Figure 1Marcel Duchamp, Nude Decending a StaircasePhiladelphia Museum of Art: Louise and WalterArensberg Collection

CHAPTER 3 DESCRIPTION OF MOTION

In the next few chapters, a similar technique will beused to describe motion. We now have devicesavailable, such as the stroboscope (called the strobe),that produce short bursts of light at regular inter-vals; with the strobe, we can photograph the succes-sive positions of an object, such as a ball moving onthe end of a string (see Figure 2). Although we do nothave the artist’s freedom of expression to convey theconcept of motion by using a strobe photograph, wedo obtain a more accurate measure of the motion.

On the facing page is a reproduction (Figure 1) ofMarcel Duchamp’s painting, Nude Descending aStaircase, which was first displayed in New York atThe International Exhibition of Modern Art, gener-ally known as the Armory Show, in 1913. The objec-tive of the painting, to convey a sense of motion, isachieved by repeating the stylized human form fivetimes as it descends the steps. At the risk of obscur-ing the artistic qualities of the painting, we mayimagine this work as a series of five flash photo-graphs taken in sequence as the model walkeddownstairs.

Chapter 3Description of Motion

Figure 2Strobe photograph showing the motion of a ball on the end of a string.

3-4 Description of Motion

The photograph in Figure (2) was taken with thestrobe flashing five times per second while the ballwas moving slowly. As a result, we see a smoothcurve and have a fairly complete idea of the ball’sentire motion. When we run the strobe at a rate offive flashes per second but move the ball morerapidly in a complicated pattern, the result is asshown in Figure (3). From this picture it is difficultto guess the ball’s path; thus Figure (3) provides uswith a poor representation of the motion of the ball.But if we turn the strobe up from 5 to 15 flashes persecond (as in Figure 4), the rapid and complicatedmotion of the ball is easily understood.

The motion of any object can be described by locat-ing its position at successive intervals of time. Astrobe photograph is particularly useful because itshows the position at equal time intervals through-

out the picture; that is, in Figure (2) at intervals of1/5 sec and in Figure (4) at intervals of 1/15 sec. Forthis text, we will use a special symbol, ∆t, to repre-sent the time interval between flashes of the strobe.The t stands for time, while the ∆ (Greek letter delta)indicates that these are short time intervals betweenflashes. Thus, ∆t = 1/5 sec in Figures (2) and (3), and∆t = 1/15 sec in Figure (4).

For objects that are moving slowly along fairlysmooth paths, we can use fairly long time intervals∆t between strobe flashes and their motion will beadequately described. As the motion becomes fasterand more complicated, we turn the strobe up to ahigher flashing rate to follow the object, as in Figure(4). To study complicated motion in more detail, welocate the position of the object after shorter andshorter time intervals ∆t.

Figure 3Strobe photograph of a moving object. In this photograph, the timebetween flashes is so long that the motion is difficult to understand.

3-5

12 3

4

5

67

8

9

s1s2 s3

s4

s5

s6s7

s8

Figure 5Displacement vectors. The displacement betweenflash number 1 and flash number 2 is representedby the displacement vector s1 and so on. Theentire path taken by the ball is represented by theseries of eight displacement vectors.

Figure 4Strobe photograph of a similar motion. In thisphotograph, the time between flashes was reducedand the motion is more easily understood.

DISPLACEMENT VECTORSWhen we represent the motion of an object bya strobe photograph, we are in fact representingthis motion by a series of displacements, thesuccessive displacements of the object in equalintervals of time. Mathematically, we can de-scribe these displacements by a series of dis-placement vectors, as shown in Figure (5). Thisillustration is a reproduction of Figure (2) withthe successive displacement vectors drawn fromthe center of the images.

3-6 Description of Motion

Figure 6

Representation of the path of a ball for various ∆t. As the shorter and shorter ∆t isused, the path of the ball is more accurately represented, as in figures (b) through (d).

(a) (b)

(c) (d)

(e) (f)

s1 s1

s1

s1

s1

s2

3-7

In a sense we are approximating the path of the ballby a series of straight lines along the path. This isreasonably accurate provided that ∆t is short enough,as shown in Figure (6).

In Figure (6), (a) is the strobe photograph shown inFigure (4), taken at a strobe interval of ∆t = 1/15 sec;(b) shows how this photograph would have looked ifwe had set the strobe for ∆t = 10/15 sec, or 2/3 sec.Only one out of ten exposures would have beenproduced. If we had represented the path of the ballby the vector s1 it would have been a gross misrep-resentation. In (c), which would be the strobepicture at ∆t = 6/15 sec, we see that the ball is nolonger moving in a straight line, but still s1 and s2provide a poor representation of the true motion.Cutting ∆t in half to get (d), ∆t = 3/15 sec, we woulddiscover that there is a kink in the path of the ball.While taking the picture, we would have had to becareful in noticing the sequence of positions in orderto draw the correct displacement vectors.

Reducing ∆t to 2/15 sec (e), would give us a moredetailed picture of the kink. This is not too differentfrom (d); moreover, we begin to suspect that the

Figure 8Strobe photograph of a steel ball projectile.The strobe flashes were 1/10 second apart.

Figure 7Experimental setup for taking strobe photographs. APolaroid camera is used record the motion of a ballmoving in front of a grid. The grid, made of stretchedfish line, is mounted in front of a black painted wall.

seven displacement vectors in (e) represent the pathfairly accurately. When we reduce ∆t to 1/15 sec (f),we get more pictures of the same kink and the curvebecomes smoother. It now appears that in mostplaces the 14 displacement vectors form a fairlyaccurate picture of the true path. We notice, how-ever, that the very bottom of the kink is cut offabruptly; here, shorter time intervals are needed toget an accurate picture of the motion.

A Coordinate SystemIn the strobe photographs discussed so far, we havea precise idea of the time scale, 1/5 second betweenflashes in Figure (2), 1/15 second in Figure (4), butno idea about the distance scale. As a result we knowthe direction of the succeeding displacement vec-tors, but do not know their magnitude.

One way to introduce a distance scale is to photo-graph the motion in front of a grid as shown in Figure(7). With this setup we obtain photographs like thatshown in Figure (8), where we see the strobe motionof a steel ball projectile superimposed on the grid.The grid is illuminated by room lights which aredimmed to balance the exposure of the grid and thestrobe flashes.

3-8 Description of Motion

Using techniques like that illustrated in Figure (9) tolocate the centers of the images, we can transfer theinformation from the strobe photograph to graphpaper and obtain the results shown in Figure (10).Figure (10) is the end result of a fair amount oftedious lab work, and the starting point for ouranalysis. For those who do not have strobe facilities,or the time to extract the information from a strobephotograph, we will include in the text a number ofexamples already transferred to graph paper in theform of Figure (10).

Using a television camera attached to an Apple IIcomputer, we can, in under 2 minutes, obtain resultsthat look like Figure (10). We will include a few ofthese computer strobe photographs in our examplesof motion. However the computer strobe is not yetcommercially available because we plan to use acomputer with more modern graphics capabilities. Itis likely that within a few years, one will be able toeasily and quickly obtain results like those in Figure

(10). The grid, which has now become the graphpaper in Figure (10), serves as our coordinate systemfor locating the images.

0 1

2

s0s1

s2

0 10 20

Figure 11Measuring the length of the vector S1 .

Figure 9Using a pin and cylinder to locate the center of theball. Move the cylinder until it just covers the image ofthe ball and then gently press down on the pin. Thepin prick will give an accurate location for the center.

Manipulation of VectorsFigure (10) represents the kind of experimental dataupon which we will base our description of motion.We have, up to now, described the motion of theprojectile in terms of a series of displacement vec-tors labeled s-1, s0, s3 as shown. To go further, tointroduce concepts like velocity and acceleration,we need to perform certain routine operations onthese displacement vectors, like adding and sub-tracting them. A number of vector operations werediscussed in Chapter 2, let us briefly review herethose that we need for the analysis of strobe photo-graphs. We will also introduce the concept of acoordinate vector which will be useful in much ofour work.

3-9

Measuring the Length of a VectorOne of the first pieces of information we need froma strobe photograph is the magnitude or length of thedisplacement vectors we have drawn. Figure (11)illustrates the practical way to obtain the lengths of

the individual vectors from a graph like Figure (10).Take a piece of scrap paper and mark off the lengthof the vector as shown in the upper part of the figure.Then rotate the paper until it is parallel to the gridlines, and note the distance between the marks.

0 10 20 30 40 50 60 70 80 10090

10

20

30

40

50

60

70

80

90

10

20

30

40

50

60

70

80

90

0 10 20 30 40 50 60 70 80 10090

-1

0 1

3

2

4

s-1

s0s1

s2

s3

Ball coordinates-1) ( 8.4, 79.3) 0) (25.9, 89.9) 1) (43.2, 90.2) 2) (60.8, 80.5) 3) (78.2, 60.2) 4) (95.9, 30.2)

Figure 10Strobe photograph transferred to graph paper. Using the pin and cylinder of Figure (9), we located thecoordinates of the center of each image in Figure (8), and then reconstructed the strobe photograph asshown. We can now perform our analysis on the large graph paper rather than the small photograph.

3-10 Description of Motion

In Figure (11), we see that the marks are 20 smallgrid spacings apart. In Figure (10), we see that eachgrid spacing represents a distance of 1 centimeter.Thus in Figure (11), the vector s1 has a magnitude of20 centimeters. We can write this formally as

s1 = 20 cm

This technique may seem rather simple, but it workswell and you will use it often.

Graphical Addition and SubtractionSince we are working with experimental data ingraphical form, we need to use graphical techniquesto add and subtract vectors. These techniques,originally introduced in Chapter 2, are reviewedhere in Figures (12) and (13). Figure (12a) and (12b)show the addition of two vectors by placing themhead to tail. Think of the vectors A and B as separatetrips; the sum A + B is our net displacement as wetake the trips A and B in succession. To subtract Bfrom A, we simply add (–B) to A as shown in Figure(12c).

To perform vector addition and subtraction, we needto move the vectors from one place to another. Thisis easily done with a triangle and a straight edge asindicated in Figure (13). The triangle and straightedge allows you to draw a parallel line; then mark apiece of paper as in Figure (11), to make the newvector have the same length as the old one.

For those who are mathematically inclined, thissimple graphical work with vectors may seem el-ementary, especially compared to the exercises en-countered in an introductory calculus course. But,as we shall see, this graphical work emphasizes thebasic concepts. We will have many opportunitieslater to extract sophisticated formulas from thesebasic graphical operations.

For these exercises, you may use the practicegraph on page 3-28, and the tear out sheet on page3-29.

Exercise 1

Find the magnitudes of the vectors s0, s1, s2, and s3

in Figure (10).

Exercise 2Explain why the vector s0 4, given by

s0 4 = s0 + s1 + s2 + s3

has a magnitude of 91.3 cm which is quite a bit lessthan the sum of the lengths s0 + s1 + s2 + s3 .

Exercise 3

Use graphical methods to find the vector s3 - s2 .(The result should point vertically downward andhave a length of about 10 cm.)

A

B

–B

(A – B)

A

B

Figure 12Addition and subtraction of vectors.

Figure 13Moving vectors around. (Thiswas discussed in Figure 2-12.)

BA +

BA

(b)

(a) (c)

3-11

Coordinate Systemand Coordinate VectorsA coordinate system allows us to convert graphicalwork into a numerical calculation that can, for ex-ample, be carried out on a computer. Figure (14)illustrates two convenient ways of describing thelocation of a point. One is to give the x and ycoordinates of the point (x,y), and the other is to usea coordinate vector R which we define as a vectorthat is drawn from the origin of the coordinatesystem to the point of interest.

Figure (15) illustrates the way an arbitrary vector S canbe expressed in terms of coordinate vectors. From thediagram we see that R2 is the vector sum of R1 + S, thuswe can solve for the vector S to get the result

S = R2 – R1.

RR ≡ (X,Y)

(X,Y)

Y

X

R

Y

X

1

R2R1

R2

+

S = –

S = SR1

R2

Figure 14The coordinate vector R , which startsat the origin, locates the point (x,y).

Figure 15Expressing the vector S in terms of coordinate vectors.

ANALYSIS OFSTROBE PHOTOGRAPHSIn our analysis of the strobe photograph of projectilemotion, Figure (10), we are representing the path ofthe ball by a series of displacement vectors S0 ... S3 (We will think of the photograph as starting at point(0). The point labeled (-1) will be used later in ourcalculation of the instantaneous velocity at point (0).In a sense, we “know” that the ball actually wentalong a smooth continuous curve, and we could haverepresented the curve more accurately by reducing∆ t as we did in Figure (6). But with many images tomark the trajectory, each displacement vector Sibecomes too short for accurate graphical work. Intaking a strobe photograph, one must reach a com-promise where the displacement vectors Si are longenough to work with, but short enough to give areasonable picture of the motion.

VelocityThe series of displacement vectors in Figure (10)show not only the trajectory of the projectile, butbecause the images are located at equal time inter-vals, we also have an idea of the speed of theprojectile along its path. A long displacement vectorindicates a higher speed than a short one. For eachof the displacement vectors we can calculate whatone would call the average speed of the projectileduring that interval.

The idea of an average speed for a trip should befairly familiar. If, for example, you went on a trip fora total distance of 90 miles, and you took 2 hours,you divide 90 miles by 2 hours to get an averagespeed of 45 miles per hour. For more detailedinformation about your speed, you break the trip upinto small segments. For example, if you wanted toknow how fast you were moving down the interstatehighway, you measure how long it takes to pass twoconsecutive mile markers. If it took one minute,then your average speed during this short timeinterval is one mile divided by 1/60 hour which is 60miles per hour. If you broke the whole trip down into1 minute intervals, measured how far you wentduring each interval, and calculated your averagespeed for each interval, you would have a fairlycomplete record of your speed during your trip. It is

3-12 Description of Motion

this kind of record that we get from a strobe photo-graph of the motion of an object.

In physics, we use a concept that contains moreinformation than simply the speed of the object. Wewant to know not only how many miles per hour orcentimeters per second an object is moving, but alsowhat direction the object is moving. This informa-tion is all contained in the concept of a velocityvector.

To construct a velocity vector for the projectileshown in Figure (10), when, for example, the ball isat position 1, we take the displacement vector S1,divide it by the strobe time interval ∆ t, to get whatwe will call the velocity vector v1:

v1 ≡ S1∆ t

(1)

In Equation (1), what we have done is multiply thevector S1 by the number (1/∆ t) to get v1. From ourearlier discussion of vectors we know that multiply-ing a vector by a number gives us a vector that pointsin the same direction, but has a new length. Thus v1is a vector that points in the same direction as S1, butit now has a length given by

v1 = S1

∆ t = 20 cm

.1 sec = 200 cm

sec

where we used S1 = 20 cm from Figure (11) andwe knew that ∆t = .1 sec for this strobe photograph.

Not only have we changed the length of S1 bymultiplying by (1/∆t), we have also changed thedimensions from that of a distance (cm) to that of aspeed (cm/sec). Thus the velocity vector v1 containstwo important pieces of information. It points in thedirection of the motion of the ball, and has a lengthor magnitude equal to the speed of the ball.

(Physics texts get rather picky over the use of thewords speed and velocity. The word speed is re-served for the magnitude of the velocity, like 200cm/sec. The word velocity is reserved for the veloc-ity vector as defined above; the velocity vector

describes both the direction of motion and the speed.We will also use this convention throughout thetext.)

In our discussion of strobe photographs, we notedthat if we used too long a time interval ∆t, we got apoor description of the motion as in Figures (6b) and(6c). As we used shorter time intervals as in Figures(6d, e, and f), we got a better and better picture of thepath.

We have the same problem in dealing with thevelocity of an object. If we use a very long ∆ t, weget a crude, average, description of the object’svelocity. As we use a shorter and shorter ∆ t, ourdescription of the velocity, Equation (1), becomesmore and more precise.

Since, in this chapter, we will be working withexperimental data obtained from strobe photographs,there is a practical limit on how short a time interval∆ t we can use and have vectors big enough to workwith. We will see that, for the kinds of motion thatwe encounter in the introductory physics lab, areasonably short ∆ t like .1 sec gives reasonablyaccurate results.

If you make more precise measurements of theposition of an object you generally find that as youuse shorter and shorter ∆ t to measure velocity, youreach a point where the velocity vector no longerchanges. What happens is that you reach a pointwhere, if you cut ∆ t in half, the particle goes in thesame direction but only half as far. Thus both thedisplacement S1 and the time interval ∆ t are both cutin half, and the ratio v1 = S1 /∆t is unchanged. Thislimiting process, where we see that the velocityvector changes less and less as ∆ t is reduced, isdemonstrated graphically in our discussion of in-stantaneous velocity at the end of the chapter.

Exercise (4)

What is the magnitude of the velocity vector v3 , for theball in Figure (10). Give your answer in cm/sec.

3-13

AccelerationIn Chapter 1 on Einstein’s special theory of relativ-ity, we limited our discussion to uniform motion,motion in a straight line at constant speed. If we tooka strobe photograph of an object undergoing uni-form motion, we would get a result like that shownin Figure (17). All the velocity vectors would pointin the same direction and have the same length. Wewill, from now on, call this motion with constantvelocity , meaning that the velocity vector is con-stant, unchanging.

From the principle of relativity we learned that thereis something very special about motion with con-stant velocity—we cannot feel it. Recall that onestatement of the principle of relativity was that thereis no experiment that you can perform to detect yourown uniform motion relative to empty space. Youcannot tell, for example, whether the room you aresitting in is at rest or hurdling through space at aspeed of 100,000 miles per hour.

Although we cannot feel or detect our own uniformmotion, we can easily detect non uniform motion.We know what happens if we slam on the brakes andcome to a sudden stop—everything in the car fallsforward. A strobe photograph of a car using thebrakes might look like that shown in Figure (18a).Each successive velocity vector gets shorter andshorter until the car comes to rest.

Equation (1) is well suited for graphical work but fornumerical calculations it is convenient to express Si interms of the coordinate vectors Ri. This is done inFigure (16), where we see that the vector sum

Ri + Si = Ri+1 thus Si = Ri+1 – Ri and Equation (1)becomes

vi = Ri+1 – Ri

∆ t(2)

If we call Ri+1 – Ri the “change in the position R

during the time ∆t ”, and denote this change by ∆R,Equation (2) becomes

vi = Ri+1 – Ri

∆ t = ∆R

∆ t(2a)

which is perhaps a more familiar notation for thosewho have already studied calculus. In a calculuscourse, one would define the velocity vi by takingthe limit as ∆ t → 0 (i.e., by turning the strobeflashing rate “all the way up”). In our experimentalwork with strobe photographs, we reduce ∆ t onlyto the point where we have a reasonable representa-tion of the path; using too short a time interval makesthe experimental analysis impossible.

i

= Ri + 1

– R iSi

R i

i + 1R

Vi = Sit

= i + 1(R R– i )t

Figure 16Expressing the velocity vector vi , in termsof the coordinate vectors Ri and Ri+1.

V1 V2 V3 V4 V5 V6 V7

Figure 17Motion with constant velocity.

V5V4V3V2V1

Figure 18aPut on the brakes, and your velocity changes.

3-14 Description of Motion

Another way the velocity of a car can change is bygoing around a corner as illustrated in Figure (18b).In that figure the speed does not change, each veloc-ity vector has the same length, but the directions arechanging. It is also easy to detect this kind of changein velocity—all the packages in the back seat of yourcar slide to one side of the seat.

The point we want to get at is, what do we feel when ourvelocity changes? Consider two examples. In the first,we are moving at constant velocity, due east at 60 milesper hour. A strobe photograph showing our initial andfinal velocity vectors vi and vf would look like that inFigure (19a). If we define the change in velocity ∆v bythe equation

∆v ≡ vf - vi

then from Figure (19b) we see that ∆v = 0 foruniform motion.

For the second example, suppose we are travelingdue south at 60 miles per hour, and a while later are

moving due east at 60 miles per hour, as indicated inFigure (20a). Now we have a non zero change invelocity ∆v as indicated in Figure (20b).

In our two examples, we find that if we have uniformmotion which we cannot feel, the change in velocity∆v is zero. If we have non uniform motion, ∆v is notzero and we can feel that. Is it ∆v, the change invelocity, that we feel?

Almost, but not quite. Let us look at our secondexample, Figure (20), more carefully. There are twodistinct ways that our velocity can change frompointing south to pointing east. In one case therecould have been a gradual curve in the road. It mayhave taken several minutes to go around the curveand we would be hardly aware of the turn.

In the other extreme, we may have been drivingsouth, bounced off a stalled truck, and within afraction of a second finding ourselves traveling dueeast. In both cases our change in velocity ∆v = vf – viis the same, as shown in Figure (20b). But the effecton us is terribly different. The difference in the twocases is that the change in velocity ∆v occurredmuch more rapidly when we struck the truck thanwhen we went around the curve. What we feel is not∆v alone, but how fast ∆v happens.

If we take the change in velocity ∆v and divide it bythe time ∆t over which the change takes place, thenthe smaller ∆t, the more rapidly the change takesplace, the bigger the result. This ratio ∆v/∆t whichmore closely represents what we feel than ∆v alone,is given the special name acceleration.

V4

V3

V2

V1

Figure 18bWhen you drive around a corner, your speed may notchange, but your velocity vector changes in direction.

Vf

Vi

(a)

Figure 20∆v ≠ 0 when we change our direction of motion.

Vf

V = 0i

Vf–

Vi–

Vi

VfV ≡∆

(a)

(b)

Figure 19We see that ∆∆v = 0 for motion with constant velocity.

V i

Vf

–V f

V =∆

(b) Vi–

3-15

The Acceleration VectorThe quantity ∆v/∆t, which we call acceleration, isusually denoted by the vector a

a = ∆v∆ t

(3)

where ∆v is the change in the velocity vector duringthe time ∆t. To see how to apply Equation (3) to astrobe photograph, suppose that Figure (21) repre-sents a photograph of a particle moving with somekind of non uniform velocity. Labeling the imagepositions 1, 2, 3, etc. and the corresponding velocityvectors v1, v2, v3 , let us consider what theparticle’s acceleration was during the time it wentfrom position 2 to position 3. At position 2 theparticle’s velocity was v2. When it got to position 3its velocity was v3. The time it took for the velocityto change from v2 to v3, a change ∆v = v3 - v2, wasthe strobe time ∆t. Thus according to Equation (3),the particle’s acceleration during the interval 2 to 3,which we will call a3, is given by

a3 ≡ ∆v23∆ t

= v3 – v2

∆ t(4a)

(One could object to using the label a3 for theacceleration during the interval 2 to 3. But a closerinspection shows that a3 is an accurate name. Actu-ally the velocity v3 is the average velocity in theinterval 3 to 4, and v2 is the average velocity in theinterval 2 to 3. Thus ∆v = v3 - v2 is a change invelocity centered on position 3. As a result Equation(4a) gives surprisingly accurate results when work-ing with experimental strobe photographs. In anycase such errors become vanishingly small when weuse sufficiently short ∆t 's.)

The physicists’ use of the word acceleration for thequantity ∆v/∆t presents a problem for students. Thedifficulty is that we have grown up using the wordacceleration, and already have some intuitive feel-ing for what that word means. Unfortunately thisintuition usually does not match what physicistsmean by acceleration. Perhaps physicists shouldhave used a different name for ∆v/∆t, but this did nothappen. The problem for the student is therefore notonly to develop a new intuition for the quantity∆v/∆t, but also to discard previous intuitive ideas ofwhat acceleration might be. This can be uncomfort-able.

The purpose of the remainder of this chapter is todevelop a new intuition for the physics definition ofacceleration. To do this we will consider threeexamples of motion; projectile motion, uniformcircular motion, and projectile motion with air resis-tance. In each of these cases, which can be carefullystudied in the introductory lab or simulated, we willuse strobe photographs to determine how the accel-eration vector ∆v/∆t behaves. In each case we willsee that there is a simple relationship between thebehavior of the acceleration vector and the forcespulling or pushing on the object. This relationshipbetween force and acceleration, which is the corner-stone of mechanics, will be discussed in a laterchapter. Here our goal is to develop a clear pictureof acceleration itself.

Determining Accelerationfrom a Strobe PhotographWe will use strobe photographs to provide an ex-plicit experimental definition of acceleration. In thenext chapter we will see how the strobe definitionsgo over to the calculus definition that you may havealready studied. We prefer to start with the strobedefinition, not only because it provides a moreintuitive approach to the concept, but also because ofits experimental origin. With an experimental defi-nition we avoid some conceptual problems inherentin calculus. It turns out, surprisingly, that some ofthe concepts involved in the calculus definition ofacceleration are inconsistent with physics. We canmore clearly understand these inconsistencies whenwe use an experimental definition of acceleration asthe foundation for our discussion.

Figure 21Determining a for non uniform motion.

1

2

34

V1

V2

V3 V4

3-16 Description of Motion

If we have a strobe photograph with many images,then by extending Equation (4a), the acceleration atposition i is

ai ≡

∆vi∆t

=vi – vi – 1

∆tstrobe definitionof acceleration (4)

We will call Equation (4) our strobe definition ofacceleration. Implicit in this definition is that weuse a short enough ∆t so that all the kinks in themotion are visible, but a long enough ∆t so that wehave vectors long enough to work with.

PROJECTILE MOTIONAs our first example in the use of our strobe defini-tion of acceleration, let us calculate the accelerationof the ball at position 2 in our strobe photograph,Figure (10), of projectile motion.

The first problem we face is that Equation (4)expresses the acceleration vector a2 in terms of thevelocity vectors v1 and v2, while the strobe photo-graph shows only the displacement vectors S1 andS2, as seen in Figure (10a), a segment of Figure (10)reproduced here.

The easiest way to handle this problem is to use theformulas

v1 = S1∆t

; v2 = S2∆t

in Equation (4a) to express a2 directly in terms of theknown vectors S1 and S2. The result is

a2 = v2 – v1

∆t =

S1/∆t – S2/∆t∆t

a2 =

S2 – S1

∆t2

experimentalmeasurmentof acceleration

(5)

Equation (5) tells us that we can calculate the accelera-tion vector a2 by first constructing the vector S2 – S1,and then dividing by ∆t2 . That means that a2 points inthe direction of the vector S2 – S1 , and has a lengthequal to the length S2 – S1 (in cm) divided by ∆∆t 2.As a result the magnitude of the acceleration vector hasthe dimensions of cm/sec2.

Let us apply Equation (5) to our projectile motionphotograph, Figure (10), to see how all this works.The first step is to use vector subtraction to constructthe vector S2 – S1. This is done in Figure (22). Firstwe draw the vectors S1 and S2, and then constructthe vector –S1 as shown. (The vector –S1 is thesame as S1 except that it points in the oppositedirection.) Then we add the vectors S2 and –S1 toget the vector S2 – S1 by the usual technique ofvector addition as shown

S2 – S1 = S2 + –S1 (6)

2

1

3

s

–

1

s1

s2 s1–s2

Figure 22The vector S2 – S1 points straight downand has a length of about 10 cm.

1

3

2s1

s2

Figure 10aA section of the projectile motion photograph, Figure(10), showing the displacement vectors S1 and S2 .

3-17

Note that even if S2 and S1 had the same length, thedifference S2 – S1 would not necessarily be zero be-cause this is vector subtraction, NOT NUMERICALSUBTRACTION.

Once we have constructed the vector S2 – S1, we knowthe direction of the acceleration vector a2 because itpoints in the same direction as S2 – S1. In Figure (22),we see that S2 – S1 points straight down, thus a2 pointsstraight down also.

Now that we have the direction of a2, all that is leftis to calculate its magnitude or length. This magni-tude is given by the formula

a2 = S2 – S1

∆ t2 =

length of vector S2 – S1

∆ t2(7)

To get the length of the S2 – S1, we can use thetechnique shown in Figure (11). Mark off the lengthof the vector S2 – S1 on a piece of scrap paper, andthen use the grid to see how many centimeters apartthe marks are. In this case, where S2 – S1 pointsstraight down, we immediately see that S2 – S1 isabout 10 cm long. Thus the magnitude of a2 is givenby

a2 = S2 – S1

∆ t2 = 10 cm

.1 sec 2 = 1000 cm

sec2(8)

where we knew ∆ t = .1 sec for the strobe photographin Figure (10).

Our conclusion is that, at position 2 in the projectilemotion photograph, the ball had an acceleration a2that pointed straight down, and had a magnitude ofabout 1000 cm/sec2.

For this exercise, you may use the tear out sheeton page 3-30.

Exercise 5

(Do this now before reading on.) Find theacceleration vectors a0,a1, and a3 for the projectilemotion in Figure (10). From your results, what can yousay about the acceleration of a projectile?

UNIFORM CIRCULAR MOTIONTo give the reader some time to think about theabove exercise on projectile motion, we will changethe topic for a while and analyze what is calleduniform circular motion. In uniform circular mo-tion, the particle travels like a speck of dust sitting ona revolving turntable.

The explicit example we would like to consider is agolf ball with a string attached, being swung in acircle over the instructor’s head, as indicated inFigure (23a). We could photograph this motion, butit is very easy to simulate a strobe photograph ofuniform circular motion by drawing a circle with acompass, and marking off equal intervals as shownin Figure (23b). In that figure we have also sketchedin the displacement vectors as we did in our analysisof the projectile motion photograph.

S1

S2

S3S4

rStrin

gV

Golf ball

Figure 23aSwinging a golf ball around at constant speed in acircle.

Figure 23bSimulating a strobe photograph of a golfball swinging at constant speed in a circle.We marked off equal distances using acompass.

3-18 Description of Motion

Figure (23b) shows the kind of errors we have to dealwith in using a strobe to study motion. Clearly thegolf ball travels along the smooth circular pathrather than the straight line segments marked by thevectors . As we use shorter and shorter ∆ t ourapproximation of the path gets better and better, butsoon the vectors get too short for accurate graphicalwork. Choosing images spaced as in Figure (23b)gives vectors a reasonable length, and a reasonableapproximation of the circular path.

(It will turn out that when we use our strobe defini-tion of acceleration, most errors caused by using afinite ∆ t cancel, and we get a very accurate answer.Thus we do not have to worry much about how farapart we draw the images.)

Now that we have the displacement vectors we canconstruct the acceleration vectors a1,a2, usingEquation (5). The construction for a2 is shown inFigure (24). To the vector S2 we add the vector –S1to get the vector S2 – S1 as shown.

The first thing we note is that the vector S2 – S1points toward the center of the circle! Thus theacceleration vector a2 given by

a2 = S2 – S1

∆t2(9)

also points toward the center of the circle.

Exercise (6)

(Do this now.) Find the direction of at least 4 moreacceleration vectors around the circle. In each caseshow that ai points toward the center of the circle.

We said earlier that the physicists’ definition ofacceleration, which becomes ai = Si – Si–1 /∆ t2,does not necessarily agree with your own intuitiveidea of acceleration. We have just discovered that,using the physicists’ definition, a particle moving atconstant speed along a circular path acceleratestoward the center of the circle. Unless you had aprevious physics course, you would be unlikely toguess this result. It may seem counter intuitive. But,as we said, we are using these examples to developan intuition for the physics definition of accelera-tion. Whether you like it or not, according to thephysics definition, a particle moving at constantspeed around a circle, is accelerating toward thecenter. In a little while, the reason for this willbecome clear.

Magnitude of the Accelerationfor Circular MotionAlthough perhaps not intuitive, we have gotten afairly simple result for the direction of the accelera-tion vector for uniform circular motion. The centeris the only unique point for a circle, and that is wherethe acceleration vector points. The next thing weneed to know is how long the acceleration vectorsare; what is the magnitude of this center pointingacceleration. From the strobe definition, the magni-tude a2 is a2 = S2 – S1 /∆ t2, a rather awkwardresult that appears to depend upon the size of ∆ t thatwe choose. However with a bit of geometricalconstruction we can re-express this result in terms ofthe particle’s speed v and the circle’s radius r. Thederivation is messy, but the result is simple. This isone case, where, when we finish the derivation, werecommend that the student memorize the answerrather than try to remember the derivation. Uniformcircular motion appears in a number of importantphysics problems, thus the formula for the magni-tude of the acceleration is important to know.

Figure 24We find that the vector S2 – S1 , and therefore theacceleration, points toward the center of the circle.

S1

S2

S3

S4

S1–

S1–S2( )

3-19

In Figure (25a) we have constructed two triangles,which are shown separately in Figures (25b) and(25c). As seen in Figure (25b), the big trianglewhich goes from the center of the circle to positions(1) and (2) has two equal sides of length r, the radiusof the circle, and one side whose length is equal tothe particle’s speed v times the strobe time ∆t.

The second triangle, shown in Figure (25c), hassides of length S2 and –S1 , but both of these areof length v∆t as shown. The third side is of length

S2 – S1 , the length we need for our calculation ofthe magnitude of the acceleration vector.

The trick of this calculation is to note that the angleslabeled θ in Figures (29b, c) are the same angle, sothat these two triangles are similar isosceles tri-angles. The proof that these angles are equal is givenin Figure (26) and its caption.

With similar triangles we can use the fact that theratios of corresponding sides are equal. Equating theratio of the short side to the long side of the triangleof Figure (25b), to the ratio of the short side to thelong side of the triangle in Figure (25c), we get

v∆t

r =S2 – S1

v∆t(10)

Multiplying Equation (10) through by v and divid-ing both sides by ∆ t gives

v2

r =S2 – S1

∆t2 = a1 (11)

we got S2 – S1 ∆t2S2 – S1 ∆t2 on the right side, but this is justthe magnitude of the acceleration vector a1. Sincethe same derivation applies to any position aroundthe circle, we get the simple and general result that,for a particle moving with uniform circular motion,the particle’s acceleration a points toward the centerof the circle, and has a magnitude

a = v2

raccelerationof a particlein uniform circular motion (12)

where v is the speed of the particle and r is the radiusof the circle. As we said, this simple result should bememorized.

Figure 25Derivation of the formula for the magnitude of theacceleration of a particle with uniform circular motion.

S1–S2( )

S1

S1–

rθ

rθ

r

θ

S1 = V∆t

S 2

S2 = V∆tS1–S2

S1 = V∆t

(a)

(b)

(c)

1

2

3

A

θ

B D90 – θ

α 2α

C

φ

Figure 26That the two angles labeled θθ in Figure (25) are thesame, may be seen in the following geometricalconstruction. Since the sum of the angles in anytriangle is 180°, we get αα + ϕϕ + 90°° = 180°° (fromtriangle BCD). Because BAC is an isosceles triangle,

90°° – θθ + αα = ϕϕ. Eliminating ϕϕ we get 2αα = θθ , whichis the result we expected.

3-20 Description of Motion

AN INTUITIVE DISCUSSIONOF ACCELERATIONWe have now studied two examples of non uniformmotion, the projectile motion seen in the strobephotograph of Figure (10), and the circular motionof a golf ball on the end of a string, a motion weillustrated in Figure (23). In each case we calculatedthe acceleration vector of the particle at differentpoints along the trajectory. Let us now review ourresults to see if we can gain some understanding ofwhy the acceleration vector behaves the way it does.

If you worked Exercise (5) correctly, you discov-ered that all the acceleration vectors are the same, atleast to within experimental accuracy. As shown inFigure (27), as the steel ball moves along its trajec-tory, its acceleration vector points downward to-ward the earth, and has a constant magnitude ofabout 1000 cm/sec2.

As shown in Figure (28), the golf ball being swungat constant speed around in a circle on the end of astring, accelerates toward the center of the circle, inthe direction of the string pulling on the ball. Themagnitude of the acceleration has the constant value.

We said that the string was pulling on the ball. To seethat this is true, try swinging a ball on the end of astring (or a shoe on the end of a shoelace) in a circle.To keep the ball (or shoe) moving in a circle, youhave to pull in on the string. In turn, the string pullsin on the ball (or shoe). If you no longer pull in onthe string, i.e., let go, the ball or shoe flies away andno longer undergoes circular motion. The stringpulling on the ball is necessary in order to havecircular motion.

What is the common feature of projectile and circu-lar motion? In both cases the object accelerates inthe direction of the force acting on the object. Whenyou throw a steel ball in the air, the ball does notescape earth’s gravity. As the ball moves throughthe air, gravity is constantly pulling down on theball. The result of this gravitational pull or force isto accelerate the ball in the direction of the gravita-tional force. That is why the projectile motionacceleration vectors point down toward the earth.

When we throw a ball a few feet up in the air, it doesnot get very far away from the surface of the earth.In other words we expect the gravitational pull to beequally strong throughout the trajectory. If the ball’sacceleration is related to the gravitational pull, thenwe expect the acceleration to also be constant through-out the trajectory. Thus it is not surprising that all thevectors have the same length in Figure (27).

a0

0 1

2

3

a1

a2

a3

Figure 27All the acceleration vectors forprojectile motion point down towardthe earth.

rstrin

g

a

golf ball

Figure 28The golf ball accelerates in thedirection of the string that is pullingon it.

3-21

In the case of circular motion, the string has to pullin on the golf ball to keep the ball moving in a circle.As a result of this pull of the string toward the centerof the circle, the ball accelerates toward the center ofthe circle. Again the acceleration is in the directionof the force on the object.

This relationship between force and acceleration,which we are just beginning to see in these twoexamples, forms the cornerstone of what is calledclassical or Newtonian mechanics. We have moredetails to work out, but we have just glimpsed thebasic idea of much of the first half of this course. Togive historical credit for these ideas, it was Galileowho first saw the importance of the concept ofacceleration that we have been discussing, and IsaacNewton who pinned down the relationship betweenforce and acceleration.

Acceleration Due to GravityTwo more topics, both related to projectile motion,will finish our discussion in this section. The first isthe fact that, if we can neglect air resistance, allprojectiles near the surface of the earth have thesame downward acceleration a. If a steel ball and afeather are dropped in a vacuum, they fall togetherwith the same acceleration. This acceleration, whichis caused by gravity, is called the acceleration dueto gravity and is denoted by the symbol g. The vectorg points down toward the earth, and, at the surface ofthe earth, has a magnitude.

g ≡ g = 980 cm/sec2 (13)

This is quite consistent with our experimental resultof about 1000 cm/sec2 that we got from the analysisof the strobe photograph in Figure (10).

If we go up away from the earth, the acceleration dueto gravity decreases. At an altitude of 1,600 miles,the acceleration is down to half its value, about500 cm/sec2. On other planets g has different val-ues. For example, on the moon, g is only about 1/6as strong as it is here on the surface of the earth, i.e.

gmoon = 167 cm/sec2 (14)

From the relationship we have seen between forceand acceleration we can understand why a projectilethat goes only a few feet above the surface of theearth should have a constant acceleration. Thegravitational force does not change much in thosefew feet, and therefore we would not expect theacceleration caused by gravity to change much ei-ther.

On the other hand there is no obvious reason, at thispoint, why in the absence of air, a steel ball and afeather should have the same acceleration. Galileobelieved that all projectiles, in the absence of airresistance, have the same acceleration. But it wasnot until Newton discovered both the laws of me-chanics (the relationship between acceleration andforce) and the law of gravity, that it became aphysical prediction that all projectiles have the samegravitational acceleration.

In the early part of the 20th century, Einstein went astep farther than Newton, and used the fact that allobjects have the same gravitational acceleration todevelop a geometrical interpretation of the theory ofgravity. The gravitational force was reinterpreted asa curvature of space, with the natural consequencethat a curvature of space affects all objects in thesame way. This theory of gravity, known asEinstein’s general theory of relativity, was a resultof Einstein’s effort to make the theory of gravityconsistent with the principle of relativity.

It is interesting how the simplest ideas, the principleof relativity, and the observation that the gravita-tional acceleration is the same for all objects, are thecornerstones of one of the most sophisticated theo-ries in physics, in this case Einstein’s general theoryof relativity. Even today, over three quarters of acentury since Einstein developed the theory, we stilldo not understand what many of the predictions orconsequences of Einstein’s theory will be. It isexciting, for these predictions may help us under-stand the behavior of the universe from its verybeginning.

acceleration due togravity at the sur-face of the earth

3-22 Description of Motion

Exercise (7)The first earth satellite, Sputnik 1, traveled in a low,nearly perfect, circular orbit around the earth asillustrated in Figure (29).

(a) What was the direction of Sputnik 1’s accelerationvector as it went around the earth?

(b) What was the direction of the force of gravity onSputnik 1 as the satellite went around the earth?

(c) How is this problem related to the problem of themotion of the golf ball on the end of a string? Give ananswer that your roommate, who has not had aphysics course, would understand.

Projectile Motionwith Air ResistanceBack to a more mundane subject, we wish to end thisdiscussion of acceleration with the example of pro-jectile motion with air resistance. Most introductoryphysics texts avoid this topic because they cannotdeal with it effectively. Using calculus, one canhandle only the simplest, most idealized examples,and even then the analysis is beyond the scope ofmost texts. But using strobe photographs it is easyto analyze projectile motion with air resistance, andwe learn quite a bit from the results.

What turned out to be difficult, was to find anexample where air resistance affected the motion ofa projectile enough to produce a noticeable effect.We found that a golf ball and a ping pong ball havealmost the same acceleration when thrown in the air,despite the considerable difference in weight ormass. Only when we used the rough surfacedStyrofoam balls used for Christmas tree ornamentsdid we finally get enough air resistance to give asignificant effect.

Sputnik 1 orbit

EARTH

Figure 29Sputnik 1's circular orbit.

Figure 30aMotion of a Styrofoam ball. This is the lightest ball wecould find.

3-23

A strobe photograph of the projectile motion of theStyrofoam ball is seen in Figure (30a), and ananalysis showing the resulting acceleration vectorsin Figure (30b). In Figure (30b) we have also drawnthe acceleration vectors g that the ball would havehad if there had not been any air resistance. We seethat the effect of air resistance is to bend back andshorten the acceleration vectors.

Figure (31) is a detailed analysis of the Styrofoam’sacceleration at point (3). (We used an enlargementof the strobe photograph to improve the accuracy ofour work, such detailed analysis is difficult usingsmall Polaroid photographs.) In Figure (31) v3 is thevelocity of the ball, g is the acceleration due togravity, and a3 the ball’s actual acceleration. Thevector aair, which represents the change in a causedby air resistance is given by the vector equation

a3 = g + aair (15)

The important feature of Figure (31) is that aair isoppositely directed to the ball’s velocity v3. Tounderstand why, imagine that you are the stick

figure riding on the ball in Figure (31). You will feela wind in your face, a wind directed oppositely to v3.This wind will push on the ball in the directionopposite to v3, i.e., in the direction of aair. Thus weconclude that the acceleration aair is created by theforce of the wind on the ball.

What we learn from this example is that if we havetwo forces simultaneously acting on an object,each force independently produces an accelera-tion, and the net acceleration is the vector sum ofthe independent accelerations. In this case theindependent accelerations are caused by gravity andthe wind. The net acceleration a3 of the ball is givenby the vector Equation (15), a3 = g + aair. As wewill see in later chapters, this vector addition ofaccelerations plays a fundamental role in mechan-ics.

-1) ( 5.2, 94.9)

0) (24.0, 101.4) 1) (40.8, 97.8)

2) (56.5, 85.3)

3) (70.8, 64.7) 4) (83.4, 37.1)

5) (95.2, 3.9)

Figure 30bAcceleration of the Styrofoam ball.

3

4

2

10

5

g

g

a0 a1

a2

a3

a4

3

"wind"

v3

ag

air

aair = –Kv

a3

Figure 31When we do a detailed comparison of a andg at point 3, we see that the air resistanceproduces an acceleration aair that points in thedirection of the wind felt by the ball.

3-24 Description of Motion

INSTANTANEOUS VELOCITYIn calculus, instantaneous velocity is defined bystarting with the equation vi = Ri+1 - Ri /∆ t andthen taking the limiting value of vi as we use shorterand shorter time steps ∆ t. This corresponds in astrobe photograph to using a higher and higherflashing rate which would give increasingly shortdisplacement vectors Si. In the end result onepictures the instantaneous velocity being defined ateach point along the continuous trajectory of theobject.

The effect of using shorter and shorter ∆ t is illus-trated in Figure (32). In each of these sketches thedotted line represents the smooth continuous trajec-

tory of the ball. In Figure (32a) where ∆ t = 0.4 secand there are only two images the only possibledefinition of v0 is the displacement between theseimages, divided by ∆ t as shown. Clearly ∆ t is toolarge here for an accurate representation of the ball’smotion.

A better description of motion is obtained in Figure(32b) where∆ t = 0.1 sec as in the original photo-graph. We used this value of in our analysis of theprojectile motion, Figure (10). Reducing ∆ t byanother factor 1/4 gives the results shown in Figure(32c). At this point the images provide a detailedpicture of the path and v0 = S0/∆ t is now tangent tothe path at (0). A further decrease in ∆ t wouldproduce a negligible change in v0.

1

=

∆t = 0.4 Sec

∆t

0

v0

v0S 0

S 0

10

=

∆t = 0.1 Sec

∆t

v0

v0S 0

S 0

10

=

∆t = 0.025 Sec

∆t

v0

v0S 0

S 0

~

~ =

∆t 0 Sec

instantaneousvelocity

∆tvi

vi

S i

Figure 32We approach the instantaneous velocityas we make ∆∆t smaller and smaller.

a)

b)

c)

d)

3-25

The instantaneous velocity at point (0) is the finalvalue of v0, the value illustrated in Figure (32d)which no longer changes as ∆ t is reduced. This isan abstract concept in that we are assuming such afinal value exists. We are assuming that we alwaysreach a point where using a stroboscope with a stillhigher flashing rate produces no observable changein the value of v0. This assumption, which hasworked quite well in the analysis of large objectssuch as ping pong balls and planets, has proven to befalse when investigated on an atomic scale. Accord-ing to the quantum theory which replaces classicalmechanics on an atomic scale when one uses asufficiently short ∆ t in an attempt to measure veloc-ity, the measurement destroys the experiment ratherthan giving a better value of v0.

3-26 Description of Motion

Instantaneous Velocityfrom a Strobe PhotographIn the case of projectile motion (i.e., motion withconstant acceleration) there is a simple yet precisemethod for determining an object’s instantaneousvelocity vi from a strobe photograph. (Vectorsrepresenting instantaneous velocity will be under-lined in order to distinguish them from the vectorsrepresenting the strobe definition of velocity.) Thismethod, which also gives quite good approximatevalues for other kinds of motion, will be used in ourcomputer calculations for determining the initialvelocity of the object.

To see what the method is, consider Figure (33)where we have drawn the vector obtained fromFigure (32d). We have also drawn a line from thecenter of image (–1) to the center of image (+1) andnotice that vi ∆t is parallel to and precisely half aslong as this line. Thus we can construct vi ∆t byconnecting the preceding and following images andtaking half of that line.

The vector constructed by the above rule is actuallythe average of the preceding velocity vector v–1 andthe following vector v0.

vi

=v–1 + v0

2(16)

as illustrated in Figure (33). (Note that the vectorsum v–1 + v0 ∆t is the same as the line 2v i ∆twhich connects the preceding and following image.)This is a reasonable estimate of the ball’s instanta-neous velocity because v–1 is the average velocityduring the time ∆ t before the ball got to (0), and v0the average velocity during the interval after leaving(0). The ball’s velocity at (0) should have a valueintermediate between v–1 and v0, which is whatEquation (16) says.

The constant acceleration formula

S = vit + 12 at2 (17)

which may be familiar from a high school physicscourse, provides a direct application of the conceptof instantaneous velocity. (Remember that this is nota general formula; it applies only to motion withconstant acceleration where the vector a changesneither in magnitude or direction.) As illustrated inFigure (34) the total displacement of the projectile

Figure 33

iV~t

Vi~ t + atS = 2

S

212at

t = 3∆t

_

12_

Figure 34Illustration of the constant accelerationformula as a vector equation.

Vi~∆t

2

Vi~∆t

= ( )V–1 V0+Vi~

Vi~ ∆t2 ( )∆tV–1 V0+=

( )∆

tV–1

V0+

∆t

V-1

V0∆t

2

3-27

during a time t (here t = 3∆t) is the vector sum of vi tand 1/2at2. To draw this figure, we used

vi t = vi 3∆t = 3 vi∆t

and obtained v i ∆t from our method of determininginstantaneous velocity. We also used

12 a t2 = 1

2 a 3∆t2

= 92 a∆t2

where we obtained a ∆t2 from the relation

a =

S2 - S1

∆t2

a∆t2 = S2 - S1

as illustrated in Figure (35).

For these exercises, use the tear out sheet onpages 3-31,32.

Exercise 8Use Equation 17 to predict the displacements of theball

(a) Starting at position (0) for a total time t = 4∆t.

(b) Starting at position (1) for a total time t = 3∆t.

Do the work graphically as we did in Figures 33-35.

Exercise 9The other constant acceleration formula is

v f = v i +at

where v i is the initial velocity, and v f the object’svelocity a time t later. Apply this equation to Figure 10to predict the ball’s instantaneous velocity v f at point(3) for a ball starting at point (0). Check your predic-tion by graphically determining the instantaneousvelocity at point (3).

Show your results on graph paper.

Exercise 10

Show that the constant acceleration formulas wouldcorrectly predict projectile motion even if time ranbackward. (For example, assume that the ball wentbackward as shown in Figure (36), and repeat Exer-cise 8b, going from position 3 to position 0.)

Vi~t =

Vi~∆t

Vi~(3 ∆t)

2 a t 2 a(3∆t) 2a ∆t= 4.5=

2a ∆t4.5

2a∆t

12_ 1

2_

S2S1–

Figure 35How to construct the vectors vit and

1 21 2 at2 from a strobe photograph.

appearence of ball moving backward in time

0

3

Figure 36Run the motion of the ball backward in time, andit looks like it was launched from the lower right.

3-28 Description of Motion

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3-30 Description of Motion

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3-32 Description of Motion

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3-33

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3-35

IndexAAcceleration

Definition of 3-13Due to gravity 3-21From a strobe photograph 3-15Intuitive discussion 3-20Uniform circular motion

Direction of 3-18Magnitude of 3-18

Vector, definition of 3-15Air Resistance

Strobe analysis for projectile motion 3-22

CCircular motion

UniformIntroduction to 3-17Magnitude of acceleration 3-18

Coordinate vectorDefinition of 3-11In definition of velocity vector 3-13

DDescription of motion 3-3Direction of time

And strobe photographs 3-27Displacement vectors

From strobe photos 3-5

EExperiments I

- 1- Graphical analysis of projectile motion 3-17

GGraph paper

For graphical analysis 3-33For projectile motion 3-29

Graphical analysisOf instantaneous velocity 3-26Of projectile motion 3-17Of projectile motion with air resistance 3-22X. See Experiments I: - 1- Graphical analysis of

projectile motionGravity

Acceleration due to 3-21

IInstantaneous velocity

Definition of 3-24From strobe photograph 3-26

MMotion

Description of, chapter on 3-3Uniform circular motion 3-17

PProjectile motion

Analysis ofGraphical 3-16

Constant acceleration formulasGraphical analysis of 3-26

Determining acceleration for 3-16Graph paper tear out pages 3-29Instantaneous velocity 3-24Strobe photograph of 3-7With air resistance

Graphical analysis 3-22

SStrobe photographs

Analyzing 3-8, 3-11Defining the acceleration vector 3-15Defining the velocity vector 3-11Taking 3-7

TTime, direction of

And strobe photographs 3-27

UUniform circular motion. See Circular motion

VVector

Measuring length of 3-9Velocity from coordinate vector 3-13

VectorsCoordinate 3-11Displacement

from Strobe Photos 3-5Graphical addition and subtraction 3-10

Velocity 3-11Instantaneous 3-24

From strobe photograph 3-26Using strobe photos 3-11

Velocity vector from coordinate vector 3-13

XX-Ch 3

Exercise 1 3-10Exercise 2 3-10Exercise 3 3-10Exercise 4 3-12Exercise 5 3-17Exercise 6 3-18Exercise 7 3-22Exercise 8 3-27Exercise 9 3-27Exercise 10 3-27