Ch 2 1 Dim Motion

MOTION ALONG ASTRAIGHT LINE 2

Blasting down a track in a dragster is an exhilaratingexample of straight-line motion, but what exactly, besides

the noise, thrills the driver?

16 CHAPTER 2 MOTION ALONG A STRAIGHT LINE

2-1 MOTION

The world, and everything in it, moves. Even seem-ingly stationary things, such as a roadway, move withthe Earth's rotation, the Earth's orbit around thesun, the sun's orbit around the center of the MilkyWay, and the galaxy's migration relative to other gal-axies. The classification and comparison of motions(called kinematics) are often challenging. What ex-actly do you measure, and how do you compare?

Here are two examples of motion. In 1977,Kitty O'Neil set records for "terminal speed" and"elapsed time" for a dragster on a 440-yd run. Froma standstill, she reached 392.54 mi/h (about 632.1km/h) in a sizzling time of 3.72 s. In 1958, Eli Beed-ing, Jr. rode a rocket sled when it was shot along atrack from a standstill to a speed of 72.5 mi/h(= 117 km/h) in the fantastic time of 0.04 s (lessthan the blink of an eye). How can we compare thetwo motions to see which must have been the morethrilling (or more frightening) - by final speeds, byelapsed time, or by some other quantity?

Before we attempt an answer, we first examinesome general properties of motion that is restrictedin three ways.

1. The motion is along a straight line only. The linemay be vertical (a falling stone), horizontal (a car ona level highway), or slanted, but it must be straight.

2. The cause of the motion will not be specifieduntil Chapter 5. In this chapter you study only themotion itself. Does the object speed up, slow down,stop, or reverse direction; and, if the motion doeschange, how is time involved in the change?3. The moving object is either a particle (a pointlikeobject such as an electron) or an object that moveslike a particle (every portion moves in the same di-rection and at the same rate). A pig slipping downa straight playground slide might be considered tobe moving like a particle; however, a rotating play-ground merry-go-round would not because differentpoints around its rim move in different directions.

2-2 POSITION AND DISPLACEMENT

To locate an object means to find its position relativeto some reference point, often the origin (or zeropoint) of an axis such as the x axis in Fig. 2-1. The

Positive directionNegative direction +----...

-3 -2 -1 0

Origin.J

2 3 4 5

FlGURE 2·1 Position is determined on an axis that ismarked in units of length and that extends indefinitely inopposite directions.

positive direction of the axis is in the direction ofincreasing numbers, which is toward the right as thefigure is drawn. The opposite direction is the nega-tive direction.

For example, a particle might be located at x =

5 m, which means that it is 5 m in the positive direc-tion from the origin. Were it at x = - 5 m, it wouldbe just as far from the origin but in the oppositedirection.

A change from one position Xl to another posi-tion x2 is called displacement Ilx, where

(2-1)

(The symbol 11, which represents a change in aquantity, means that the initial value of that quantityis to be subtracted from the final value.) When num-bers are inserted for the position values, a displace-ment in the positive direction (toward the right inFig. 2-1) always comes out positive, and one in theopposite direction (left in the figure) negative. Forexample, if the particle moves from Xl = 5 m toX2 = 12 m, then Ilx = (12 m) - (5 m) = + 7 m.The plus sign indicates that the motion is in the pos-itive direction. If we ignore the sign (and thus thedirection), we have the magnitude of Ilx, which is7 m. If the particle then returns to X = 5 m, the dis-placement for the full trip is zero. The actual num-ber of meters covered is immaterial; displacementinvolves only the original and final positions.

Displacement is an example of a vector quantity,which is a quantity that has both a direction and amagnitude. We explore vectors more fully in Chap-ter 3 (in fact, some of you may have already read thatchapter), but here all we need is the idea that dis-placement has two features: (1) its magnitude is thedistance (such as the number of meters) betweenthe original and final positions, and (2) its directionon an axis, from an original position to a final posi-tion, is represented by a plus or minus sign.

2-3 AVERAGE VELOCITY ANDAVERAGE SPEED

A compact way to describe position is with a graph ofposition xplotted as a function of time t-a graph ofx( t). As a simple example, Fig. 2-2 shows x( t) for ajack rabbit (which we treat as a particle) that is sta-tionary at x = - 2 m.

Figure 2-3a, also for a rabbit, is more interest-ing, because it involves motion. The rabbit is appar-ently first noticed at t = 0 when it is at the positionx = - 5 m. It moves toward x = 0, passes through

!------+---+-+-!-----!-----!---- t (s)

FIGURE 2-2 The graph of x( t) for a jack rabbit that isstationary at x = - 2 m. The value of x is - 2 m for alltimes t.

I---+----:+---+--+-~~-+_-----'t (s)

(a)

~k ,ji!, ~,/}~>~;"---------""~;:,,,~,.~(-"--~:,,,;,?;-x (m)

~ 0 2o 3 4 Time t (s)

(b)

FIGURE 2-3 (a) The graph of x(t) for a moving jackrabbit. (b) The path associated with the graph. The scalebelow the x axis shows the times at which the rabbitreaches various x values.

2-3 AVERAGE VELOCITY AND AVERAGE SPEED 17

If a member of this troupe were now to march back towhere she started, her average velocity would be zero, be-cause her net displacement would then be zero.

that point at t = 3 s, and then moves on to increas-ingly larger positive values of x.

Figure 2-3b depicts the actual motion of the rab-bit and is something like what you would see. Thegraph is more abstract and quite unlike what yousee, but it is richer in information. It also reveals howfast the rabbit moves. Several quantities are asso-ciated with the phrase "how fast." One of them isthe average velocity V, which is the ratio of the dis-placement Lix that occurs during a particular timeinterval Lit to that interval:*

(2-2)

On a graph of x versus t, v is the slope of the straightline that connects two points on the x(t) curve: onepoint corresponds to X2 and t2, and the other pointcorresponds to Xl and t1. Like displacement, v hasboth magnitude and direction. (Average velocity isanother example of a vector quantity.) Its magni-tude is the magnitude of the line's slope. A positivev (and slope) tells us that the line slants upwardtoward the right; a negative v (and slope), that theline slants upward to the left. The average velocityalways has the same sign as the displacement be-cause Lit is a positive number.

*In this book, a bar over a symbol usually means an average valueof the quantity that the symbol represents.


FIGURE 2-4 Calculations of average velocity betweent = 1 sand t = 4 s.

Figure 2-4 shows the calculation of v for therabbit of Fig. 2-3 for the time interval t = 1 s to t =

4 s. The average velocity during that time interval isv = + 6 m/3 s = + 2 mz's, which is the slope of thestraight line that connects the point on the curve atthe beginning of the interval with the point on thecurve at the end of the interval.

SAMPLE PROBLEM 2-1

You drive a beat-up pickup truck down a straight roadfor 5.2 mi at 43 mi/h, at which point you run out offuel. You walk 1.2 mi farther, to the nearest gas station,in 27 min (= 0.450 h). What is your average velocityfrom the time that you started your truck to the timethat you arrived at the station? Find the answer bothnumerically and graphically.

SOLUTION To calculate v we need your displacement!:!.x, from start to finish, and the elapsed time !:!.t. As-sume, for convenience, that your starting point is at theorigin of an x axis (so Xl = 0) and that you move in thepositive direction. You end up at X2 = 5.2 mi +1.2 mi = + 6.4 mi, and so, !:!.x = X2 - Xl = + 6.4 mi.To get the driving time, we rearrange Eq. 2-2 and insertthe data about the driving:

!:!.x 5.2 mi!:!.t = -=- = = 0.121 h.

v 43 mi/h

So the total time, start to finish, is

!:!.t = 0.121 h + 0.450 h = 0.571 h.

x

Stalled

Gasstation 761----t---

p

FIGURE 2-5 Sample Problem 2-1. The lines marked"Driving" and "Walking" are the position-time plotsfor the driver-walker in Sample Problem 2-1. Theslope of the straight line joining the origin and point Pis the average velocity for the trip.

Finally, we insert !:!.x and !:!.t into Eq. 2-2:

_ !:!.x 6.4 miv = - = --- "" + 11 mi/h. (Answer)!:!.t 0.571 h

To find v graphically, we must first plot x( t), as inFig. 2-5 where the start and finish points on the curveare the origin and P, respectively. Your average velocityis the slope of the straight line connecting those points.The dashed lines show that the slope also gives v =6.4 mi/0.57 h = + 11 mi/h.

SAMPLE PROBLEM 2-2

Suppose that you next carry the fuel back to the truck,making the return trip in 35 min. What is your averagevelocity for the full journey, from the start of your driv-ing to your arrival back at the truck with the fuel?

SOLUTION As previously, we must find your displace-ment !:!.x from start to finish and then divide it by thetime interval Az between start and finish. In this prob-lem, however, the finish is back at the truck. Youstarted at Xl = O. Back at the truck you are at positionX2 = 5.2 mi. And so !:!.X is 5.2 - 0 = 5.2 mi. The totaltime !:!.t you take in going from start to finish is

5.2 mi!:!.t = + 27 min + 35 min43 mi/h

= 0.121 h + 0.450 h + 0.583 h = l.l5 h.

-----------------------------------------~--------------------------------------------------------------------------------

So

_ Llx 5.2 miv = A = -5 h = + 4.5 mi /h. (Answer)ut 1.1

This is slower than the average velocity computed inSample Problem 2-1 because here the displacement issmaller and the time interval longer.

Average speed s is a different way of describing"how fast" a particle moves. Whereas average veloc-ity involves the particle's displacement ,lx, the aver-age speed involves the total distance covered (forexample, the number of meters run), independentof direction. That is,

total distance,It (2-3)s=

Average speed also differs from average velocity inthat it does not include direction and thus lacks anyalgebraic sign. Sometimes s is the same (except forthe absence of a sign) as v. But, as demonstrated inSample Problem 2-3, when an object doubles backon its path, the results can be quite different.

SAMPLE PROBLEM 2-3

In Sample Problem 2-2, what is your average speed?

SOLUTION From the beginning of your drive to yourreturn to the truck with the fuel, you covered a total of5.2 mi + 1.2 mi + 1.2 mi = 7.6 mi, taking 1.15 h, andso

_ 7.6 mis = -- = 6.6 mi /h.

1.15 h(Answer)

PROBLEM SOLVING

--'M,.....--TACTIC 1: READ THE PROBLEM CAREFULLY

For beginning problem solvers, no difficulty is morecommon than simply not understanding the problem.The best test of understanding is this: Can you explainthe problem, in your own words, to a friend? Give it atry.

TACTIC 2: UNDERSTAND WHAT IS GIVENAND WHAT IS REQUESTED

Write down the given data, with units, using thesymbols of the chapter. (In Sample Problems 2-1 and

2-3 AVERAGEVELOCITYAND AVERAGESPEED 19

2-2, the given data allow you to find your net dis-placement Llx and the corresponding time interval Llt.)Identify the unknown and its symbol. (In these sampleproblems, the unknown is your average velocity, symbolv.) Then find the connection between the unknownand the data. (The connection is Eq. 2-2, the definitionof average velocity.)

TACTIC 3: WATCH THE UNITSBe sure to use a consistent set of units when puttingnumbers into the equations. In Sample Problems 2-1and 2-2, which involve a truck, the logical units interms of the given data are miles for distances, hoursfor time intervals, and miles per hour for velocities. Youmay need to make conversions.

TACTIC 4: THINK ABOUT YOUR ANSWERLook at your answer and ask yourself whether it makessense. Is it far too large or far too small? Is the signcorrect? Are the units appropriate? In Sample Problem2-1, for example, the correct answer is 11 miz'h. If youfind 0.00011 miz h, -11 miz h, 11 mils, or 11,000 milh, you should realize at once that you have done some-thing wrong. The error may lie in your method, in youralgebra, or in your arithmetic. Check the problem care-fully, being sure to start at the very beginning.

In Sample Problem 2-1, your answer must begreater than your speed of walking (2-3 mi/h) but lessthan the speed of the truck (43 miz'h). Finally, the an-swer to Sample Problem 2-2 must be less than that toSample Problem 2-1 for two reasons: the displace-ment's magnitude is smaller in 2-2 and the time re-quired is longer.

TACTIC 5: READING A GRAPHFigures 2-2, 2-3a, 2-4, and 2-5 are examples of graphsthat you should be able to read easily. In each graph,the variable on the horizontal axis is the time t, the di-rection of increasing time being to the right. In each,the variable on the vertical axis is the position x of themoving particle with respect to the origin, the directionof increasing x being upward.

Always note the units (seconds or minutes; meters,kilometers, or miles) in which the variables are ex-pressed, and note whether the variables are positive ornegative.

TACTIC 6: SIGNIFICANT FIGURESIf you were going to divide 137 jelly beans among 3people, you would not think of giving each person ex-actly 137/3 or 45.66666666 ... beans. You would giveeach person 45 beans and draw straws to see who wouldnot get one of the remaining two. We need to developthat same kind of common sense in dealing with nu-merical calculations in physics.

In Sample Problem 2-1, for example, if you calcu-late the average velocity with your calculator wide open,you get v = 11.20840631 miy'h. This number has 10


significant figures. The original data in the problem haveonly tw.osignificant figures.

In general, no final result should have more signif-icant figures than the original data from which itwas derived.

If multiple steps of calculation are involved, you shouldretain more significant figures than the original datahave. However, when you come to the final result, youshould round off according to the original data withthe least significant figures. We did that in SampleProblem 2-1 to get 11 = 11 mi/h. (Hereafter, the an-swer to a sample problem might be presented with thesymbol = instead of =, but rounding off may still beinvolved.)

It is hard to escape the feeling that you are throw-ing away good data when you round off in this way but,in fact, you are doing the opposite; you are throwingaway useless and misleading numbers. You may be ableto set your calculator to do this rounding for you. Re-gardless of how you set it, your calculator continues tocompute wide open internally, displaying only therounded result that you ask it to show you.

When a number such as 3.15 or 3.15 X 103 is pro-vided in a problem, the number of significant figures isapparent. But how about the number 3000? Is it knownto only one significant figure (it could be written as3 X 103)? Or is it known to as many as four significantfigures (it could be written as 3.000 X 103)? In thisbook, we assume that all the zeros in such providednumbers as 3000 are significant, but you had better notmake that assumption elsewhere.

TABLE 2-1THE LIMITING PROCESS

TACTIC 7: SIGNIFICANTFIGURESANDDECIMALPLACES

Don't confuse these. Consider the lengths 35.6 m,3.56 m, 0.356 m, and 0.00356 m. They all have threesignificant figures but, in sequence, they have one, two,three, and five decimal places.

2-4 INSTANTANEOUS VELOCITYAND SPEED

You have now seen two ways to describe how fastsomething moves: average velocity and averagespeed, both of which are measured over a time in-terval Az. But the phrase "how fast" more commonlyrefers to how fast a particle is moving at a giveninstant-its instantaneous velocity v (or simply ve-locity) .

The velocity at any instant is obtained from theaverage velocity by shrinking the time interval I:ltcloser and closer to O.As I:lt dwindles, the averagevelocity approaches a limiting value, which is the ve-locity at that instant:

I:lx dxv= lim -=-

tit->O I:lt dt(2-4)

Velocity is another vector and thus has an associateddirection.

Table 2-1 shows an example of the limiting pro-cess. The first column gives the position x of a parti-

BEGINNINGPOINT ENDPOINT INTERVALS VELOCITY

XI (m) tl (s) "2 (m) o (s) L1x (m) L1t (s) L1x/ L1t (m/s)

5.00 1.00 9.00 3.00 4.00 2.00 + 2.05.00 1.00 8.75 2.50 3.75 1.50 +2.55.00 1.00 8.00 2.00 3.00 1.00 + 3.05.00 1.00 6.75 1.50 1.75 .!2 0.50 +3.5

<=:

5.00 1.00 5.760 1.200 0.760.;::

0.200 +3.8il5.00 1.00 5.388 1.100 0.388 <i 0.100 + 3.95.00 1.00 5.196 1.050 0.196 0.050 +3.95.00 1.00 5.158 1.040 0.158 0.040 +4.0 } a limiting

5.00 1.00 5.119 1.030 0.119 0.030 +4.0value hasbeen reached

cle at t = 1 s, which is when a time interval D..tbegins. The third and fourth columns give the valuesof x and t at the end of D..t. And the fifth and sixthcolumns give the displacement D..x and the intervalD..t (which we are shrinking). As D..t shrinks, v (= D..x/D..t, in the last column) gradually changes until itreaches a limiting value of + 4.0 m/s. That value isthe instantaneous velocity vat t = 1 s.

In the language of calculus, the instantaneousvelocity is the rate at which a particle's position x ischanging with time at a given instant. According toEq. 2-4, the velocity of a particle at any instant is theslope of its position curve at the point representingthat instant.

Speed is the magnitude of a velocity; that is,speed is velocity that has been stripped of any indi-cation of direction, either in words or via an alge-braic sign.* A velocity of + 5 m/s and one of - 5m/ s both have an associated speed of 5 m/ s. Thespeedometer in a car measures the speed, not thevelocity, because it cannot ascertain anything aboutthe direction of motion.

SAMPLE PROBLEM 2-4

Figure 2-6a is an x(t) plot for an elevator cab that isinitially stationary, then moves upward (which we taketo be the positive direction), and then stops. Plot v(t)as a function of time.

SOLUTION The slope, and so also the velocity, is zeroin the intervals containing points a and d, when the cabis stationary. During the interval be the cab moves with aconstant velocity, arid the slope of x(t) is

ilx 24 m - 4.0 mv=-= = +4.0m/s.

ilt 8.0 s - 3.0 s

The plus sign indicates that the cab is moving in thepositive x direction. These values are plotted in Fig.2-6b. In addition, as the cab initially begins to move andthen later slows to a stop, v varies as indicated in theintervals 1 s to 3 sand 8 s to 9 s. (Figure 2-6e is consid-ered later.)

Given a v(t) graph such as Fig. 2-6b, we could"work backward" to produce the shape of the asso-ciated x(t) graph (Fig. 2-6a). However, without addi-tional information, we would not know the actual

*Speed and average speed can be quite different, so you must becareful solvingproblems that involveeither quantity.

2-4 INSTANTANEOUS VELOCITY AND SPEED 21

Ii -25'----~~--~---4---X~=~2~4+m--~,i...~---6II_c~~~~-=-~a:

at t=8.0s V ;~ 20 -I

S I I Vi::¥ 15 ! I~l' -:-ct:,.-x-+----l

~ 10 _ ! x~r'\ ~ , :---+---1x- 4.0 m I /( "I::"i'J/tb+-+~-+-j---+---1, t

o 1 2 345 6 7 8Time (8)

(a)

9

v

4

"'<,

5 3c-'0 20

~

00

Slopeof x(t)

s:I 1\

I---+--+il ,-- I 1 j I \ ~,/\' I . +-+-7-\+-1v, j,--f---1-- ! i ; u I

a , ! \ I Iii I ~ d t2 345 6 7 8 9

Time (s)

a

Slopeof v(t)

(b)

3_Acc~~e~ra~ti~·~'~ln~4-__ +_'__ + ~_. ! + "_..c

~ 2 ' '\ IIS 1~--l~-+---I----l-~+---~--~-+---+--~§ 0 a i b aCt) d,'~ -1 2 8 19vOJ -2 ~---!---+-v< -31·---+·----··--+---- t. +1··············+1··········,·,,···+,··-- 1----1---+ 1

-4 I -f --+-I Decele :ationl---.-/

(c)

FIGURE 2-6 Sample Problem 2-4. (a) The x(t) curvefor an elevator cab that moves upward along an x axis.(b) The v(t) curve for the cab. Note that it is the deriva-tive of the x( t) curve (v = dx/ dt). (c) The a(t) curvefor the cab, It is the derivative of the v(t) curve (a =dv/ dt). The sketches suggest how a passenger's bodymight respond to the accelerations.


values for x, because the v( t) graph indicates onlychanges in x. To find the change in x during any inter-val, we must, in the language of calculus, calculate thearea "under the curve" on the v(t) graph for the sameinterval. For example, during the interval in which thecab has a velocity of 4.0 ml s, the change in x is given bythe "area" under the v(t) curve:

area = (4.0 mls)(8.0 s - 3.0 s) = + 20 m.

(This area is positive because the v(t) curve is abovethe t axis.) Figure 2-6a shows that x does indeed in-crease by 20 m in the interval.

The instantaneous velocity of one of these speedboats isits velocity at the instant the photograph was taken. Theboat may have had a different instantaneous velocity be-fore or after that instan t.

SAMPLE PROBLEM 2-5

The position of a particle moving on the x axis is givenby

x = 7.8 + 9.2t - 2.1t3. (2-5)

What is its velocity at t = 3.5 s? Is the velocity constant,or is it continuously changing?

SOLUTION For simplicity, the units have been omit-ted but you can insert them if you like by changing thecoefficients to 7.8 m, 9.2 rn/s, and - 2.1 m/s3. To solvethe problem, we use Eq. 2-4 with the right side of Eq.2-5 substituted for x:

dx dv = - = - (7.8 + 9.2t - 2.1t3),

dt dt

which becomes

v = 0 + 9.2 - (3) (2.1) t2 = 9.2 - 6.3t2• (2-6)

At t = 3.5 s,

v = 9.2 - (6.3) (3.5)2 = - 68 m/s. (Answer)

At t = 3.5 s, the particle is moving toward decreasing x(note the minus sign) with a speed of 68 m/s. Since thequantity t appears in Eq. 2-6, the velocity v depends ont and so is continuously changing.

PROBLEM SOLVING

--t...N\...---TACTIC 8: NEGATIVENUMBERS

The line below is an x axis with its origin (x = 0) at thecenter. Using this scale, make sure you understandthat, for example, - 40 m is less than - 10 m and thatboth are less than 20 m. Note also that 10 m is greaterthan -30 m.

The four arrows pointing to the right all representincreases in x, that is, positive values for .:lx, the changein x. The four arrows pointing to the left represent de-creases in x, that is, negative values for .:lx.

-50 -40 -30 -20 -10 0 10 20 30 40 50Position (m)

TACTIC 9: DERIVATIVESAND SLOPESEvery derivative is the slope of a curve. In Sample Prob-lem 2-4, for example, the velocity of the cab at any in-stant (a derivative; see Eq. 2-4) is the slope of the x(t)

x

5

4

~c 3.g'§P-<

2

Time (s)

fiGURE 2-7 The derivative of a curve at any point isthe slope of its tangent line at that point. At t = l.0 s,the slope of the tangent line (and thus dx/ dt, the in-stantaneous velocity) is t:.x/ t:.t = + 2.1 m/s.

curve of Fig. 2-6a at that instant. Here's how you canfind a slope (and thus a derivative) graphically.

Figure 2-7 showsan x( t) plot for a moving particle.To find the velocityof the particle at t = 1 s, put a doton the curve at that point. Then draw a line tangent tothe curve through the dot (tangent means touching; thetangent line touches the curve at a single point, thedot) judging carefully by eye. Then construct the righttriangle ABC. (Although the slope is the same no mat-ter what the size of this triangle, the larger the trianglethe more precise willbe your graphical measurement.)Find t:.x and t:.t, using the vertical and horizontal scalesto provide the magnitude, the unit, and the sign. InFig. 2-7 you find the slope (derivative) from the follow-ing equation:

t:.x 5.5 m - 2.3 m 3.2 mslope = - = = -- = +2.1 m/s.t:.t l.8 s - 0.3 s l.5 s

As Eq. 2-4 tells you, this slope is the velocityof the par-ticle at t = 1 s.

If you change the scale on the x or the t axis ofFig. 2-7, the appearance of the curve and the angle ()will change but the value you find for the velocity att = 1 s will not.

If you have a mathematical expression for thefunction x(t), as in Sample Problem 2-5, you can findthe derivative dx/ dt by the methods of calculus andavoid this graphical method.

2-5 ACCELERATION 23

2-5 ACCELERATION

When a particle's velocity changes, the particle issaid to undergo acceleration (or to accelerate). Theaverage acceleration a over an interval tJ..t is com-puted as

a=V2 - VI tJ..v

t2 - tl - Tt· (2-7)

The instantaneous acceleration (or simply accelera-tion) is the derivative of the velocity:

dva=- dt . (2-8)

In words, the acceleration of a particle at any instantis the rate at which its velocity is changing at thatinstant. According to Eq. 2-8, the acceleration at anypoint is the slope of the curve of v(t) at that point.

A common unit of acceleration is meter per sec-ond per second: m/(s' s) or m/s2. You will see otherunits in the problems, but they will each be in theform of distanceyt time- time) or distancey'time". Ac-celeration has both magnitude and direction (it isyet another vector quantity). The algebraic sign rep-resents the direction on an axis just as it does fordisplacement and velocity.

Figure 2-6c is a plot of the acceleration of thecab discussed in Sample Problem 2-4. Compare thecurve with the v(t) curve-each point on the a(t)curve is the derivative (slope) of the correspondingpoint on the v(t) curve. When v is constant (at ei-ther 0 or 4 m/ s), the derivative is zero and so also isthe acceleration. When the cab first begins to move,the v( t) curve has a positive derivative (the slope ispositive), which means that a( t) is positive. Whenthe cab slows to a stop, the derivative and slope ofthe v( t) curve are negative; that is, a( t) is negative.

Next compare the slopes of the v(t) curve dur-ing the two acceleration periods. The one associatedwith the cab's stopping (commonly called "deceler-ation") is steeper, because the cab stops in half thetime it took to get up to speed. The steeper slopemeans that the magnitude of the deceleration islarger than that of the acceleration, as indicated inFig.2-6c.

The sensations you would feel while riding inthe cab of Fig. 2-6 are indicated by the sketched fig-ures. When the car first accelerates, you feel as


FIGURE 2-8 Colonel J. P. Stapp in a rocket sled as it is brought up to highspeed (acceleration out of the page) and then very rapidly braked (accelerationinto the page).

though you are pressed downward; when later thecab is braked to a stop, you seem to be stretchedupward. In between, you feel nothing special. Yourbody reacts to accelerations (it is an accelerometer)but not to velocities (it is not a speedometer). Whenyou are in a car traveling at 60 mi/h or an airplanetraveling at 600 mi/h, you have no bodily awarenessof the motion. But if the car or plane quicklychanges velocity, you may become keenly aware ofthe change, perhaps even frightened by it. Part ofthe thrill of an amusement park ride is due to thequick changes of velocity that you undergo. A moreextreme example is shown in the photographs ofFig. 2-8, which were taken while a rocket sled wasrapidly accelerated and then rapidly braked to astop.

SAMPLE PROBLEM 2-6

a. When Kitty O'Neil set the dragster records for thegreatest speed and least elapsed time, she reached392.54 mi/h in 3.72 s. What was her average accelera-tion?

SOLUTION From Eq. 2-7, O'Neil's average accelera-tion was

_ !::.v 392.54 mi/h - 0a = Tt = 3.72 s - 0

rm= +106-

h·s'(Answer)

where the motion is taken to be in the positive x direc-tion. In more conventional units, her acceleration was47.1 m/s2. Often, large accelerations are expressed in"g" units, where 19 = 9.8 m/ s2 (= 32 ft/ s2), as will be

explained in Section 2-8. O'Neil's average accelerationwas 4.8g.

b. What was the average acceleration when Eli Beed-ing,jr. reached 72.5 mi/h in 0.04 s on a rocket sled?

SOLUTION Again from Eq. 2-7,

!::.v 72.5 mi/h - 0a=-=

!::.t 0.04 s - 0

mi= + 1.8 X 103 - "'" + 800 m/s2, (Answer)

h·s

or about 80g.Recall our question in Section 2-1, where O'Neil

and Beeding were introduced. How can we tell whohad the more thrilling ride-by final speeds, byelapsed times, or by some other quantity? You now cananswer that question. Because the human body sensesacceleration rather than speed, you should compareaccelerations, and so Beeding wins out, even thoughhis final speed was considerably slower than O'Neil's.In fact, Beeding's acceleration could have been lethalhad it continued for much longer.

PROBLEM SOLVING-_ .....W\.,..--

TACTIC 10: AN ACCELERATION'S SIGNLook again at the algebraic sign for the accelerationsthat are calculated in Sample Problem 2-6. In manycommon examples of acceleration, the sign has a com-mon-sense meaning: positive acceleration means thatthe speed of an object (such as a car) is increasing, andnegative acceleration means that the speed is decreas-ing (the object is undergoing deceleration).

2-6 CONSTANT ACCELERATION: A SPECIAL CASE 25

Such meanings cannot be interpreted withoutsome thought, however. For example, if a car with aninitial velocity v = - 27 m/s (= - 60 mi/h) is brakedto a stop in 5.0 s, a = + 5.4 m/s2. The acceleration ispositive, but the car has slowed. The reason is the differ-ence in signs: the direction of the acceleration is oppo-site that of the velocity.

Here then is a better way to interpret the signs: ifthe signs of the velocity and acceleration are the same,a particle picks up speed; if the signs are opposite, theparticle slows. The interpretation will have more mean-ing when we later explore the vector nature of velocityand acceleration.

SAMPLE PROBLEM 2-7

A particle's position is given by

x = 4 - 27t + t3,

where the units of the coefficients are m, mZs, and m/s3, respectively, and the x axis is shown in Fig. 2-1.

a. Find v( t) and a( t).

SOLUTION To get v( t), we differentiate x( t) with re-spect to t:

v = - 27 + 3t2• (Answer)

a = + 6t.

To get a( t), we differentiate v( t) with respect to t:

(Answer)

b. Is there ever a time when v = O?

SOLUTION Setting v( t) = 0 yields

0= - 27 + 3t2,

which has the solution t = ::'::3 s. (Answer)

c. Describe the particle's motion for t 2: O.

SOLUTION To answer, we examine the expressionsfor x(t), v(t), and a(t).

At t = 0 the particle is at x = + 4 m, is moving left-ward with a velocity of - 27 my's, and is, at that instant,not accelerating.

For 0 < t < 3 s, the particle continues to move tothe left, but at decreasing speed, because it is accelerat-ing to the right. (Check v(t) and a(t) for, say, t = 2 s.)The rate of the acceleration is increasing.

At t = 3 s, the particle stops momentarily (v = 0)and is as far to the left as it will ever get (x = - 50 m).It continues to accelerate to the right at an increasingrate.

For t> 3 s, its acceleration to the right continuesto increase, and its velocity, which is now also to theright, increases rapidly. (Note that the signs of v and amatch.) The particle moves to the right without bound.

2-6 CONSTANT ACCELERATION:A SPECIAL CASE

In many common types of motion, the accelerationis either constant or approximately so. For example,you might accelerate a car at an approximately con-stant rate when a traffic light turns from red togreen. (Graphs of your position, velocity, and accel-eration would resemble those in Fig. 2-9.) If youlater had to brake the car to a stop, the decelerationduring the braking might also be approximatelyconstant.

Such cases are so ubiquitous that a special set ofequations has been derived for dealing with them.One approach to the derivation of the equations isgiven in this section. A second approach is given inthe next section. Throughout both sections andlater when you work on the homework problems,keep in mind that the equations are valid only for con-


x

o(a)

x

(b)

'" a

.! I

"~ L- _o

a(t)

Slope = 0

(e)

FIGURE 2-9 (a) The position x(t) of a particle movingwith constant acceleration. (b) Its velocity v( t), given ateach point by the slope of the curve in (a). (c) Its (con-stant) acceleration, equal to the (constant) slope of vet).

stant acceleration (or situations in which you can approxi-mate the acceleration as being constant).

When the acceleration is constant, the distinc-tion between average acceleration and instanta-neous acceleration loses its meaning and we canwrite Eq. 2-7, with some changes in notation, as

v - Vaa=---t - 0 .

Here Vo is the velocity at time t = 0 and V is the ve-locity at any later time t. We can recast the equationas

V = Vo + at. (2-9)

As a check, note that this equation reduces to V = Vofor t = 0, as it must. As a further check, take the de-rivative of Eq. 2-9. Doing so yields dv/ dt = a, whichis the definition of a. Figure 2-9b shows a plot of Eq.2-9, the vet) function.

In similar manner we can rewrite Eq. 2-2 (with afew changes in notation) as

x = Xo + vt, (2-10)

in which Xo is the position of the particle at t = 0,and v is the average velocity between t = 0 and alater time t.

If you plot v against t using Eq. 2-9, a straightline results. Under these conditions, the average ve-locity over any time interval (say, t = 0 to a later timet) is the average of the velocity at the beginning ofthe interval (= vo) and the velocity at.the end of theinterval (= v). For the interval t = 0 to the later timet then, the average velocity is

v = Hvo + v). (2-11)

Substituting for v from Eq. 2-9 yields, after a littlerearrangement,

v = Vo + tat. (2-12)

Finally, substituting Eq. 2-12 into Eq. 2-10 yields

(2-13)

As a check, note that putting t = 0 yields x = xo, as itmust. As a further check, taking the derivative ofEq.2-13 yields Eq. 2-9, again as it must. Figure 2-9a is aplot ofEq. 2-13.

Five quantities can possibly be involved in anygiven problem regarding constant acceleration,namely, x - Xo, v, t, a, and Vo' Usually, one of thesequantities is not involved in the problem, either as agiven or as an unknown. We are then presented withthree of the remaining quantities and asked to findthe fourth.

Equations 2-9 and 2-13 each contain four ofthese quantities, but not the same four. In Eq. 2-9,the "missing ingredient" is the displacement, x -xo' In Eq. 2-13, it is the velocity v. These two equa-tions can also be combined in three ways to yieldthree additional equations, each of which involves adifferent "missing ingredient." Thus

v2 = VB + 2a(x - xo)· (2-14)

This equation is useful if we do not know t and arenot required to find it. We can, instead, eliminatethe acceleration a between these same two equationsto produce an equation in which a does not appear:

x - Xo = t(Vo + v) t.

Finally, we can eliminate vo, obtaining

x - Xo = vt - tat2•

(2-15)

(2-16)

Note the subtle difference between this equationand Eq. 2-13. One involves the initial velocity vo; theother involves the velocity vat time t.

TABLE 2-2EQUATIONS FOR MOTION WITH CONSTANTACCELERATIONa

EQUATIONNUMBER EQUATION

MISSINGQUANTITY

2-92-132-142-152-16

v = Vo + atx - Xo = vot + tat2

if = vij + 2a(x - Xo)X-Xo=t(vo+v)tx - Xo = vt - tat2

x - Xov

aVo

"Make sure that the acceleration is indeed constant before usingthe equations in this table. Note that if you differentiate Eq. 2-13you get Eq. 2-9. The other three equations are found byeliminat-ing one or another of the variables between Eqs. 2-9 and 2-13.

Table 2-2 lists Eqs. 2-9, 2-13, 2-14, 2-15, and 2-16and shows which one of the five possible quantities ismissing from each. To solve a constant accelerationproblem, you must decide which of the five quanti-ties is not involved in the problem, either as a givenor as an unknown. Select the correct equation fromTable 2-2 and substitute for the three given quanti-ties to find the unknown. Instead of using the table,you might find a solution more easily if you use onlyEqs. 2-9 and 2-13, solving them as simultaneousequations when needed.

SAMPLE PROBLEM 2-8

Spotting a police. car, you brake a Porsche from75 km/h to 45 km/h over a displacement of 88 m.

a. What is the acceleration, assumed to be constant?

SOLUTION In this problem the time is not involved,being neither given nor requested. Table 2-2 thenleads us to Eq. 2-14. Solving this equation for a yields

(45 km/h)2 - (75 km/h)2(2) (0.088 km)

= - 2.05 X 104 km/h2 = - l.6 m/s2. (Answer)

(In converting hours to seconds in the last step, wemust convert both the hour units.) Note that the veloci-ties are positive and the acceleration is negative, whichis consistent with a slowing of the car.

b. What is the elapsed time?

SOLUTION Now time is not the missing ingredient,but the acceleration is. Table 2-2 suggests Eq. 2-15.

2-6 CONSTANT ACCELERATION: A SPECIAL CASE 27

Solving that equation for t, we obtain

2(x - xo) (2) (0.088 km)t = = --'-'-------'--Vo + v (75 + 45) km/h

= l.5 X 10-3 h = 5.4 s. (Answer)

c. If you continue to slow down with the accelerationcalculated in (a) above, how much time would elapsein bringing the car to rest from 75 km/h?

SOLUTION The quantity not given or asked for here isthe displacement, x - Xo' Table 2-2 then suggests thatwe use Eq. 2-9. Solving for t gives

v - Vo 0 - (75 km/h)t = -- = ----'---~~~

a (- 2.05 X 104 km/h2)

= 3.7 X 1O-3h = 13s. (Answer)

d. In (c) above, what distance would be covered?

SOLUTION From Eq. 2-13, we have, for the displace-ment of the car,

x - Xo = vot + tat2

= (75 km/h) (3.7 X 10-3 h)

+ H- 2.05 X 104 krrr/ h") (3.7 X 10-3 h)2

= 0.137 km = 140 m. (Answer)

(Neglecting the sign on the acceleration would give awrong result. When you work problems, you shouldalways be alert to signs.)

e. Suppose that, on a second trial with the accelerationcalculated in (a) above and a different initial velocity,you bring your car to rest after traversing 200 m. Whatwas the total braking time?

SOLUTION The missing quantity here is the initial ve-locity, so we use Eq. 2-16. Noting that v (the final veloc-ity) is zero and solving this equation for t, we obtain

t = (_ (2) (x - xo) )1/2 = (_ (2) (200 ffi) )1/2a - l.6 m/s2

= 16 s. (Answer)

PROBLEM SOLVING

--A.N\,..---TACTIC 11: CHECK THE DIMENSIONS

The dimension of a velocity is (L/T), that is, length (L)divided by time (T). The dimension of acceleration is(L/T2); and so on. In any physical equation, the di-mensions of all terms must be the same. If you are indoubt about an equation, check its dimensions.


To check the dimensions of Eq. 2-13 (x - Xo =

vot + tat2), we note that every term must be a length,because that is the dimension of x and of Xo. The di-mension of the term vot is (LIT) (T), which is (L). Thedimension of tat2 is (L/T2) (T2), which is also (L). Thisequation checks out. A pure number such ast or 7T hasno dimension.

2·7 ANOTHER LOOK AT CONSTANTACCELERATION *

The first two equations in Table 2-2 are the basicequations from which the others are derived. Thosetwo can be obtained by integration of the accelera-tion with the condition that a is constant. The defi-nition of a (in Eq. 2-8) is

dva=-

dt '

which can be rewritten as

dv = a dt.

If we take the indefinite integral (or antiderivative) ofboth sides, we get

J dv = J a dt,

which reduces to

v = J a dt + C,

where C is a constantof integration. Since accelerationa is constant, it can be taken outside the integration.Then

v = a J dt + C = at + C. (2-17)

To evaluate the constant C, we let t = 0, at whichtime v = Vo. Substituting these values into Eq. 2-17(which must hold for all values of t, including t = 0)yields

Vo = (a) (0) + C = C.

With this substitution, Eq. 2-17 takes the same formas Eq. 2-9.

To derive the other basic equation in Table 2-2,we rewrite the definition of velocity (Eq. 2-4) as

dx = v dt

*This section is intended for those students who have had inte-gral calculus.

and then take the indefinite integral of both sides to

obtain Jx = v dt + C',

where C' is another constant of integration. Thistime there is no reason to believe that v is constant,so we cannot move it outside the integration. But wecan substitute for v with Eq. 2-9:

x = J (vo + at) dt + C'.

Since Vo is a constant, this can be rewritten as

x = Vo J dt + a J t dt + C'.

Integration yields

x = vot + ~at2 + C'. (2-18)

At time t = 0, we have x = Xo. Substituting thesevalues in Eq. 2-18 yields Xo = C'. Replacing C' withXo in Eq. 2-18 gives us Eq. 2-13.

2~8 FREE~FALLACCELERATION

If you tossed an object either up or down and couldsomehow eliminate the effects of air on its flight, you

FIGURE 2-10 A feather and an apple, undergoing freefall in a vacuum, move downwardat the same accelerationg. The acceleration causes the increase in distance be-tween images during the fall.

TABLE 2-3EQUATIONSFORFREEFALL

EQUATIONNUMBER EQUATION

MISSINGQUANTITY

2-192-202-212-222-23

v = Vo - gtY - Yo = vot - tgt2

V2 = VB - 2g(y - Yo)y - Yo = HVo + V) tY - Yo = vt + tgt2

y - YoV

g

would find that the object accelerates downward at aparticular rate. That rate is called the free-fall accel-eration g. The acceleration g is independent of theobject's characteristics, such as mass, density, orshape.

Two examples of free-fall acceleration areshown in the photograph of Fig. 2-10, which is astroboscopic series of photos of a feather and anapple. As these objects descend, they acceleratedownward at the rate g, picking up speed. The valueof g varies slightly with latitude and also with eleva-tion. At sea level in the mid-latitudes the value is9.8 m/s2 (or 32 ft /s"), which is what you should usefor the problems in this chapter.

The equations of motion in Table 2-2 for con-stant acceleration apply to free fall near the Earth'ssurface. That is, they apply to an object in verticalflight, either up or down, when the effects of the aircan be neglected. However, we can make themsimpler to use with two minor changes. (1) The di-rections of motion. are along the vertical y axis in-stead of the x axis, with the positive direction of yupward. (This change will reduce confusion in laterchapters when combined horizontal and verticalmotions are examined.) (2) The free-fall accelera-tion is then negative, that is, downward on the yaxis,and so we replace a with - g in the equations.

With these small changes, the equations ofTable 2-2 become, for free fall, the equations mTable 2-3.

SAMPLE PROBLEM 2-9

A worker drops a wrench down the elevator shaft of atall building.

a. Where is the wrench 1.5 slater?

2-8 FREE-FALLACCELERATION29

SOLUTION The missing ingredient is the velocity V,

which is neither given nor requested. This suggestsEq.2-20 of Table 2-3. Choose the release point of thewrench to be the origin of the y axis. Setting Yo = 0,Vo = 0, and t = 1.5 s in Eq. 2-20 gives

y = vot - tgt2

= (0) (1.5 s) - t(9.8 mjs2) (1.5 S)2

= -11 m. (Answer)

The minus sign means that the wrench is below its re-lease point, which we certainly expect.

b. How fast is the wrench fallingjust then?

SOLUTION The velocityof the wrench is given by Eq.2-19:

V = Vo - gt = 0 - (9.8 m/s2) (1.5 s)

= -15 m/s. (Answer)

Here the minus sign means that the wrench is fallingdownward. Again, no great surprise. Figure 2-11 dis-plays the important features of the motion up to t =4 s.

y V a

y (s) (m) (m/s) (m/s2)I

0 =fs 0 0 0 -9.8III= -4.9 -9.8 -9.8IIIII= 2 -19.6 -19.6 -9.8IIII

TIII

~ 3 -44.1 -29.4 -9.8IIIIIIIIIII=fs 4 -78.4 -39.2 -9.8,

FIGURE 2-11 Sample Problem 2-9. The position, ve-locity, and acceleration of a freely falling object.


SAMPLE PROBLEM 2-10

In 1939, Joe Sprinz of the San Francisco Baseball Clubattempted to break the record for catching a baseballdropped from the greatest height. Members of theCleveland Indians had set the record the precedingyear when they caught baseballs dropped about 700 ftfrom atop a building. Sprinz used a blimp at 800 ft. Ig-nore the effects of air on the ball and assume that theball falls 800 ft.

a. Find its time of fall.

SOLUTION Mentally erect a vertical y axis with its ori-gin at the point of the ball's release in the blimp, whichmeans that Yo = O. The initial velocity Vo is zero. Themissing ingredient is v, and so Eq. 2-20 is required:

y - Yo = vot - tgt2

- 800 ft = Ot - t(32 ft/s2) t2

16t2 = 800

t = 7.1 s. (Answer)

When taking a square root, we have the option of at-taching a plus or minus sign to the square root. Herewe choose the plus sign, since the ball reaches theground after it is released.

b. What IS the velocity of the ball just before it iscaught?

SOLUTION To get the velocity from the original data,rather than from the result of (a), we use Eq. 2-21:

v2 = va - 2g(y - Yo)

= 0 - (2) (32 ft/s2) (- 800 ft)

= 5.12 X 104 ft2/s2

v = - 226 ft/s (= -154 mi/h). (Answer)

Since the ball is moving downward, we choose theminus sign in our option of signs in taking the squareroot.

Neglecting the effects of air is actually un-warranted in such a fall. If you included them, youwould find that the fall time was longer and the finalspeed was smaller than the values calculated above.Still, the speed must have been considerable, becausewhen Sprinz finally managed to get a ball in his glove(on his fifth attempt) the impact slammed the gloveand hand into his face, fracturing the upper jaw in 12places, breaking five teeth, and knocking him uncon-scious. And he dropped the ball.

SAMPLE PROBLEM 2-11

A pitcher tosses a baseball straight up, with an initialspeed of 12 mys. See Fig. 2-12.

a. How long does it take to reach its highest point?

SOLUTION The ball is at its highest point when its ve-locity v becomes zero. From Eq. 2-19, we have

Vo - v 12 m/s - 0t = --- = = l.2 s. (Answer)

g 9.8 m/s2

b. How high does the ball rise above its release point?

SOLUTION We take the release point of the ball to bethe origin of the yaxis. Putting Yo = 0 in Eq. 2-21 andsolving for y, we obtain

va - v2y=--=

2g

= 7.3m.

(12 m/s)2 - (0)2(2) (9.8 m/s2)

(Answer)

If we wanted to take advantage of the fact that wealso know the time of flight, having found it in (a), wecould also calculate the height of rise from Eq. 2-23.Check it out.

~

all

IIV= 0 at 1"1

highest point I ::.IIIIII

: I

IIIIII

+1During ascent, ~ :a=-g and II

speed decreases I:IIII:.IIIIII!.I

y

Duringdescent,a=-gandspeedincreases

y= 0

~\ ..

1J;;1;~iJ;r,~~~;!i!);."j~!);.ie:tl;;'h~J;l;l

FIGURE 2-12 Sample Problem 2-1l. A pitcher tossesa baseball straight up into the air. The equations offreefall apply for rising as well as for falling objects, pro-vided that any effects from the air can be neglected.

c. How long will it take for the ball to reach a point5.0 m above its release point?

SOLUTION Inspection of Eqs. 2-19 to 2-23 suggeststhat we try Eq. 2-20. With Yo = 0, we have

y = vot - ~gt2.

5.0 m = (12 m/s) t - m (9.8 m/s2) t2.

If we temporarily omit the units (having noted thatthey are consistent), we can rewrite this as

4.9t2 - 12t + 5.0 = O.

Solving this quadratic equation for t yields*

t = 0.53 sand t = 1.9 s. (Answer)

There are two such times! This is not really surprisingbecause the ball passes twice through y = 5.0 m, onceon the way up and once on the way down.

We can check our findings because the time atwhich the ball reaches its maximum height should liehalfway between these two times, or at

t = ~(0.53 s + 1.9 s) = 1.2 s.

This is exactly what we found in (a) for the time toreach maximum height.

PROBLEM SOLVING

---A.N\~---TACTIC 12: MINUS SIGNS

In Sample Problems 2-9,2-10, and 2-11, many answersemerged automatically with minus signs. It is importantto know what these signs mean. For falling body prob-lems, we established a vertical axis (the yaxis) and wechose-quite arbitrarily-its upward direction to bepositive.

We then choose the origin of the y axis (that is, they = 0 position) to suit the problem. In Sample Problem2-9, the origin was the worker's hand; in Sample Prob-lem 2-10 it was at the blimp; in Sample Problem 2-11 itwas the pitcher's hand. A negative value of y means thatthe body is below the chosen origin.

A negative velocity means that the body is movingin the direction of decreasing y, that is, downward. Thisis true no matter where the body is located.

We have taken the acceleration (= - 9.8 m/s2) tobe negative in all problems dealing with falling bodies.A negative acceleration means that, as time goes on,the velocity of the body becomes either less positive or

*See Appendix G for the formula used to solve a quadratic equa-tion.

2-9 THE PARTICLESOF PHYSICS 31

more negative. This is true no matter where the body islocated and no matter how fast or in what direction it ismoving. In Sample Problem 2-11, the acceleration ofthe ball is negative throughout its flight, whether theball is rising or falling.

TACTIC 13: UNEXPECTEDANSWERSMathematics often generates answers that you mightnot have thought of as possibilities, as in Sample Prob-lem 2-11c. If you get more answers than you expect, donot discard out of hand the ones that do not seem tofit. Examine them carefully for physical meaning; it isoften there.

If time is your variable, even a negative value canmean something; negative time simply refers to timebefore t = 0, the (arbitrary) time at which you decidedto start your stopwatch.

2-9 THE PARTICLES OF PHYSICS

As we progress through the book, we plan to stepaside occasionally from the familiar world of large,tangible objects and look at nature on a much finerscale. The "particles" that we have dealt with in thischapter, for example, have included pigs, baseballs,and dragsters. In the spirit of our plan, we ask: "Howsmall can a particle be? What are the ultimate parti-cles of nature?" Particle physics-for so the fieldthat relates to our inquiry is called-attracts the at-tention of many of the best and brightest of today'sphysicists.

The realization that matter, on its finest scale, isnot continuous but is made up of atoms was the be-ginning of understanding for physics and chemistry.With the scanning tunneling microscope, we cannow "photograph" these atoms, as Fig. 2-13 makesclear. It is also possible to keep single atoms in tinyelectromagnetic "traps" and monitor them at lei-sure. A single electron was kept in such a trap at theUniversity of Washington for lO months before-bymisadventure -it hit a wall and escaped.

We describe the "lumpiness" of matter by say-ing that matter is quantized, the word coming fromthe Latin word quantus, meaning "how much."Quantization is a central feature of nature, and youwill see other physical quantities as we go along thatare quantized when we look at them on a fineenough scale. This pervasiveness of quantization isreflected in the name we give to physics at theatomic and subatomic level-quantum physics, the

--------------------------------------------

32 CHAPTER 2 MOTION ALONG A STRMGHT LINE

FIGURE 2-13 A hexagonal array of uranium atoms is re-vealed in this image from a scanning transmission electronmicroscope. The color has been added by a computer.

physics that deals with the ultimate particles of na-ture.

There is no sharp discontinuity between thequantum world and the world of large-scale objects.The quantum world and the laws that govern it areuniversal but, as we move from electrons and atomsto baseballs and automobiles, the fact of quantiza-tion becomes less noticeable and finally totally un-detectable. The "graininess" effectively disappears,and the laws of classical physics that govern the mo-tions of large objects emerge as special limitingforms of the more general laws of quantum physics.

The Structure of Atoms

An atom consists of a central, almost unimaginablycompact and dense nucleus that is surrounded byone or more light-weight electrons. An atom is usu-ally considered to be spherical; so is the nucleus.The radius of a typical atom is on the order of10-10 m; the radius of a nucleus is 100,000 timessmaller, about lO-15 m. An atom is held together byelectrical attraction between the electrons, which areelectrically negative, and protons, which are electri-cally positive and reside within the nucleus. The na-ture of that attraction is explored later in this book,

but for now you might realize that were it notpresent, atoms could not exist, and so neither couldyou.

The Structure of Nuclei

The simplest nucleus, that of common hydrogen,has a single proton. There are two other, rare, ver-sions of hydrogen: they differ from the commonversion by the presence of one or two neutrons(electrically neutral particles) inside the nucleus.Hydrogen, in any of its versions, is an example of anelement; each element is distinguished from all theothers by the number of protons in the nucleus.When there is only one proton, the element is hy-drogen. When, instead, there are six, the element iscarbon. The various versions of each element arecalled isotopes; they are distinguished by the num-ber of neutrons.

Roughly speaking, the purpose of the neutronsis to glue together the protons, which, being all elec-trically positive and closely packed, strongly repelone another. If the neutrons did not provide theglue, the only type of atom that could exist would becommon hydrogen; all others would blow apart.

Such instability can be found in many isotopesof common elements, but thankfully not the ele-ments on. which your existence depends. For exam-ple, of the 17 isotopes of copper, all but two are un-stable and undergo transformations to becomeother elements. The stable isotopes are the onesused in electronics and other technology.

The Structure of the ParticlesWithin Atoms

The electron is simple but perplexingly so. It ap-pears to be infinitesimal in size; that is, it has no sizeand no structure. It is a member of a family of otherpointlike particles called leptons; there are six basictypes, each with an antiparticle version.

Protons and neutrons are believed to be differ-ent from electrons and the other leptons, becauseeach of the former appears to be a bundle of threesimpler particles called quarks, * "up" or "down"

*On a quirk, the word "quark" was lifted from Finnegans Wake byJames Joyce:

Three quarks for Muster Mark.Sure he hasn't got much of a barkAnd sure any he has it's all beside the mark.

d/ ""-dNeutron

Nucleus

FIGURE 2-14 A representation of the nucleus of anatom, showing the neutrons and protons that make it up.These particles, in turn, are composed of "up" and"down" quarks.

REVIEW & SUMMARY 33

quarks. A proton consists of two "up" quarks andone "down" quark, and a neutron the reverse (Fig.2-14). Other, more exotic particles that were firstthought to be fundamental appear to be similarbundles.

Provocatively, quarks come in six basic types*(each with its antiparticle) just as do leptons. Herephysicists wonder: Is there a basic reason for thematch in number of types? Or is the match simplycoincidence? We do not know.

"The other types are called charm, strange, top, and bottom(even physicistshave their moments).

REVIEW & SUMMARY

PositionThe position x of a particle on an axis locates it with respectto the origin of the axis. The position is either positive ornegative, according to which side of the origin the particleis on, or zero if the particle is at the origin. The positivedirection on an axis is the direction of increasing positivenumbers; the opposite direction is the negative direction.

DisplacementThe displacement Ilx of a particle is the change in its posi-tion:

(2-1)

Displacement is a vector quantity. It is positive if the parti-cle has moved in the positive direction of the x axis, andnegative if it has moved in the negative direction.

Average VelocityWhen a particle has moved from position Xl to X2 during atime interval ar = t2 - tl, its average velocity is

_ Ilxv = t;i. (2-2)

The algebraic sign of v indicates the direction of motion(ii is a vector quantity). Average velocity does not dependon the actual distance a particle covers, but instead de-pends on its original and final positions. Sample Problems2-1 and 2-2 illustrate the calculation of average velocity.

On a graph of X versus t, the average velocity for atime interval Az is the slope of the straight line connectingthe points on the curve that represent the ends of the in-terval.

Average SpeedThe average speed s of a particle depends on the full dis-tance it covers in a time interval Az:

total distances=

III(2-3)

Instantaneous VelocityIfwe allow Ilt to approach zero in Eq. 2-2, then Ilx will alsoapproach zero; however, their ratio, which is ii, will ap-proach a limiting value v, the instantaneous velocity (or sim-ply velocity) of the particle at the time in question, or

Ilx dxv = Jim -=-.

~t ....•o Ilt dt (2-4)

The instantaneous velocity (at a particular time) may berepresented as the slope (at that particular time) of thegraph of x versus t. See Sample Problem 2-3 and Fig. 2-9.Sample Problem 2-5 illustrates how we can find velocity bydifferentiating a function x(t). Speed is the magnitude ofinstantaneous velocity.

Average AccelerationAverage acceleration is the ratio of the change in velocity Ilvthat occurs within a time interval Ar to that time interval:

Ilva = t;i. (2-7)

The algebraic sign indicates the direction of a. See SampleProblem 2-6 for examples.

Instantaneous AccelerationInstantaneous acceleration (or simply acceleration) is therate of change of velocity,

Ilv dva= Jim -=-.

~t ....•o Ilt dt (2-8)


Sample Problem 2-7 shows how to differentiate v(t) to geta( t). On a graph of v versus t, a( t) is the slope of the curve.

Constant AccelerationFigure 2-9 shows x( t), v( t), and a( t) for the importantcase in which a is constant. In this circumstance, the fiveequations in Table 2-2 describe the motion:

v = Vo + at, (2-9)

(2-13)

(2-14)

(2-15)

(2-16)

if = V5 + 2a(x - xo),

x - Xo = t(vo + v)t,

x - Xo = vt - tat2.

These equations are not valid when the acceleration isnot constant. Sample Problem 2-8 illustrates the use ofthese equations.

Free-Fall AccelerationAn important example of straight-line motion with con-stant acceleration is that of an object rising or falling freelynear the Earth's surface. The constant acceleration equa-tions describe this motion, but we make two changes innotation: (1) we refer the motion to the vertical y axis with

+ Y vertically up; (2) we replace a with - g, where g is themagnitude of the free-fall acceleration. Near the Earth'ssurface, g = 9.8 m/s2 (= 32 ft/s2). The free-fall equations,with these conventions, are shown as Eqs. 2-19 to 2-23.Sample Problems 2-9, 2-10, and 2-11 show how theseequations can be used.

The Structure of MatterAll ordinary matter is composed of atoms, which, in a sim-ple model, consist of electrons that surround a highlycompact central core, the nucleus. Neutrons and protonsreside inside nuclei. Each element is distinguished by thenumber of protons in its nucleus. Variations of an ele-ment, differing in the number of neutrons, are isotopes ofthe element.

QJ.tarksand LeptonsElectrons appear to be pointlike particles with no size orinternal structure, but protons and neutrons appear tohave size and to contain more elementary particles, calledquarks. There are six basic types of quarks, each with anantiparticle version. Electrons are members of a family ofparticles, leptons, that also come in six basic types, eachwith an antiparticle version.

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Ch 2 1 Dim Motion