' .. ' ....__

Solution, The figure below shows that the displacement is down, so we call down the 'positivedirection. Therefore, a = +g and Vo= -8 ml s (because up is the negative direction). We're given a. v..and i, and we need to find /!'s. Since v is missing, we use Big Five #3:

In this example,the magnitude of thedisplacement equ als

the height from whichthe objec t w as thrown,

,,

,,

,,,,,,

,,f

1 2/!,s = vot + '2at

= (-8)(3) + t(+9.8)(32

)

= 20 m

J

PROJECTILE MOTIONIn general, an object that moves near the surface of the earth will not follow a straight-line pa th (forexample, a baseball hit by a bat, a golf ball stru ck by a club, or a tenni s ball hit from the baseline). Ifwe launch an object at an angle other than straight upward and consider only the effect of accelera­tion due to gravity, then the object will travel along a parabolic trajectory,

26 • CRACKING THE AP : PHYSICS E.XAM

parabolic at the top,trajec~?ry v y = 0

"" ---0 -- ,~"~ ,. / / , --' -

oVot:&0,;, launch angle

..,,

To simplify the analysis of parabolic mo tion, we analyze the horizontal and vertical motions sepa­rately, using The Big Five. This is the key to doing projectile motio n problems. Calling down thenegative direction, we have

Horizontal motion : Vertical motion :

1III =v",t t.y =vOyt +'2(-gW

VX=v", (constant ') v

y=vOy+ (-g)t

ax=0 ay=-g

The quantity VOx' which is the horizontal (or x) component of the initial velocity, is equal to Vocos eO'where eoIS the launch angle, the angle that the initial velocity vector, v0' makes with the horizontal. .

Similarly, the quantity VOy' the vertical (or y) component of the initial velocity, is equal to Vosin eo'

EX~;I~2.18 Anobjectis thr;~horizontally~ilianinifuIspeedofl0 ';: . __ .", .",

m/s .How far will it drop in 4 seconds? .. "'.:':. ,. . .", .: '/':' ': >.'c'.. :'. ,.~, - ' " . c.:.." , ' ., :~_ , ...•.•~. ,0;

Solution. The first step is to decide whe ther this is a horizontal question or a vertical question, since youmus t consider these motions separately. The question How far will it drop? is a vertical question, so theset of equations we will consider are those listed above under vertical motion. Next, How far...? impliesthat we will use the first of the vertical-motion equations, the one that gives vertical displacement, 6.y.

Now, since the object is thrown horizontally, there is no vertical component to its initial velocityvector vrY that is, VOy = O. Therefore,

,,,

•

6.Y = VOy t + ~(_g)t2 -7 t.y =~(_g)t2 (because VOy=0)

= ~(-9.8)(42 )

=- 78 m

KIN EMATICS . 27

The fact that !J.y is negative means that the displacement is down. Also, notice that the informationgiven about v" is irrelevant to the question.

... '~d • • "

Example 2.19 From a height of 100m, a ball is thrown horizontally withan initial speed of 15 m / s. How far does it travel horizontally in the first2 seconds? .: .

Solution. The question, How far does it travel horizontally...? immediately tells us that we should usethe first of the horizontal -motion equations listed above:

!J.x =v" t =(15m/ s)(2 s) =30 m.The information that the initial vertical position is 100m above the ground is irrelevant (except for thefact tha t it's high enough that the ball doesn't strike the ground before the two seconds have elapsed) .

Example 2.20 A projectile is traveling 'in a parabolic path for a total of 6 ' ; : .,,--, : ; seconds. How does its horizontal velocity 1 s after launch compare to its ;.': ,: .

horizontal velocity 4 s afterlaunch? . .... .; .... .. ... _.

Solu tion. The only acceleration experienced by the projectile is due to gravity, which is purelyvertical, so that there is no horizontal acceleration . If there 's no horizontal acceleration, then thehorizonta l velocity cannot change during flight, and the projectile's horizontal velocity 1 s after it'slaunched is the same as its horizontal velocity 3 s later.

Example 2.21An object is projected upward with a 30° launch angle andan initial speed of39 m/ s. How long will it take for the object to reach the

. top of its trajectory? How high is this?

Solution. When the projectile reaches the top of its trajectory, its velocity vector is momentarilyhorizontal; that is, v,= O. Using the vertical-motion equation for v" we can set it equal to 0 and solvefor t:

Vy':O ~ VOy + (-g )t=0

_ VOy _ vosin8o _ (39 m /s)sin30' 2 .t- 8 - g - 9.8 m/s2 ~ s

At this time, the projectile's vertical displacement is

!J.y = voyt+ ~(_g)t2 = (vosin 8o)t + ~(_g)t22 . 2

= [(39 m /s) sin 30·](2s)+~(-9.8 m / s2)(2 S) 2, 2

=19 m

. Example 2.22An object is projected upward with a 30° launch angle andan initial speed of 59 m/ s. For how many seconds will it be in the air?How far will it travel horizontally? .

28 • CRACKING THE AP: PHYSICS EXAM

)

. Solution. The total time the object spends in the air is equal to twice the time required to reach thetop of the trajectory (because the parabola is symmetrical). So, as we did in the previous example, wefind the time required to reach the top by setting v y equal to 0, and now double that amount of time:

t =VOy =V o sineog g

(59 m /s)sin30'

9.8 m/ S23 s

Therefore, the total flight time (that is, up and down) is T = 2t = 2 x (3 s) = 6 s.Now, using the first horizontal-motion 'equation, we can calculate the horizontal displacement

after 6 seconds:

III =vo.T = (vocoseo)T =[(59 m / s) cos 30·](6 s) = 310 m

By the way, the full horizontal displacement of a projectile is called the projectile'~ range .

[C] KINEMATICS WITH CALCULUSAs we mentioned earlier, velocity is the time derivative of position, and acceleration is the timederivative of velocity. Equivalently, the integral of acceleration is velocity, and the integral of velocityis position. These relationships can be summarized in the following diagram:

dilierentiale ) accelerationposition

x(t)

differentiate

integrate

) velocity

v(t) integrate a(t)

,.< . ,. ' ,, ': ..

~ . '

" t-; ' . , Example 2,23 The position of an object (measured in meters from theorigin, where x = 0) moving along a straight line is given as a function oftime 1(measured in seconds) by the equation x(l ) = 41' - 61- 40, Find

(a) its velocity at time I,(b) it's acceleration at time I, ' . ..' '~, ',.(c) the time at which the object is at the origin, and

, < . ,',:" .'" ,:., (d) the object's velocity and acceleration at the time calculated in (c),~~..'~:. ~'.·"~:'i;;;' ,: ~'':':', . . , ~ , . ;,,; .," , "_~"":;",""""""i-' _~_"'._""i.;~"''''i'''.~

Solution,

(a) The velocity as a function of 1is the derivative of the position function:

v(l) =x(l)= :1 (412 - 61- 40) =8t-6 (in m / s)

(b) The acceleration as a function of 1is the derivative of the velocity function:

a(l)= v(l)= ~(81 -6) =8 m/s2

dl

KINEMA TICS . 29

•

Notice that the acceleration is constant (because it doesn't depend on I).

(c) The object is at the origin when xU) is equal to 0; that is, when

41' - 61-40 = 0

2(21' - 31 - 20) = 0

2(21 + 5)(t'- 4) = 0

5I =-- or 4

2Disregarding the negative value for I, we can say that the object passes through the origin at

I = 4 s. At this time, the object's velocity is

v(4)=v(I)[t=4=(81- 6t 4=8·4 -6=26 mls

The object's acceleration is a constant 8 mis' throughout its motion, so, in particular, at I =4 s, theacceleration is 8 m/ s' .

..f·',' , .:.'~:' Example 2.24 An'-object is moving alo~g' a straight axis witl~a co~stant : ':' :',",.;"--: :.,.:; . - :•. , acceleration of 3 m/s' . If the object's initial velocity is 2 m/ sand its initial - <.s :..' .,'':,':;.' "... ': position is x =3 m, where will this object be at time I =4 s? . . '. .

Solu tion.

By integrating the acceleration with respect to time, we find the velocity function; then, integratingthis with respect to time, we find the position function.

v(I)=Ja(l) dl= J3 dl =31+c

To get a value for c, the arbitrary constant of integration, we use the fact that the object's initialvelocity was 2 ml s: .

V(O) =2 => (31 +c)[t=o=2 => c= 2

(c is just v/Y the initial velocity.) So the velocity function is given by the equation v(l) = 31 + 2(in m/ s). Integratin g again gives the position function: .

X(I)=f V(I ) dl= J(31+2) dl = ~ I'+ 21 +C'. . 2

To evaluate c', the arbit rary constant in this integration, we use the fact tha t the object's initialposition was x = 3 m: .

(Note that c' is just xc' the initial position.) So the velocity function is given by the equation3 .

x(l) = 2: 1' +21+3 (in m)

and the object's position at time I = 4 s can be determined:

x(4) ss x(I)lt=4 =(~I ' + 21+ 3)1 =~. 4' + 2 ·4 + 3 = 35 m2 1=4 2

30 • CRAC KI NG THE AP : PH YSICS EX AM

)

)

CHAPTER 2 REVIEW QUESTIONS

SEatON I: MUlTIPLECH OICE

1. An object that's moving with constantspeed travels once around a circular path.Whic h of th e following is / ar e tr ue con­cerning this motion?

I. The displacement is zero.II . The aver age speed is ze ro.

III. The acceleration is zero.

(A) I only(B) I an d II only(C) I and III on ly(0 ) III only(E) II and III only

2. At time I = I" an object's velocity is givenby the vector v I sh own below :

A short time later, at I = I" the object'svelocity is the vector v2:

If v, = VI' which one of th e followingvectors best illustrates the object's averageacceleratio n between t:= tt and t = t2?

(A ) »:"

(B) I(C) \

(O)~

(E)~

3. Which of the following is / ar e true?

I. If an object's acceleration isco nsta nt, then it must move in astraight line.

II. If an object's acceleratio n is zero,then its speed must rem ainconstant.

III. If an object's sp eed remainsconstant, then its accelerationmust be zero.

(A) I and II only(B) I and III only(C) II only(0) III only(E) II and !II only

4. A baseball is thrown straight upward.What is the ba ll's acceleration at itshi ghest point?

(A) 0

(B)1- g, downward2

(C) s-downward

(0)1-g, u pward2

(E) g, upward

5. How long would ittake a car, startingfrom rest and acceleratin g uniformly in astraight line at 5 m/s', to cover a d istanceof 200 m?

(A) 9.0 s(B) 10.5 s(C ) 12.0 s(0) 15.5 s(E) 20.0 s

KINEMATI CS . 31

6. A rock is dropped off a cliff and st rikes 9. A stone is thrown horizontally with anthe ground with'an impact velocity of 30 initial speed of 30 m l s from a bridge. Findm / s. How high was the cliff? the stone's total speed when it enters the

(A) 15 m wa ter 4 seconds later. (Ignore air resis-

(B) 20m tance.)

(C) 30 m (A) 30m/s(D) 4Sm (B) 40 mls(E) 60 m (C) SOmis

(D) 60 mls

7. A stone is thrown horizontally with an(E) 70 m ls

ini tial speed of 10 m l s from a bridge. Ifair resistance could be ignored , how long 10. Which one of the following statements iswould it take the stone to strike th e water true.con cerning the motion of an ideal80 m below the bridge? projec tile launched at an an gle of 45° to

(A) 1 s the ho rizontal?

(B) 2s (A) The accele ration vecto r points(C) 4s opposite to the veloci ty vec tor on(D) 6s the way up and in th e same(E) 8 s direction as the velocity vector on

the way down.

8. A soccer ball, a t res t on th e ground , is(B) The speed at the top of the trajectory

is zero.kicked w ith an initial ve locity of 10 mls at

(C) The object's tot al speed rem ainsa launch angle of 30°. Calcula te its to tal constant"during th e en tire fligh t.flight time, assuming that air resistance is (D) The horizontal speed decreases on thenegligible. way up and increases on the way(A) 0.5 s down. ,(B) 1 s (E) The vertica l speed decreases on the(C) 1.7 s way u p an d increases on the way(D) 2s down.(E) 4s

32 • CRA C~ ING THE AP : PHYSICS EXAM

SEGION II: FREE RESPONSE

1. This question concerns the motion of a car on a straight track: the car's velocity as a functio nof time is plotted below.

20

ui 10<,

ED 0.~

1 2 3 4u0-'" - 10;>

-20

time (5)

7

)

(a) Describe what happened to the car at time t = 1 s.

(b) How does the car's average ve locity between time t =oand t = 1 s compare to itsav erage velocity between times t = 1 sand t = 5 s?

(c) What is the displacemen t of the car from time i =0 to time t =7 s?

(d ) Plot the car' s acceleration during this interval as a function of time.

(e) Plot the object's position during thi s interval as a func tion of time. Assume that the carbegins at s = O.

2. Consider a projectile moving in a parabolic trajectory under cons tant gravita tional accelera­tion. Its initial velocity has magnitude vo' and its launch an gle (with the horizontal) is e o' .

(a) Calc ul ate th e ma ximum height, H, of the projectile.

(b) Calculate the (ho riz ontal) range, R, of the p rojectile.

(c) For w hat value of e ow ill th e range be maximized?

(d) If 0 < h < H, compute the time that elapses betw een pa ssin g thro ugh the horizo n tal lineof height h in both directions (ascending and descending); th at is, compute the timerequired for the projectile 'to pass through the two points shown in this figure :

. . , ___ Ira jectory. : ~, ~

.« ,' ",•..:/ -------r·---_.:.,:~, ' ,, , ,

, , ', , ', , '

" :h "! b

KINEMA TICS . 33

3. A cannonball is sho t wi th an initial speed of 50 mls a t a launch an gle of 40° toward a castlewall 220 m away. The height of the wall is 30 m. Assume tha t effec ts d ue to the air are negli­gible .

(a) Show that the cannonball will str ike the castle wall.

(b) How long will it take for the cannonball to strike the wa ll?

(c) At what height above the base of the wall will the cannonba ll strike?

(d) Assuming that neither the location of the cannon nor the launch spe ed of the cannonballcan be varied-but the launch ang le can- determine the launch angles that will allowthe cannonball to just clea r the top of the castle wall.

4. [C] A particle moves along a straigh t axis in such a way that its acceleration a t time t is giv enby the equa tio n a(t) = 6t (rrr/s-). If the particle's initial velocity is 2 m l s and its initial positionis x :;:; 4 m , determine

(a) the time at which th e particle's velocity is 14 m i s, and

(b) the particle' s po sition at time t = 3 s.

34 • CR ACKING THE AP : PHYSICS EXAM

)

I1

)

SOLUTIONS TO THE

CHAPTER REVIEW

) QUESTIONS

CHAPTER 2 REVIEW QUESTIONS

SEGION I: MUlTIPLE CHOICE1. (A) Traveling once around a circular pa th means that the final position coincides with the

initial position. Therefore, the displacement is zero. The average speed, which is total distancetraveled divided by elapsed time, cannot be zero. And since the velocity changed (because itsdirection changed), there was a nonzero acceleration. Therefore, only Statement I is true .

2. (C) By definition, a=f'"v I f'"t. We determine f'" v =v, - VI =v, + (-VI) geometrically as follows:

-v• 1

Since f',.t is a positive scalar, the direc tion of a is the same as the direction of f'" v, which isdisplayed above; choice (C) is best.

3. (C) Statement I is false since a projectile experiencing only the constant acceleration due togravity can travel in a parabolic trajectory. Statement II is true: Zero acceleration means nochange in speed (or direction). Statement ill is false: An object whose speed remains constantbut whose velocity vector is changing dir ection is accelerating.

4. (C) The baseball is still under the influence of Earth 's gravity. Its acceleration throughout theentire flight is constant, equal to g downward.

5. (A) Use Big Five #3 with Vo=0:

)

~ t =P f'"s = 2(200 m)a 5 m is'

9s

v' =v~ + 2af'"s =2af'"s ~

6. (D) Use Big Five #5 with Vo=0 (calling down the positive direction):

v' v' (30 m I s)'f'"s= -=-= - 4S m2a 2g 2(9.8 m Is')

7. (C) Apply Big Five #3 to the vertical motion, calling down the positive direction:

Notice that the stone's initial horizontal speed (vo, =10 m/s) is irrelevant.

8. (B) First we determine the time required for the ball to reach th e top of its parabolic trajectory(which is the time required for the vertical velocity to drop to zero).

426 • CRACKING THE AP PHYSICS EXAM

l The total flight time is equal to twice this value:

T = 21 = 2 VOy = 2 Vo sin80 2(10m I s)sin30° 1 sg g 9.8m / s'

9. (C) After 4 seconds, the stone's vertical speed has changed by t>v = a1= (10m/s')(4 s) = 40 m/s.Since vCly =0, the value of vyat 1=4 is 40 m/s. The horizontal speed does not change. Therefore,when the rock hits the water, its velocity has a horizontal component of 30 ml s and a verticalcomponent of 40 ml s.

V x = 30 rrr /s. --_ .. _- -- - - ----- .,.

Vy = 40 mls i•

By the Pythagorean theorem, the magnitude of the total velocity, v, is 50 ml s.

10. (E) Since the acceleration of the projectile is always downward (because it's gravitationalacceleration), the vertical speed decreases as the projectile rises and increases as the projectilefalls. Statements A, B, C, and D are all false.

) SEmON II: FREE RESPONSE1. (a) At time 1= 1 s, the car's velocity starts to decrease as the acceleration (which is the slope of

the given v vs. 1graph) changes from positive to negative.

(b) The average velocity between 1= 0 and 1= 1 s is t(v,.o +v,) = t (0+20 m/s) = 10mis, and

the average velocity between 1= 1 and 1= 5 is t (vI: 1 +VI.,,) = t (20 ml s +0) = 10 ml s. The two

average velocities are the same.

(c) The displacement is equal to the area bounded by the graph and the 1 axis, taking areasabove the 1axis as positive and those below as negative. In this case, the displacement from 1=0 to 1=5 s is equal to the area of the triangular region whose base is the segment along the1axis from 1=0 to 1=5 s:

o»(I = 0 to 1= 5 s) = t x base x height = t (5 s)(20 m/s) = 50 m

The displacement from 1=5 s to 1=7 s is equal to the negative of the area of the triangularregion whose base is the segment along the 1axis from 1=5 s to 1=7 s:

t> s (t = 5 s to 1= 7 s) = - t x base x height = - t (2 s)(10 m/s )= - 10 m

SOLUTI ON S TO THE CHA PTE R REVIEW QUE STIONS . 427

Therefore, the displacement from I =0 to I =7 s is

c,s (I = 0 to I = 5 s) + C,S (I = 5 s to I = 7 s) = 50 m + (- 10 m) = 40 m

(d) The acceleration is the slope of the v vs. I graph. The segment of the graph from

1= 0 to I = 1 s has a slope of a = c,vl M = (20 mls - 0)/ (1 s - 0) = 20 mis', and the segment ofthe graph from I = 1 s to I = 7 s has a slope of a = c,vl C,I = (-10 mls - 20 s)/ (7 s - 1 s) =-5 mls'. Therefore, the graph of a vs. I is

20+-- -

)

f----+---+--j--+---j---+--I- time (s)

~glO§

:P~~

OJ -5~.. - 10

-20

1 2 3 4 5 6 7

(e) One way to determine the displacement is to determine equations for v(l) and integrate to gets(I). The segment from I = 0 to I = 1 s connects the points (0,0) and (1, 20); the slope is 20, and weget v = 201. The segment from I = 1 to I = 7 connects the points (1,20) and (7,-10); the slope is -5,and we get v = - 51+25. In summary,

{

201v(t) = -51 + 25

Therefore, since s(l ) =fv(l )dl, we find that

O~I ~ l

1 ~1 ~7

428 • CRACKING THE AP PHYSICS EXA M

{

1012

s(l) = 2- t l +251 - '2'O ~ I ~l

1 ~1~7

\L--+---+----1- -+--+- ---+- -j_ time (s)

50 --------------.-----. ----.. .--.. -....~. -o>__ _

,.. _- - - -- _. _-- --- - - - -- -- -- - - - - -. _- - - --- - - - ~ ---- - - - - - - --- - - -40

~

S 30§

'.jj'iil 200c,

10

01 2 3 4 5 6 7

2. (a) The maximum height of the projectile occurs at the time at which its ver tical velocity dropsto zero:

( set VOyV =0 =) Vo - gt=O =) t=-

Y Y g

The vertical displacement of the projectile at this time is computed as follows:

~ )

' z , .,A _ t _ .1gt2 H- Voy _ ~ VOy _vOy _vo sm liouy - Vo 2 =) - Vo z - -

y y g g 2g 2g

(b) The total flight time is equal to twice the time computed in part (a):

VoT=2t=2-yg

The horizontal displacement at this time gives the projectile's range:

g

SOLUTIONS TO TH E CHAPTER REVIEW QUE STION S . 429

(c) For any given valu e of v", the range,

will be maximized when sin 2()0 is maximized. This occurs when 2e o=90°, that is, when

8 0=45°.

(d) Set the general expression for the projectile's vertical displacement equal to h and solve forthe two values of t:

Applying the quadratic formula, we find that

VOy ± ~(-VOy)2 - 4(~ g)(h) V Oy ±~V~y - 2ght =-'--'---'-,--- -"-----

2(~g) g

Therefore, the two times at which the projectile crosses the horizontal line at height h are

and

so the amount of time that elapses between these events is

3. (a) The cannonball will certainly reach the wall (which is only 220 m away) since the ball'srange is

R= v~ sin 2eog

(50 m /s)2sin 2(400)

9.8 m /s2 251m

We simply need to make sure that the cannonball's height is less than 30 m at the point whereits horizontal displacement is 220 m (so that the ball actually hits the wall rather than flyingover it). To do this, we find the time at which x = 220 m by first writing

(1)xx

t=-=-- -vox Vocos eo

430 • CRAC KI NG THEAP PHYSIC S EXAM