Constructing Two-Dimensional Motion from One-Dimensional … · 50 Chapter 3 Constructing...

CHAPTER THREE

Constructing Two-Dimensional Motion from One-Dimensional Motions

49“. . . From the graceful arc of a leaping dolphin . . . to the dizzying gyrationsyou experience on a carnival ride.”

This chapter deals with a more varied selection of the motions foundin the physical universe—from the graceful arc of a leaping dolphin tothe corkscrew path of the tip of a boat propeller churning through thewater, from the orbit of a telecommunications satellite or a planet to thedizzying gyrations you experience on a carnival ride.

In the previous chapter, we developed vocabularyand equations for describing motion along a straight line. In this chapter, building on the one-dimensional description, we develop the conceptual and mathe-matical tools you need for describing and analyzing the richer array of motions that occur in two or more dimensions.

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50 ◆ Chapter 3 Constructing Two-Dimensional Motion from One-Dimensional Motions

3-1 Constructing Complex Motions from Simpler Motions

Complicated motions can be built up from two or more simpler motions. In theactivity that follows, you will produce a two-dimensional motion by combiningtwo one-dimensional motions.

On-The-Spot Activity 3-1Part A Follow steps 1–3 of the figure. The marks left by your pen point showwhere the pen point has been, but the later marks cover over the earlier ones.

Part B Now repeat your pen motion while moving your ruler slowly to the right(see step 4 of the figure). This time the mark left by your pen provides a clearerrecord of the pen’s motion. It is a composite motion, combining the purely verticalmotion of your pen along the ruler with the purely horizontal motion of the ruler.In the language of physics, we say the actual motion has a horizontal and a verti-cal component. Go to WebLink 3-1 to explore these ideas further.

When motions combine, we sometimes say they have been superimposed,or that the resulting complex motion is a superposition of the simpler motions.We may also find it useful to speak of the complex motion as the resultant oftwo or more simpler component motions. Many kinds of complex motions canbe built from simpler component motions.

Here are some real-world examples. You may have seen devices, in hospi-tals, laboratories, or elsewhere, in which an electronically controlled pen movesup and down a track, just as your pen moves along the ruler in Activity 3-1, tokeep a record of some quantity that is varying over time (Figure 3-1). The heightof the pen above its lowest point is proportional to the quantity being measured.While the pen goes up and down, the paper scrolls to the left, so that the recordor trace left by the pen advances toward the right end of the paper. (This isaccomplished by sliding your ruler to the right in Activity 3-1.)

Such devices are called xy-recorders because the trace is like a graph drawnon an xy coordinate frame. In a device called an oscilloscope (see Figure 3-2), a

Motion of pen(up and down ruler)

1. Fasten paper to flat surface (table, book, etc.)

2. Line up ruler with left edge of paper.

3. Run your pen up and down length of ruler several times.

4. Move ruler to the right while you repeat pen motion.

Part A

Part B

Motion of ruler(left to right)

Motion of ruler

For WebLink 3-1:Combining Two Motions, go towww.wiley.com/college/touger

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3-1 Constructing Complex Motions from Simpler Motions ◆ 51

point of light on a screen substitutes for the pen point. Its up-and-down motionis controlled electronically, much as the vertical motion of the pen along its trackis controlled electronically. The motion across the screen, though controlled elec-tronically, is like the horizontal motion of the paper scroll or your ruler. The tracesproduced by the xy-recorder and the oscilloscope are position-versus-time graphs;the displacement along the horizontal axis represents time. Because the pen orother recording device sweeps horizontally at constant speed, the horizontal dis-tance x is proportional to the time interval t.

The component motions from which complex motions are built are not alwaysstraight line motions. Figure 3-3 shows how the spiral motion of a propeller tipcan be produced by combining circular motion with motion along a straight lineperpendicular to the circle. A common type of amusement park ride (Figure 3-4)puts together two circular motions. The rider feels “whipped around” at eachpoint labeled W along the resulting path. For animated versions of Figures 3-3and 3-4, see WebLinks 3-2 and 3-3.

In these examples, one or both components are circular motions. Later on(Chapter 13), you will see that even circular motion can be constructed from twostraight line motions, though of a very special sort.

In developing a quantitative approach to two-dimensional motion, we canbuild on what we already know about one-dimensional motions if we treat two-dimensional motion as a composite of two one-dimensional motions (as in Activity3-1 and the xy-recorders). The two-dimensional motion in the following case com-bines one-dimensional motions that we treated quantitatively in Chapter 2.

¢¢

Figure 3-1 Examples of xy recorders. (a) In a barograph used by meteorologists, thepen rises and falls with air pressure and leaves a trace on the rotating roll of paper. (b) Anelectroencephalogram (EEG) shows the electrical activity of the brain (brain waves). Theseveral traces produced by the xy recorder display information from pairs of electrodesplaced at different locations on the patient’s scalp. An EEG is used to detect abnormalitiesin brain activity.

(a) (b)

Figure 3-2 An oscilloscope. An os-cilloscope is basically an xy recorderthat records electrical signals andproduces its trace electronically on avideo screen.

Figure 3-3 Components ofmotion for the tip of apropeller blade.

For WebLink 3-2:Outboard Propeller

and Weblink 3-3: The Octopus,go towww.wiley.com/college/touger

The first motion + The second motion

Boat moving forward,propeller stationaryBoat stationary,

propeller rotating

The combined motion

Boat moving forward,while propeller turns

=

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Figure 3-4 The octopus. (a) The ride as you see it. (b) A geometricdescription of the motion. The motion that the rider experiences iscompounded of two simple circular motions.

Here the two one-dimensional motions will be providedby a carnival game (Figure 3-5a) and a flatbed truck.In the carnival game, the player attempts to ring thebell by using a sledgehammer to hit a lever. The lever,in turn, sends a weight up a slide wire. If, as usuallyhappens, the weight falls short of hitting the bell, it issubject to gravitational acceleration throughout its riseand fall. Therefore, its motion can be described by theconstant acceleration equations (2-9 to 2-12 in Section2-7) with and x replaced by y.

We now load the entire game onto the back of aflatbed truck, which is then driven at constant speed(described by Equation 2-6a) while the weight goes upand down the slide wire. (This is like the xy-recorder,with the horizontal sweep now provided by the truck.)You can see the trajectory of the weight—that is,the path it actually follows—by tracing its successivepositions in the time sequence depicted in Figure 3-5b.The trajectory is in effect a graph of y versus x.

a � �g

Figure 3-5 The ring-the-bell carnival game in time andspace. (a) A sequence in time: Here the carnival game re-mains stationary. Spreading a sequence of stop action picturesout horizontally along a t axis results in a graph of y versus t.(b) A sequence in space. Here the carnival game is loaded onthe back of a truck moving at constant speed The motion

of the weight is a composite of the vertical motion in (a) andthe horizontal motion of the truck. The truck’s motion spreadsthe successive pictures along the x axis, resulting in the actualtrajectory or path (dashed line) of the weight—in effect, agraph of y versus x.

v.

Case 3-1 ◆ The Carnival Game on the Flatbed Truck

(a)a0 t

x

y

(b)

Directionof mainplatform

The car and riderscircle the tentacle tip

(a) The ride as you see it (b) Geometric description of the motion

=

+

Path oftentacle tip

Path of a rideraround tentacle tip

C W

W

Resultingpath

P


3-1 Constructing Complex Motions from Simpler Motions ◆ 53

STOP&Think Compare this trajectory to the graph of y versus t for an objectthrown vertically upward (Figure 3-6). Why do they look so similar? ◆

The motion graphed in Figure 3-6b, which we considered in Chapter 2, ispurely vertical. But in plotting y against t, we represent values of t asdisplacements along the horizontal axis. For the flatbed driven at constant speed,on the other hand, the x axis of the graph of y versus x (Figure 3-6a) shows thetruck’s actual horizontal displacement. If we simply relabel the horizontal axis,replacing each horizontal position reading with the clock reading at which itoccurs (we can do this when x and t are proportional), we end up with a graphof y versus t.

To explore the importance of Case 3-1 further, turn to WebLink 3-4.Suppose a lit bulb was mounted on the weight as it followed the path in

Figure 3-5b. Figure 3-7 shows what you would see if you watched the motionof the bulb on a dark night when only the bulb was visible. If you had no knowl-edge of the elaborate set-up producing its motion, what would you assume youwere seeing? Probably just a glowing object—a flare or a firework perhaps—thatsomeone had thrown or fired through the air (Figure 3-8). In short, it would looklike just like some kind of projectile.

A projectile is an object thrown, launched, or otherwise projected so thatonce released, if air resistance is negligible, its path is affected only by Earth’sgravitational attraction. A pop fly in baseball is an example of projectile motion.STOP&Think Once the ball leaves the bat, does the bat have any influence onthe subsequent motion of the ball? ◆

When we say a projectile is affected only by Earth’s gravitational attraction,it means that the ball becomes a projectile only when it ceases to be in contactwith the bat. We’ll return to this point in Chapter 4.

Figure 3-6 Why do these graphslook similar? Compare (a) a plot ofy versus x for the weight in Figure3-5 and (b) a graph of y versus t foran object thrown vertically upward(reproduced from Figure 2-21a).

x

y y

t

01 Start clock at t = 0

when ball leaves hand23

(a) (b)

Figure 3-7 Key ideas from Weblink 3-4. (a) Combining the components of the weight’smotion. (b) The trajectory of the illuminated weight as observed at night.

x

y

(a) (b)

x

y

Figure 3-8 Trajectories of fire-works. A fireworks display showsa tracery of parabolic paths thathave been altered somewhat by airresistance.

For WebLink 3-4:Carnival Game

on Flatbed, go to www.wiley.com/college/touger

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3-2 Breaking Down Two-Dimensional Motionsinto One-Dimensional Components: Projectile Motion

In Case 3-1 and also in WebLink 3-4, a two-dimensional motion was constructedfrom one-dimensional component motions. We could as easily have workedbackward, starting with the resultant motion and breaking it down into its twocomponent motions. It turns out that a powerful technique for analyzing any two-dimensional motion is to reduce it to two one-dimensional component motions.To see how to do this, we again focus our attention on projectile motion.

Picture a ball thrown into the air, but not straight up as in the previouschapter. The trajectory (Figure 3-9) of the ball, shown as it appears when airresistance is negligible, is two-dimensional. Watching the ball travel, you do notactually see the component parts of the motion separately, but it is possible toseparate them mentally. Consider the following.

Suppose that, as in Figure 3-9, you are standing in aroom with your back to a wall as you throw the ball.Imagine that as the ball travels, it casts two idealized“shadows,” one directly below it on the ground, theother directly behind it (and you) on the wall. (Theshadows are idealized because real shadows wouldn’talways be directly below and directly behind the ball.)If coordinate axes are drawn on the floor and wall, thetwo shadows mark the x and y coordinates of the ballitself. Mathematicians sometimes speak of the coordi-nates of a point as the projections of the point ontothe x and y axes, just as, using two distant spotlights,we might project our two shadows. It follows that ifyou know the motions of the shadows, which are one-

dimensional, you will know how the ball’s coordinateschange; that is, you will know the motion of the ball.

Case 3-2 ◆ The Moving “Shadows” of a Thrown Ball

x

y

Figure 3-9 Analyzing the trajectory of a thrown ball. Theball’s motion is tracked by the sequence of positions of itstwo shadows. The positions of the ball—and therefore of theshadows it casts on the wall and on the ground—are sepa-rated by equal time intervals.

In the next situation, we compare measured and calculated values for thecoordinates. Figure 3-10 shows a multiple exposure flash photograph of a golfball—the yellow one—that has been launched horizontally. Coordinate axes havebeen added to the photo, as have the ball’s horizontal and vertical “shadows” foreach exposure. On the photo, we can measure the positions x and y of theseshadows. Table 3-1 records these measured values. The table shows that the mea-sured y coordinates have the same values that we obtain by calculation using

for a body dropped vertically from rest. This is corroborated by thered ball in the figure, which was dropped vertically at the instant the yellow ballwas launched, and has exactly the same vertical motion as the yellow ball. Thetable also shows that if we neglect a very small decrease due to air resistance,the shadow on the ground advances the same distance every s—in other words,it moves at constant speed.

These are the same component motions that we had for the flatbed—themotion of the weight along the slide wire affected only by gravitationalacceleration, the horizontal motion provided only by the truck’s constant speed.For the carnival game on the flatbed, you can run either component alone. It isharder to separate the golf ball’s component motions in reality, but the shadowslet us picture or conceptualize the two component motions separately. As Fig-ures 3-7 and 3-9 show, the resulting two-dimensional trajectories are the same in

130

y � �12 gt2


3-3 Vectors ◆ 55

Figure 3-10 Measuring the hori-zontal and vertical “shadow” mo-tions of a ball that is launchedhorizontally. In the photo, the redball is given no horizontal motion,but is released from rest at the sametime that the yellow ball is launchedhorizontally. Note that both ballshave the same vertical “shadow”motion. (The balls are shown at intervals.)

130 s

y

x

2.9cm

Table 3-1 Comparison of Calculated (Theoretical) and Measured Positions of Golf Ball in Figure 3-10

y coordinate of bodyMeasured x (cm) Measured y (cm) dropped vertically from rest

(position of (position of (calculated byt (s) horizontal shadow) vertical shadow) (cm)

0 0 0 0

first exposures of ball at -s intervals not visible

26.8

34.2

41.1

47.8

54.9

62.6

69.3

76.4

83.1

(Measured values are cm)��0.3

�78.4�77.21230

�65.9�64.61130

�54.4�53.01030

�44.1�42.8930

�34.8�33.4830

�26.7�26.0730

�19.6�18.6630

�13.6�13.2530

�8.7�8.8430

130

y � �12 gt2

x (m)0 300 500

∆xx1

x2

(x2 = x1 + ∆x = 300 m + 200 m = 500 m)(x2 = x1 + ∆x = –300 m + (–200 m) = –500 m)

∆x x1

x2

300500Figure 3-11 Positions and dis-placements of two cyclists.

the two cases. Because the object occupies the same sequence of positions inthe two cases, their mathematical descriptions are the same.

In principle, any two-dimensional motion can be described by its two shadowmotions. All we need now is a way of translating back and forth between thetwo shadow descriptions and the composite two-dimensional description. For this,we turn to the mathematics of vectors.

3-3 Vectors✦VECTORS IN ONE DIMENSION In Figure 2-4, the displacements of twocyclists were shown as arrows. In Figure 3-11, we expand on this idea and usearrows to represent the cyclists’ positions as well as their displacements. Eachcyclist in the figure has an earlier position and a later position Because aposition is a displacement from the origin it also has a direc-tion. For each of the cyclists, whether the values are positive (asfor the cyclist going right) or negative (as for the cyclist going left). For eachcyclist, we can draw a new arrow from the tail of the first to the tip ofthe second to represent this addition.

A vector is a quantity that, like these arrows, must be described by givingits direction as well as the magnitude (numerical size). The arrows are one wayof representing vectors. We denote a vector by placing an arrow above the sym-bol for the quantity. In our cyclist example, when we draw the arrow from thetail of the first vector to the tip of the second, we can write the addition that itrepresents as

(3-1)xS2 � xS1 � ¢ xS

1¢x2 1x121x22x2 � x1 � ¢x,

1¢x � x � 0 � x2,x2.x1



The procedure for combining any two vectors and to obtain a vector sumor resultant vector is called vector addition. We always write torepresent this procedure, although in two dimensions it will be different thanordinary addition of numbers.

AS

� BS

� RS

RS

BS

AS

Procedure 3-1 tells us how to draw the arrow representation of Figure 3-12shows this procedure for several examples. To find the magnitude or absolutevalue of you would measure its length and use your scale to convert back tothe original units.

RS,

RS

.

✦NEGATIVES OF VECTORS Note that in Figure 3-12d, We canrewrite this as Figure 3-12d then shows that the negative of a vector isa vector equal to it in magnitude but opposite in direction. As an example ofthis reasoning, we can rewrite Equation 3-1 (depicted in Figure 3-13a) as

Subtracting a vector simply means adding the negative of that vector. This isshown in Figure 3-13b.

✦VECTORS AND SCALARS We have concentrated on positions and displace-ments as examples of vectors because they are easy to visualize. But velocitiesalso must be described by both a magnitude and a direction in space, and thisis also true of many other quantities we shall encounter. Such quantities arecalled vector quantities, and all can be treated mathematically by the same rules(such as Procedure 3-1). In contrast, quantities that do not have a spatial direc-tion (such as mass or time) and can be described fully by a single number areoften called scalars or scalar quantities.

xS2 � 1�xS12 � ¢ xS or simply as xS2 � xS1 � ¢ xS

BS

� �AS.

AS

� BS

� 0.

Figure 3-13 Picturing two equiva-lent vector equations. In (a) weadd tail-to-head to In (b) wetake which is equal and op-posite to and add it tail-to-headto xS2.

xS1,�xS1,

xS1.¢xS

Figure 3-12 Examples of in one dimension.AS

� BS

� RS

PROCEDURE 3-1Picturing Vector Addition: The “Tail-to-Head” Method of

Finding 1. Draw to some scale (say, 1 cm represents 1 m) in the proper direction.2. Draw to the same scale in its proper direction, starting its tail at the tip

of This is called adding to 3. Draw the resultant from the tail of to the tip of B

S.A

SRS

AS.B

SAS.

BSAS

RS

� AS

� BS

(A + B = R)

Vector representation

(A + B = R)

A A: Walk 5 m eastB: ...then walk another 2 m east.R: End up 7 m east

5 m + 2 m = 7 mB

R

A A: Walk 5 m eastB: ...then walk 2 m west.R: End up 3 m east

5 m + (–2 m) = 3 m

BR

A A: Walk 5 m eastB: ...then walk 7 m west.R: End up 2 m west

5 m + (–7 m) = –2 m

B

R

A A: Walk 5 m eastB: ...then walk 5 m west.R: End up back at origin

5 m + (–5 m) = 0

B(R = 0)

Arithmetic representationRepresentation in words

(d)

(c)

(b)

(a)

∆xx1

x2

x1 + ∆x = x2

∆x –x1

x2

x2 + (–x1) = ∆x

(b)(a)

➥Notation: Writing longhand, youwill always use the arrow. In mostprinted material, the vector isprinted in bold and the arrow omit-ted. For ease of recognition, this textwill use both the arrow and bold-face to identify vectors.


3-3 Vectors ◆ 57

✦VECTORS IN TWO DIMENSIONS In two dimensions, as in one, we can drawan arrow representing the magnitude and direction of a vector quantity. For exam-ple, the vector in Figure 3-14 indicates that Dallas is (224 miles) fromHouston in a direction W of N (west of north). Implicitly, we have picked theorigin (Houston) and orientation (fixing the directions N, S, E, and W) of a coor-dinate system. The magnitude is ; the direction is stated as an angle.Similarly, a pilot who reports “heading of E at 268 m/s [600 miles/hour]” isproviding a magnitude and direction description of the plane’s velocity.

We use the symbol to denote a two- or three-dimensional position vector,and we use either r or to denote its magnitude. As the absolute value barsindicate, the magnitude of a vector is always taken to be positive. In two dimen-sions, a vector’s direction is given by its directional angle (Figure 3-15), so thereis no need to use sign to indicate direction. As the figure shows, vectors in oppo-site directions are described by angles that differ by 180°.

0 rS 0rS

30° N3.6 � 105 m

22°3.6 � 105 m

If you stand at the origin facing along the axis, and then rotate coun-terclockwise (l) until you are facing along the position vector, the angle throughwhich you have rotated is If you rotate clockwise (k), the value of is negative.

✦ABOUT COORDINATE FRAMES Note that the first step in Procedure 3-2 isto choose a coordinate frame. As in one dimension (Figure 2-1), the descriptionof a position depends on the choice of origin. In two dimensions, you are alsofree to choose the orientation of the x and y axes, as long as the axes remainperpendicular to each other. In Figure 3-16, if we choose coordinate frame A, themotions of the targets on the horizontal walkway have no component motion inthe y direction. If we choose coordinate frame B, the targets on the ramp haveno component motion in the y direction. If we want to analyze the motion of aparticular object, we can often simplify our analysis by picking the frame in whichthe object has just one component motion. In making our choice, the axes neednot be in the horizontal and vertical directions.

uu.

�x

PROCEDURE 3-2Determining the Magnitude and Direction of the Position Vector

of a Point in Two-Dimensional Space1. Choose an xy coordinate frame (origin, orientation, and scale).2. Draw the vector as an arrow from the origin to the point in question.3. Measure its length to find the magnitude 4. Measure the angle the vector makes with the axis. We designate this

directional angle by (the Greek letter theta).u

�x1 0 rS 0 or r2.r

S

(a) A good choice for describing the targetson the horizontal walkway

(b) A good choice for describingthe targets on the ramp

y y

x

x

Houston

Dallas

224 miles

22°

E

S

W

N

Figure 3-14 A position vector forDallas, if the origin is at Houston.

Figure 3-15 Position vectors withequal magnitudes but oppositedirections.

Figure 3-16 Choosing a coordi-nate frame in two dimensions.(a) A good choice for describing thetargets on the horizontal walkway.(b) A good choice for describing thetargets on the ramp.

x

P1y

θ = 30°

x

P2

y

θ = 210°



✦ADDING VECTORS IN TWO DIMENSIONS Suppose you walk 4 m in aparticular direction, then 3 m in another direction (Figure 3-17a). How far areyou from your original starting point? in what direction? The figure shows the twosuccessive displacements and and the resulting displacement from thestarting point. As in one dimension, is the vector sum (or resultant) of and

As Figure 3-17a illustrates, the resultant can be found by thedrawing method of Procedure 3-1. WebLinks 3-5 and 3-6 will help you picturehow this works in two dimensions.

Before you begin calculating the sums of vectors in the next section, youneed a good mental picture of what vector addition means. You should be ableto sketch roughly what the sum of any two vectors looks like. Even if the mag-nitudes of two vectors remain the same, both the magnitude and direction of theirvector sum will change if the directions of the individual vectors change (Figures3-17b –d). Note that you can get the magnitude of the resultant by doing ordi-nary arithmetic addition of the individual magnitudes only (as in Figure 3-17c) if

and are in the same direction.

On-The-Spot Activity 3-2In each step below, you will combine the two motions shown in the figure: (1) moving a ruler 6 cm to the right while keeping it parallel to the edge of thepaper, and (2) moving your pen 8 cm up the ruler. What will differ from one step tothe next is the order in which you execute these motions. But first, draw a line6 cm in from the left edge of the paper, as shown, so you will know when yourruler has moved 6 cm.

a. Line the ruler up top-to-bottom on a sheet of paper, as you did in Activity 3-1.Place your pencil point against the ruler at a point near the bottom and label thatpoint O. Run your pencil point 8 cm up the ruler, then, keeping the pencil point

BS

AS

BS 1R

S� A

S� B

S2.

AS

RS

RS

BS

AS

A = 4.00

4 m

3 m

(R = 5 m)

Meter stick used tomeasure magnitude of R(scale: 1 cm = 1m)

R = 6.93

R drawn from tailof A to head of B

37°

53°B = 3.00

y

x

4 m

3 m

y

x

R

4 m

3 m

(R = 2 m)

y

x

R

(R = 7 m)

R = 44°(found by measuringwith a protractor)

θ

Rθ

(d )(c)

(b)(a)

Rθ

Figure 3-17 The resultant oftwo vectors with magnitudes

is obtained bythe tail-to-head method for dif-ferent directions of and (a) Shows how the resultant may befound for one such pair of vectors.(b)–(d) Show how the resultant mayvary when and are in differentdirections.

BS

AS

BS

.AS

A � 4 and B � 3

RS

For WebLink 3-5:Picturing VectorAddition and WebLink 3-6: Possible Sumsof Vectors with Equal Magnitude,go to www.wiley.com/college/touger

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3-3 Vectors ◆ 59

Pen moves 8 cm up ruler

Rulerslides6 cmto right

Dotted linedrawn 6 cmfrom left edgeof paper

pressed against the ruler, slide the ruler 6 cm to the right. Mark the point on thepaper where your pencil point ends up.

b. Now start at O again. This time, with the pencil point pressed against the ruler,you should first slide the ruler 6 cm to the right and then run your pencil point8 cm up the ruler. Now where does the pencil point end up?

c. Finally, start at O and try to run your pencil point 8 cm up the ruler while you aresliding the ruler 6 cm to the right. Where does the pencil point end up this time?

In Activity 3-2, you should find that the pencil point ends up in the sameplace each time. If “8 cm upward” is and “6 cm to the right” is the firsttwo trials illustrate that is the same as (see Figure 3-18). When and are concurrent (one occurs while the other does), you still end up in thesame place. In that case, you can draw them originating from the same point(see Figure 3-18c). If you then complete the parallelogram (in this case a rec-tangle), you can again get the same resultant by drawing the diagonal. It doesn’tmatter whether you call this or whether the vectors occur in oneorder or the other or concurrently, you get the same result.

Adding three vectors simply means adding a third vector to the resultant ofthe first two, and so on for additional vectors. Each time you add another vec-tor to a previous resultant, you follow the same procedure.

BS

� AS

;AS

� BS

RS

BS

AS

BS

� AS

AS

� BS

BS

,AS

B = 3

A = 4

R R

R

RR

B = 3

A = 4

B = 3

First completeparallelogram...

...then drawdiagonal

A = 4

B = 3

A = 4

B = 3

A = 4

B = 3

A = 4

A + B = R

A ⊥ B

A not ⊥ B

B + A = R Parallelogram method

( f )(e)(d )

(c)(b)(a)

Figure 3-18 Three ways of addingtwo vectors and B

S.A

S


3-4 Working with Vector Components✦COMPONENTS OF A VECTOR In Figure 3-7, you saw the one-dimensionalcomponent motions for the two-dimensional motion of a projectile. Figure 3-19ashows the position vectors for these two component motions. Now we draw inthe first two-dimensional position vector for the actual motion (Figure 3-19b).We then see that the position vectors for the component motions are actually the“shadows” or projections of the two-dimensional vector, just as the object itselfcasts idealized “shadows” on the two axes. They are called the componentvectors of the two-dimensional vector. As an object moves, its position vectorchanges, and the two component vectors and change correspondingly.

The numerical values x and y (the coordinates of the objects) associated withthe component position vector are called its scalar components. When we justspeak of components, we will mean scalar components. The sign of a scalar com-ponent indicates the direction of the associated vector component. For example, inFigure 3-20a, where is to the right and upward, x and y are both positive. InFigure 3-20b, where is to the left and downward, x and y are both negative.ySxS

ySxS

ySxS

rS1


r1

(b) Magnitude-and-directiondescription

y

x

y

x

(a) Vector component description

y2 y1

x1

x2

y

x

θ

y y

x

x

r r

3

–3

–4

4

0

y = 4 y = –4

x = 3

x = –3

r = 5 r = 5

0

0

y

x 0

y

x

y

x

y

x

Vector componentrepresentation

(Scalar) componentrepresentation

Vector componentrepresentation

(Scalar) componentrepresentation

(b)(a)

Example 3-1 Finding the Position Vector of a Shipwreck

The map in Figure 3-21 shows the location of the sunken fishing vessel Maki.For the coordinate frame in Figure 3-21a, use a protractor and a cm ruler to

Figure 3-19 The position vectorchanges along the trajectory of aprojectile. (a) Vector componentdescription. (b) Magnitude-and-direction description.

Figure 3-20 Component vectors and and scalar components (or simply compo-nents) x and y.

yS,xS

Wreck ofthe “Maki”

NE

WS

CapitalCity

37°N

E

W

O

S

Hidden Harboror

Wreck ofthe “Maki”Capital

City

Hidden HarborO

Scale: 1 cm = 1000 m

0 800 1600


NE

WS

+y

+x

+y

+x

+y

+x

CapitalCity

Hidden HarborO

(c)(b)(a)

x

y r

Figure 3-21 Finding a position vector by measurement. See Example 3-1.

On-The-Spot Activity 3-3Try drawing a position vector that has a negative x component and a positive y com-ponent. Repeat for a positive x component and a negative y component.


3-4 Working with Vector Components ◆ 61

Notice that a different choice of coordinate frame (Figure 3-22) changes both thewreck’s two-dimensional position vector and its components.

Most of what we have said about position vectors holds true for any kind ofvector, such as velocity, force, or other vector quantities that you will encounterlater. To completely describe any vector in two dimensions, you must either

1. give its magnitude and an angle representing orientation or direc-tion, or

2. give the values of its components in the x and y directions. We speak of theseas the x and y components of the vector, and we label them and Wecall (1) the magnitude and direction description of the vector, and we call(2) the component description. In either case, the fact that the vector is two-dimensional means it requires two numbers for its complete description. Incontrast, a scalar quantity such as time or density is fully described by a singlenumber.

Vy.Vx

uV 1or 0VS 0 2VS

find (a) the magnitude and direction of the wreck’s position vector and (b) the(scalar) components of its position vector.

Solutiona. Following Procedure 3-1, we draw an arrow (see Figure 3-21b) from the

origin at Hidden Harbor to the wreck and measure its length to be 2.0 cm.Using the scale (1 cm represents 1000 m), the magnitude Remember—vector magnitudes are always positive.

If we place a protractor as shown in Figure 3-21c, we get a reading ofBut because we are measuring clockwise from the axis, we must

take this as negative, so that As the figure shows, it is equallycorrect to say that

b. The component vectors and have been drawn in Figure 3-21b. Theirdirections must be conveyed by the signs of the components x and y: xpositive because is to the right, y negative because is downward. Thusthe (scalar) components are and (again using1 cm to represent 1000 m, so that the marks on the axes are 400 m apart).

You should pay careful attention to the signs on all results in thisexample.

y � �1200 mx � 1600 mySxS

ySxSu � �323�.

u � �37�.�x37°.

r � 2000 m.

Summary of Notation for Vectors• Vectors in general are represented in bold type (in printed texts) or with an

arrow above. In this text we use both: for position, for force, etc. Whenwriting longhand, use the arrow.

• The magnitude of a vector is represented by V (not in bold type) or The absolute value bars remind you that the magnitude is always positive.

• The directional angle is represented by If a situation involves several vec-tors etc., their directional angles will be represented by etc.

• The component vectors of in the x and y directions are represented byand

• The x and y (scalar) components of a vector are represented by andThe x and y components may be positive or negative, depending on the

vector’s direction.

• The components of a point’s position vector are usually written simply asx and y to remind you that these particular components are just the point’scoordinates.

rS

Vy.VxV

SVS

y.VS

x

VS

uA, uB,AS

, BS,

u.

0VS 0 .VS

FS

rS


(No y component now)

x

y

CapitalCity

Hidden HarborO

NE

WS

Figure 3-22 A different choice ofcoordinate frame. Compare withFigure 3-21.

◆ Related homework: Problems 3-25 and 3-29.

0540T_c03_49-90.qxd 08/04/2004 18:28 Page 61 EQA

✦CONVERTING BETWEEN DESCRIPTIONS The next step is for you to learnto convert back and forth between the two vector descriptions. In Figure 3-23,the features of both the magnitude and direction and component descriptions areshown together. The resulting diagram is simply a right triangle with sides and

and hypotenuse V (the magnitude or size of ). We now apply the relation-ships that hold true for all right triangles. For vectors, the Pythagorean theorembecomes

(3-2)

and the ratios of sides (see figure) become

(3-3)

(3-4)

(3-5)

Multiplying both sides by V in Equations 3-3 and 3-4, you get expressionsfor and :

Finding the component description of a vector from themagnitude-and-direction description:

(3-6)

(3-7)

( must be the angle the vector makes with the axis; if a different angle is given, you must first find )

✦FINDING COMPONENTS BY INSPECTION When a vector of known mag-nitude is purely in the ( ) x or y direction, it is easier to find its componentsjust by looking at it (called finding components by inspection) than by using theequations. For instance, a vector of magnitude 5 directed toward the left pointsentirely in the negative x direction and has no y component, so andVy � 0.

Vx � �5VS

��

u.�xu

Vy � V sin u

Vx � V cos u

1Vx and Vy2VyVx

tan u �Vy

Vx

sin u �Vy

V

cos u �Vx

V

V 2x � V y

2� V

2

VS

Vy

Vx


Figure 3-23 A right triangleconnects the magnitude-and-direction description and thecomponent description.

Magnitude-and-direction description

Componentdescription

VVy

Vx

Vθ

θ

θ

θ

θ

Vy

Vx

sin = =opphyp

Vy

V

cos = =adjhyp

Vx

V

tan = =oppadj

Vy

Vx

Vx2 + Vy

2 = V2

➥Using your calculator: For thematerial that follows, make sure youare familiar with using the sin andcos keys of your calculator. Supposeyou want to find cos Makingsure that your calculator is dealingwith angles in degrees (a typical cal-culator may have a DRG key thatyou can push until DEG, fordegrees, appears on your readoutdisplay), key in the number 140 andthen press the cos key. (The orderof these last two steps is reversedon some calculators.) You shouldget cos 140° � �0.766.

140°.

Figure 3-24 Vector for Example 3-2. (a) The given vector. (b) Itscomponents.

Example 3-2 From Magnitude and Direction to Components

The vector in Figure 3-24a is shown with its magnitude and direction descrip-tion. Find its component description and draw it.

SolutionCalculating the components. Use your calculator to obtain values of cos and sin and insert them in Equations 3-6 and 3-7:

and

Note that when you draw the component vectors (Figure 3-24b), is to theleft and is upward, because the vector in Figure 3-25a is simultaneously tothe left and upward.

◆ Related homework: Problem 3-30.

Vy

Vx

Vy � V sin u � 120.0 m2 sin 140° � 120.0 m2 10.6432 � 12.9 m

Vx � V cos u � 120.0 m2 cos 140° � 120.0 m2 1�0.7662 � �15.3 m

140°,140°

V = 20.0 m

(a) The given vector

(b) its components

= 140°θ



With Equations 3-2 and 3-5, you can calculate V and (and thus find )if you know and In other words,

to go from a component description to a magnitude and direction description,use

(3-2)

and (3-5)

From the signs of and decide what quadrant is in, then find from

is, take the inverse function u � tan�1 1Vy

Vx2.tan u—that

uuVy,Vx

tan u �Vy

Vx

V 2

� V x2

� V y2

Vy.Vx

utan u

✦CALCULATING VECTOR SUMS In Section 3-3 we developed a tail-to-headpicture of what vector addition means. We now develop a procedure for

(b)

= –69.6°θ

Example 3-3 From Components to Magnitude and Direction

In Figure 3-25a, Ms. Mulkeen’s window is 5.2 m to the left and 14.0 m upfrom the base of Firefighter Towle’s ladder. In what direction should the lad-der be angled and how far along it must Towle climb to rescue Ms. Mulkeen(assuming the truck must remain stationary)?

SolutionInterpreting the question. Taking the base of the ladder as the origin, the com-ponents of Ms. Mulkeen’s position vector are and Thequestion is answered by finding the magnitude and direction of this vector.For position vectors, becomes becomes y, and V becomes r.

Magnitude. From Equation 3-1, it follows that

Substituting the given values, you get

Direction. Applying Equation 3-4, we get

To find first key in then press the key (or the inv and tankeys; this varies from one calculator to another). Your calculator will tell youthat the angle that has this tangent is Does this (Figure 3-25b)look right? Your calculator always gives an angle between and for the given tangent value. But the negative tangent value can represent

either as in Figure 3-25b, or as inFigure 3-25c. The angles for the two situations are apart. As Figure 3-25c shows, we want the situation where the ladder points upwardto the left—a positive y component and a negative x component—so


u � �69.6° � 180° � 110.4�

180°

positive y componentnegative x component,

negative y componentpositive x component ,

�90°�90°u � �69.6°.

tan�1�0.371,u,

tan u �14.0 m

�5.2 m� �2.69

r � 21�5.2 m22 � 114.0 m22 � 2223.04 m2� 14.9 m

r � 2x2� y2

x, VyVx

y � 14.0 m.x � �5.2 m

(a)a

VyVV = y

n

(c)c

–69.6° 180°0.4= + °

θθ = –69.6°

180°

positive x

negative x

po

siti

ve y

neg

ativ

e y

x

y

Figure 3-25 Picturing Example 3-3. (a) The set-up. (b) Afirst attempt at a solution. (c) Thecorrected solution.

0540T_c03_49-90.qxd 08/24/2004 18:58 Page 63 EQA


calculating the resultant or vector sum of two or more vectors. Figure 3-26 showshow the component description of vectors is involved in this procedure. To developthe idea in the figure in fuller detail, work through WebLink 3-7. The figure showsthat the lengths of and placed end on end, add up to the length of In other words, In the same way, To add scalarcomponents, including those that are zero or negative, we do ordinary arithmeticaddition.

Adding more than two vectors just means adding each new vector to the pre-vious resultant. As Figure 3-27 shows, this just means arithmetically adding morecomponents in each direction, so that in general,

(3-8x)

and (3-8y)(Individual components may be positive, zero, or negative.)

By doing these additions, we are in fact finding the vector as given by itscomponent description. We can think of the vector addition equation

(3-8)

as a shorthand summary of the two arithmetic addition equations (3-8x, 3-8y).But what if you want a magnitude-and-direction description of As we saw

earlier, once you know either description of you can find the other. The com-ponent description, however, is better suited for calculating a resultant, becausethe actual steps of combining vectors (Equations 3-8x and 3-8y) require onlysimple arithmetic addition.

RS

,RS

?

AS

� BS

� CS

� . . . � RS

RS

,

Ay � By � Cy � . . . � Ry

Ax � Bx � Cx � . . . � Rx

Ay � By � Ry.Ax � Bx � Rx.RS

x.BS

x,AS

x

Figure 3-26 Developing a compo-nent description for the additionof two vectors. (a) Adding and

by the tail-to-head method to ob-tain the resultant (b) The x and ycomponents are shown for eachvector. (c) Adding the two vectorsmeans we can add the componentsin each direction to get the compo-nents of the resultant.

RS

.BS

AS

PROCEDURE 3-3

Calculating the Resultant of Two or More Vectors by Components

1. Use (3-6)

and (3-7)

to obtain a component description of each given vector (or find componentsby inspection if the vector is purely in the or direction). ( must beu�� y�� x

Vy � V sin u

Vx � V cos u

R

A

B

(a)

Ry

Rx

Ax

AyBy

Bx

(b)

Ay + By = Ry

Ax + Bx = Rx

+

+

=

=

or

or

(c)

Ax + Bx + Cx = Rx

Cx

Cy Cy

Ay + By + Cy = Ry +

++

=

=

or

or

Cx

C

Figure 3-27 Adding three vectors. By the same method as in Fig-ure 3-26, we add the third vector to the resultant of the first two.Again we find the components are additive.

If you start out with magnitude-and-direction descriptions of individual vec-tors, and you would like to calculate a magnitude and vector description of theirresultant, there are three main steps:

For WebLink 3-7:Vector Addition (Component Description),go towww.wiley.com/college/touger

0540T_c03_49-90.qxd 08/24/2004 17:34 Page 64 EQA


the angle the vector makes with the axis; if a different angle is given,you must first find )

2. Add components in each direction (Equations 3-8x and 3-8y) to obtain and the components of the resultant.

3. Once you know and use

(3-2 with )

and (3-5 with )

to find the magnitude R and the directional angle of the resultant.uR

VS

� RS

tan u �opposite

adjacent�

Ry

Rx

VS

� RS

R2x � R 2

y � R 2

RyRx

Ry,Rx

u.�x

Example 3-4 Calculating the Resultant of Two Vectors

For a guided interactive solution, go to Web Example 3-4 atwww.wiley.com/college/touger

A fishing trawler radios that after leaving harbor, it sailed 15 km in a directionof E, then headed due south for 21 km before its engines died. How

far and in what direction must a Coast Guard vessel based in the harbor goto provide assistance?

37° N

Brief SolutionSketch a diagram. Sketch the two legs of the trawler’s trip as vectors ( and

shown above). The course followed by the Coast Guard vessel should betheir vector sum Sketch the vector addition (shown above in b) carefullyenough so you can anticipate what should look like.

Now follow Procedure 3-3.

Convert to component description. Because is downward, we can find its com-ponents by inspection. has no x component and a negative y component,so and We verify this when we find the components of bothvectors by applying Equations 3-6 and 3-7:

Add components in each direction

Adding gives the resultant vector in component form.

Convert back to magnitude and direction description. Applying Equations 3-2and 3-5 gives

to two significant figures, and so that (Thesigns on and tell you that is indeed in the fourth quadrant.)


RS

RyRx

uR � �45�.tan uR �Ry

Rx� �12

12 � �1,

R � 2R2x � R2

y � 211222 � 1�1222 � 17

Ry � �12Rx � 12

By � B sin uB � 21 sin 270° � 1212 1�12 � �21 Bx � B cos uB � 21 cos 270° � 1202 102 � 0

Ay � A sin uA � 15 sin 37° � 1152 10.602 � 9 Ax � A cos uA � 15 cos 37° � 1152 10.802 � 12

By � �21.Bx � 0BS

BS

RS

RS

.BS

AS

A = 15

A

A = 37°B = 21

B

θAA

(b) Sketching A +A B(a)a The given vectors



Figure 3-28 rS1 � ¢ rS � rS2.

Example 3-5 Calculating Another Resultant of Two Vectors


NASA scientists are giving the Mars rover Sojourner its first test run on theMartian surface. They set it in motion, sending it 40.0 m in a direction Nof E (north of east), and then another 20.0 m in a direction of W beforestopping. Taking its initial deployment position as the origin, find the magni-tude and direction of Sojourner’s position vector when it stops.

Brief SolutionSketch a diagram. You should get roughly the diagram in Figure 3-29. Notethat N of W is counterclockwise from the direction.Now follow Procedure 3-3.

Convert to component description. For this we use Equations 3-6 and 3-7. has components

has components

Add components in each direction. We apply Equations 3-9x and 3-9y.

Adding x components Adding y components

y2 � 29.2 m x2 � 31.8 mresultant rS2

components of

¢y � 18.8 m ¢x � �6.84 mcomponents of ¢ rS y1 � 10.36 m x1 � 38.64 mcomponents of rS1

¢y � ¢r sin u � 120.0 m2 sin 110° � 120.0 m2 1�0.9402 � �18.8 m.

¢x � ¢r cos u � 120.0 m2 cos 110° � 120.0 m2 1�0.3422 � �6.84 m

¢ rS y1 � r1 sin u � 140.0 m2 sin 15° � 140.0 m2 10.2592 � 10.36 m

x1 � r1 cos u � 140.0 m2 cos 15° � 140.0 m2 10.9662 � 38.64 m

rS1

�x110°70°

70° N15°

Figure 3-29 Progress of the Marsrover Sojourner. A vector additiondiagram sketched for Example 3-5,and a NASA artist’s rendition of Soj-ourner rumbling across the Martianlandscape.

✦DISPLACEMENT VECTORS Although it is perfectly correct to use andto represent any two vectors and their resultant, we more commonly choose

symbols that tell us what physical quantities the vectors represent— for veloc-ity, for position, and so on. In two dimensions, Equation 3-1 becomes

In Figure 3-28a, as in Figure 2-4, an object moves from the tipof the earlier position vector to the tip of the later position vector So is a change in position vector; that is, a displacement vector.

(3-9)In words: earlier position displacement later position

vector vector vector

The component vectors of a position vector are simply labeled and If wedraw perpendiculars to the axes to locate the “shadows”, we see that

Vector addition applies equally for the one-dimensional component motions thatmake up our two-dimensional motion. The component equations and now become

(3-9x)

(3-9y) y1 � ¢y � y2

x1 � ¢x � x2

Ay � By � Ry

Ax � Bx � Rx

xS1 � ¢xS � xS2 and yS1 � ¢yS � yS2

yS.xSrS

��

rS1 � ¢ rS � rS2

¢ rSrS2.rS1

rS1 � ¢ rS � rS2.rS

vSRS

AS

, BS

,

r2

∆r

y

x

r1

(a)

(b)

r2y2 y1

∆y1

∆x1x1

x2

∆r

r1

y

x

0540T_c03_49-90.qxd 08/04/2004 18:28 Page 66 EQA

3-5 Velocity and Acceleration Vectors ◆ 67

Convert back to magnitude and direction description. (by Equations 3-7 and3-10)

and


tan u �y2

x2�

29.2 m

31.8 m� 0.918 so that u � 42.6�

r2 � 2x22 � y2

2 � 2131.8 m22 � 129.2 m22 � 43.2 m

–Ay

Ay –Ax

A

–A

Ax

Ax

Ay

3Ax

3Ay

3A

A

Figure 3-31 and �AS

.AS

Figure 3-32 Multiplying the vec-tor by a scalar. The multiplica-tion is depicted here for the scalarc � 3.

AS

For WebLink 3-8:Some Basic Rules

for Vectors, go to www.wiley.com/college/touger

It follows from these rules that multiplying a vector by a positive scalar onlychanges the scale of the vector (hence the term scalar).

3-5 Velocity and Acceleration VectorsIn two dimensions, as in one, we can define an average velocity vector overa time interval In equation form, the

(3-11)

with average velocity components and

You can find the average velocity of an object using a meter stick and a clockby following Procedure 2-1 to determine each of its components.

a3-11x

3-11ybvy �

¢y

¢tvx �

¢x

¢t

average velocity vS

�¢ rS

¢t

¢t.vS

Any vector obeys these rules• The magnitudes of and ( ) are equal.

• The directions of and ( ) are opposite.

• The scalar components of and ( ) are equal in absolute value but haveopposite signs.

• When you multiply or divide by a scalar c,—its components will also get multiplied or divided by c—its magnitude will be multiplied or divided by the absolute value of c.—the resulting vector will be in the same direction as the original vector if

c is positive, and opposite if c is negative.

To develop a fuller understanding of these rules, work through WebLink 3-8.

0c 0 ,AS

�AS

AS

�AS

AS

�AS

AS

AS

Figure 3-30 Visualizing two equiv-alent vector equations.

✦DISPLACEMENT VECTORS AND VECTOR SUBTRACTION As noted ear-lier, subtracting a vector means adding the negative of the vector (a vector withthe same magnitude but opposite direction). From Equations 3-9x and 3-9y, it fol-lows that

(3-10x)

and (3-10y)

Just as Equation 3-9 summarizes Equations 3-9x and 3-9y, Equations 3-10x and3-10y can be summarized by

(3-10)

Figure 3-30 compares the vector addition diagrams for Equations 3-9 and 3-10.The vector has components and

Some general rules that vectors obey are summarized below (see Figures 3-31and 3-32):

�y1.�x1�rS1

¢ rS � rS2 � 1�rS12 � rS2 � rS1.

¢y � y2 � 1�y12 � y2 � y1.

¢x � x2 � 1�x12 � x2 � x1

r1 + ∆r = r2

x1

r1 r1

y1 y

x

∆r

r2

(a)

(b)

r2 + (–r1) = r2 – r1 = ∆r

–r1

∆r

r2 –r1

y

x

–y1

–x1

0540T_c03_49-90.qxd 08/04/2004 18:28 Page 67 EQA


Figure 3-33 Finding the changein a velocity vector.¢vS

Example 3-6 Finding Your Change in Velocity


On an amusement park ride, your velocity changes from to (see part aof figure). Find the magnitude and direction of your change in velocity ¢vS.

vS2vS1

143º

v1 = 10 m/sv2 = 14 m/s 30º

(a)a The given vectors

(b) Sketching v2 – v1

Brief SolutionDoing the subtraction means adding to We sketch thisaddition (part b of figure) to anticipate what should look like. Thenwe apply Procedure 3-3 to calculate its magnitude and direction.

¢vSvS2

vS2.�vS1¢vS � vS2 � vS1

✦INSTANTANEOUS VELOCITY As in the one-dimensional case, the averagecomponents and close in on instantaneous values and as is shrunkdown to a single instant t. In two dimensions, instantaneous velocity is thevector that has and as its components. If velocity vectors are not specif-ically labeled average, they are instantaneous.

Notice that and are the rates at which the coordinates of a point objectare changing. In Case 3-2, they would be the velocities of the two “shadows” ofthe thrown ball. STOP&Think Look at the shadows in Figure 3-10. Is increas-ing, decreasing, or remaining about the same? What about ◆

The speed of an object (also instantaneous unless otherwise stated) is definedto be the magnitude of its velocity vector. You calculate it much as you calculateany vector magnitude:

(3-12)

By this definition, two objects may be traveling at the same speed, but if they aregoing in different directions, their velocities are different.

If an object’s instantaneous velocity changes from at instant to atinstant (Figure 3-33), the change in velocity is a vector as well.As always, the vector equations are summaries of the component equations ineach direction:

(3-13)

(3-13x)

(3-13y) ¢vy � v2y � v1y

¢vx � v2x � v1x

¢vS � vS2 � vS1

¢vS � vS2 � vS1t2

vS2t1vS1

speed v or 0 vS 0 � 2v2x � v2

y

vy ?vx

vyvx

vyvxvS

¢tvyvxvyvx

Actualpath

v2(at t2)

v1(at t1)

(a)

v1yv1y

v2xv2x

v1xv1x

∆vy = v2y – v1y

v1 + ∆v = v2

v2 (at t2)

v1(at t1)

(equivalent to ∆v= v2 – v1)

∆v(over interval t2– t1)

(b)

∆vx = v2x – v1x ∆vx = v2x – v1x

v2y

0540T_c03_49-90.qxd 08/04/2004 22:17 Page 68 EQA


STOP&Think If you experienced the velocity change that we calculated in Exam-ple 3-6, how would it feel? Would it feel different if it occurs over a 5-s intervalthan over a 0.5-s interval? ◆

The following case addresses this question.

Convert to component description. By Equations 3-6 and 3-7, has compo-nents

and has components

Because Equations 3-12x and 3-12y are equivalent to Equations 3-13x and3-13y, step 2 of Procedure 3-3 becomes . . .

Subtract components in each direction.

Subtracting x components: Subtracting y components:

This gives us the components of

Convert back to magnitude and direction description. (Equations 3-7 and 3-10)

and


tan u �¢vy

¢vx�

3.0 m/s

23.6 m/s� 0.13, so that u � 7.4�

¢v � 21¢vx22 � 1¢vy22 � 2123.6 m/s22 � 13.0 m/s22 � 23.8 m/s

¢vS.

¢vy � v2y � v1y � 3.0 m/s ¢vx � v2x � v1x � 23.6 m/s

�v1y � �6.0 m/s �v1x � �1�8.0 m/s2 v2y � 9.0 m/s v2x � 15.6 m/s

v2y � v2 sin u � 118 m/s2 sin 30° � 118 m/s2 10.502 � 9.0 m/s

v2x � v2 cos u � 118 m/s2 cos 30° � 118 m/s2 10.872 � 15.6 m/s

vS2

v1y � v1 sin u � 110 m/s2 sin 143° � 110 m/s2 10.602 � 6.0 m/s

v1x � v1 cos u � 110 m/s2 cos 143° � 110 m/s2 1�0.802 � �8.0 m/s

vS1

The blue car in Figure 3-34a travels halfway around atraffic circle at a speed of 10 m/s (about 22 miles/hour).The red car in Figure 3-34b is traveling at 10 m/s, andthen is propelled backward at 10 m/s after colliding

Case 3-3 ◆ Two Direction Reversals

v1 = 10 m/sv1 = 10 m/s

v2 = 10 m/s v2 = 10 m/s

before:

after:

= 10 m/s= v2v1

∆v

v2 v1

(c)(b)(a)

Figure 3-34 Changes in velocity during reversals ofdirection.

with an oncoming truck. In both cases, the direction isreversed, so the velocity has changed even though theinitial and final speeds are equal. Since (Fig-ure 3-34c), Solving for then gives

In both cases has thesame magnitude, 20 m/s. But for the blue car, thechange is gradual, occurring over a few seconds; forthe red car it is sudden, occurring over a small fractionof a second. What you experience differently in thetwo cases is the rate of change which—as in onedimension—we will call acceleration.

To understand more fully what happens to thevelocity of an object going around a circle at constantspeed, work through WebLink 3-9.

¢ vS

¢t ,

¢vS¢vS � �vS1 � vS1 � �2 vS1.¢vSvS1 � ¢vS � �vS1.vS2 � �vS1

For WebLink 3-9:Circle,

go towww.wiley.com/college/touger



✦ACCELERATION IN TWO DIMENSIONS When the velocity of a bodychanges (Equation 3-12) over a time interval its components in each direc-tion correspondingly change (Equations 3-12x and 3-12y). Then by Procedure 2-2,we can find an average acceleration in each direction:

(3-14x, 3-14y)

This procedure defines the

(3-14)

by prescribing the steps necessary to find its components. As in one dimension,when you calculate must be instantaneous velocities.STOP&Think By this definition, is an object accelerating if it travels along a cir-cular path at constant speed? Jot down your answer and your reasoning.

Did you recall that constant speed means only that the magnitude of thevelocity is constant, not the direction? Does that alter your answer?

Now consider Figure 3-35a, showing the path of this object. The velocity vec-tors of the object at points A and B are shown equal in magnitude, as required.Is the horizontal component vx the same for these two vectors? Is equal tozero? And thus, is equal to zero? What about and Once more:Is an object accelerating if it travels along a circular path at constant speed? Ifthis is still unclear, it may help to look again at WebLink 3-9. ◆

If the average acceleration vector is also nonzero and is inthe same direction as There is a change in the velocity as long as eitherthe magnitude or the direction of the velocity changes, and the direction contin-uously changes when an object travels in a circle at constant speed.

¢vS¢vS.aS

� ¢ vS

¢t¢vS � 0,

ay?¢vyax � ¢vx

¢t

¢vx

¢vS � vS2 � vS1, vS

1 and vS2

average acceleration vector aS

�¢vS

¢t

ax �¢vx

¢t and ay �

¢vy

¢t

¢t,vS

A

B

C

Dv2 = v1

135° P

P

y

x

v2

v1

v

v

a

(b)(a)

Figure 3-35 Traveling in a circleat constant speed. Is the velocitychanging? Is the car accelerating?(Point P is halfway along the arcconnecting positions C and D.)

Example 3-7 Calculating the Average Acceleration of a Car Following a Circular Path

Figure 3-35b shows a car taking a semicircular U-turn at constant speed. Itsvelocity vectors at points C and D, apart, are also shown. Suppose its speedis 10.0 m/s and it takes 4.0 s to get from C to D. Find the components of theaverage acceleration over this time interval, and then find the magnitude anddirection of the average acceleration for the same interval.

SolutionChoice of approach. You cannot be sure you are calculating accelerationcorrectly unless you are clear on its definition. Once you find its componentsfrom the definition, you can convert to a magnitude and direction description.

90°



STOP&Think Without doing any further calculation, what is the direction of in Example 3-7? How do you know? If you are not sure, try this On-The-Spot

Activity: From the head of in Figure 3-36, draw the vector that, when addedto in this way, gives you the correct as the vector sum. ◆

✦INSTANTANEOUS ACCELERATION As was true in one dimension, and(averages over an interval ) approach limiting values— and —when the

interval shrinks to a single instant t. The instantaneous acceleration is thevector that has and as components at that instant. (In the previous example,the instantaneous values and at point P are equal to the average values and for the trip from C to D.)ay

axayax

ayax

aSayax¢tay

ax

vS2vS1

¢vSvS1

¢vS

What you know/what you don’t.

(because at point C, the velocity is entirelyin the direction)

(because at point D, the velocity is entirelyin the direction)

( )

The mathematical solution. We can now follow Procedure 2-2:

and

So

and

Making sense of the result. Like is to the left, because multiplying ordividing a vector by a positive scalar such as doesn’t change its direction.Likewise, must be upward because is upward.

Now apply step 3 of Procedure 3-3 to find

Important: Although the speed does not change, the velocity does becausethe direction is changing, so the value we get for the acceleration is not zero.

This corresponds to

in the second quadrant (and ) because the signs on its componentstell us that points up and to the left. The instantaneous acceleration hasits average value at the midpoint P of the car’s path; at this point (see insetin Figure 3-35), the direction is toward the center of the circle.


u � 135°aS

aS

aS

not � 45°

U � 135°

tan u �ay

ax�

2.50 m/s2

�2.50 m/s2 � �1.00

� 3.54 m�s2

a � 2a2x � ay

2 � 21�2.50 m/s222 � 12.50 m/s222a and u:

¢vSyaS

y

¢t¢vSx, a

Sx

ay �¢vy

¢t�

10.0 m/s

4.0 s� �2.5 m�s2

ax �¢vx

¢t� �

10.0 m/s

4.0 s� �2.5 m�s2

¢vy � v2y � v1y � 10.0 m/s � 0 � 10.0 m/s 1¢vSy is upward2

¢vx � v2x � v1x � 0 � 10.0 m/s � �10.0 m/s 1¢vSx is to the left2

ax � ? ay � ? a � ? u � ?

¢vx � ? ¢vy � ?so no need to know t1 and t2¢t � 4.0 s

�yv2x � 0 and v2y � 10.0 m/s

�xv1x � 10.0 m/s and v1y � 0

v2

v1

Figure 3-36 Complete the vectordiagram so that it shows the vectoraddition Draw thevector so that the resultant goes from the tail of the first vector to the head of the second vector ¢vS.

vS1

vS2¢vSvS1 � �vS � vS2.



3-6 Solving Motion Problems in Two Dimensions:Projectile Motion Revisited

The mathematics of vectors lets us separate a two-dimensional motion probleminto problems about one-dimensional component or “shadow” motions, to whichwe can apply the methods of Chapter 2.

In one dimension In two dimensions we used . . . we now use . . .

and (3-11x, y)

and (3-14x, y)

and (3-15x, y)

Alternatively, when the acceleration is constant, we can use the set of equations(2-9, 2-11, and 2-12) that we derived for that special case (as in Solution 2 ofExample 2-11). However, now we need a set of equations for each of the twoone-dimensional component motions.

Equations of Motion in Two Dimensions for Constant Acceleration

(3-16x) (3-16y)

(3-17x) (3-17y)

(3-18x) (3-18y)

The equations of motion in each direction involve only the components in thatdirection. They become simpler when you can assume the object is at the originat so that and They are also simpler if you choose yourcoordinate system so that one axis is in the direction of the acceleration.STOP&Think Why? How do the equations become simpler? We do this now forprojectile motion. ◆

✦PROJECTILE MOTION Figures 3-7 and 3-9 (and also WebLink 3-4) showedthat a projectile (such as a thrown ball) follows the same kind of trajectory asthe weight in the truck-mounted carnival game, assuming that we can neglect airresistance. Like the constant horizontal velocity component provided by the truck,the projectile’s horizontal component is constant (recall Figure 3-10), so In both cases, (the direction is upward). So the constant accelerationequations become

horizontal motion vertical motion

When the terms with disappear. The equations for and then justsay that the horizontal velocity at any instant remains equal to the initialhorizontal velocity (We cross these out to indicate they provide no newinformation.)

If you solved for t in the first of the horizontal motion equations, giving youand substituted this into the first of the vertical motions, you would

find that y is equal to a quadratic expression in x, such as whereax2� bx � c,

t �x � xo

vox,

vox.vx

v2xvxaxax � 0,

v2y � v2

oy � 2g1y � yo2v2x � v2

ox

vy � voy � gtvx � vox

y � yo � voy t � 1

2 gt2x � xo � vox t

�yay � �gax � 0.vx

yo � 0.xo � 0t � 0,

v2y � v2

oy � 2ay 1y � yo2v2x � v2

ox � 2ax 1x � xo2vy � voy � ay

tvx � vox � ax t

y � yo � voy t � 1

2 ay t2x � xo � vox

t � 12 ax

t2

vy �voy � vy

2vx �

vox � vx

2v �

vo � v

2Condition foruniform motion

ay �¢vy

¢tax �

¢vx

¢ta �

¢v

¢tDefinition ofaverage acceleration

vy �¢y

¢tvx �

¢x

¢tv �

¢x

¢tDefinition ofaverage velocity

at the sameinstant t as

at thesame instant

t as


3-6 Solving Motion Problems in Two Dimensions: Projectile Motion Revisited ◆ 73

a, b, and c are constants. The graph of a quadratic expression is a parabola.A graph of y versus x shows the two-dimensional path that the object actuallyfollows. If you are familiar with parabolas from your math courses, you shouldrecognize that the paths followed by the objects in Figures 3-5b and 3-9 areparabolic.

As an object travels along a two-dimensional trajectory, its position and veloc-ity vectors and vary. As they do so, their x and y components likewise vary.Each of Equations 3-16x through 3-18y is a relationship among the vectorcomponents in a single direction. To use these equations, you must first have acomponent description of each vector.

vSrS

PROCEDURE 3-4A Strategy for Solving Motion Problems

1. In identifying what you know and what you don’t, give a component descrip-tion of each vector. Use Equations 3-6 and 3-7 to find components whennecessary.

2. In your solution, use separate equations for the motions in the x and ydirections. This lets you solve for the components of a vector quantity.

3. [if necessary] You can find the magnitude and direction of a vector quantityby applying Equations 3-2 and 3-5 once you know its components.

Example 3-8 A Symmetrical Trajectory

For a guided interactive solution, go to Web Example 3-5 at www.wiley.com/college/touger

A test rocket, due to technical failure, experiences no further thrust after leav-ing a ground-level launching pad at a speed of 44.0 m/s and at an angle of

above the ground. Find and the speed at 1-s intervals untilthe rocket lands. Then plot x, y, and against time to see graphically thesymmetry in the numerical results.

Brief SolutionThe mathematical solution. First you must find the components of

The equations and values are presented in tabular form to make patterns in thenumbers more evident and to facilitate graphing. The numerical calculations thatlead to the results are displayed for one representative instant

t(s) (m) (m) (m/s) (m/s) (m/s)

0 0 0 20.0 39.2 44.0

1 20.0 34.3 20.0 29.4 35.6

2 20.0

3 60.0 73.5 20.0 9.8 22.3

4 80.0 78.4 20.0 0 20.0

5 100. 73.5 20.0 22.3

6 120. 58.8 20.0 28.0

7 140. 34.3 20.0 35.6

8 160. 0 20.0 44.0�39.2

�29.4

�19.6

�9.8

�28.0�19.6�58.8�40.02120.022 � 119.62239.2 � 19.82 122139.22 122 �1

2 19.822220.0 � 2

v � 2vx2

� vy2vy � voy � gtvx � voxy � voyt � 1

2 gt2x � voxt

1t � 2 s2.

voy � vo sin 63° � 144.0 m/s2 10.8912 � 39.2 m/s

vox � vo cos 63° � 144.0 m/s2 10.4542 � 20.0 m/s

vo.

vyvx,vx, y, vx, vy,63°



Figure 3-38 shows what happens to the velocity vector and its components dur-ing the rocket’s symmetrical trajectory. Figure 3-38d illustrates the fact that we getthe velocity vector at the end of each like time interval by adding (in a vectorsense) the same to the velocity vector at the beginning of the interval. Each

is the same because does not change (Figure 3-38e).aS¢vS � aS ¢t¢vS

The quantities x, y, and are plotted against time in Figure 3-37, as wellas a graph of y versus x that displays the actual shape of the trajectory.STOP&Think Is the velocity zero at the rocket’s highest point? (Why, or whynot?) Does at this point? (Why, or why not?) What about In whatdirection is the rocket moving at its highest point? ◆

◆ Related homework: Problems 3-52, 3-53, 3-54, and 3-55.

vx?vy � 0

vyvx,

Figure 3-37 Graphs of quantities describing the motion of the rocket in Example3-8. Notice what is plotted in each graph, and make sure you understand why eachgraph looks the way it does.

80

60

40

v y in

m/s t in seconds

40

20

0

-20

-40

y in

met

ers

x in meters

(actual path of rocket)

20

00 40 80 120 160

(e)(d )

80

60

40

y in

met

ers

t in seconds

20

00 1 2 3 4 5 6 7 8

160

120

80

40

00 2 4 6 8

x in

met

ers

t in seconds

0

10

20

30

40

0 4 8

v x in

m/s

t in seconds

(c)(b)(a)

Figure 3-38 Vector aspects of a symmetrical parabolic trajectory. The following vec-tors are shown at equal time intervals during the course of the trajectory. (a) The hori-zontal vector component of the velocity. (b) The vertical vector component of the velocity.(c) The resultant velocity (with its horizontal and vertical components shown in a lightershade). (d) Each resultant velocity is obtained by adding the same to the previousvelocity (e) The acceleration vector. Note that because each is the samein (d) and is directed vertically downward because the acceleration is a constant down-ward vector.

aS¢vS¢vS � aS ¢t,vS1.

¢vSvS2

¢t

vx

vx

vy = 0 v

v

v

vx

v2

v2

v2

vv1

v1v1

v1

v1

v1

v1

v1

a

v1

v1

v1

v2v2

v2v2

v2

v2

v2

v2

v2(d) (e)(c)(b)(a)

∆

v∆v∆

v∆

v∆

v∆

v∆

v∆

v∆

v∆

v∆

vy

vy

a

a

EXAMPLE 3-8 continued

0540T_c03_49-90.qxd 08/24/2004 17:34 Page 74 EQA

3-6 Solving Motion Problems in Two Dimensions: Projectile Motion Revisited ◆ 75

Example 3-9 Finding the Final Velocity of a Projectile


A car in a movie chase scene is traveling on level ground at a speed of 45m/s when it hurtles off a 20-m-high sea cliff. With what speed and in whatdirection does the car hit the water below?

Brief SolutionRestating the problem. If we let the diver’s starting position be the water’s surface is at “Speed” is the magnitude of the velocityvector. The quantities in the equations of motion are components of vectors,but the question is asking for the magnitude and direction of the final velocityvector:

Choice of approach. We will work from basic definitions and the condition foruniform acceleration (Equations 3-15x, y). It is also possible (see Web version)to use the derived equations of motion to obtain the same result.

What you know/what you don’t. State as components

(the road is level where the car leaves the cliff )

The mathematical solution. First find the components of Inthe x direction, because In the y direction, the con-dition for uniform acceleration is because Then thedefinition gives us

The definition of average acceleration gives us so that

Then, Thus and

Because the vertical velocity is down-ward, we must choose the negative solution:

Now we find the magnitude (speed) and direction of the final velocityfrom its components: sothat the speed

and so that


u � �24� 1or u � 336�2tan u �vy

vx� �20 m/s

45 m/s � �0.44,

v � 49 m�s

v2� vx

2� vy

2� 145 m/s22 � 1�20 m/s22 � 2425 m2/s2,

vy � �20 m/s.��2�219.8 m/s22 1�20 m2 � ��20m/s.

vy � 1�2gy �v2y � �2g ¢y¢y �

vy

2 ¢t �vy

2 1 vy

�g 2.¢t �vy

�g.

�g �vy � 0

¢t ,ay � ¢vy

¢t

vy ¢t �vy

2 ¢t � ¢y

vy � ¢y¢t

voy � 0.vy �voy � vy

2 �vy

2

ax � 0.vx � vox � 45 m/sv 1vx � ?, vy � ?2.

v � ? and u � ? when y � �20 m 1or when ¢y � �20 m2vox � 45 m/s voy � 0

xo � yo � 0 ax � 0 ay � �g � �9.8 m/s2

v � ? and u � ? when y � �20 m.

y � �20 m.xo � yo � 0,

In Example 3-9, we got two solutions for but only the negative one wascorrect for the descending car. When you get two solutions for or for t at agiven value of the height y, both have meaning for the real-world situation onlyif the projectile actually passes that height during both ascent and descent. Thismay not happen if the projectile’s initial and final y coordinates are different, aswhen a cannonball is fired from a cliff. In contrast to the cannonball in

vy

vy,



y

y = h

P

(b)(a)

Figure 3-39 Trajectories of twocannonballs. One cannonball isfired from ground level, the otherfrom a height. The dashed line in(b) is an unbounded parabola—onethat has neither beginning nor end.This is the trajectory represented bythe equations of motion. The trajec-tories of the two cannonballs areboth bounded segments of thisparabola.

Example 3-10 Another Projectile


A kicked soccer ball leaves the soccer player’s foot at a speed of 19.21 m/sin a direction above the horizontal. How high does the ball rise? Howfar from the point where it was kicked does the ball land?

Brief SolutionChoice of approach. See Example 3-8. Assume the ball was kicked at the origin.

Restating the problem. Now the vertical “shadow” motion has zero velocity atthe highest point, so “how high?” becomes: when How far is ask-ing for the final horizontal position x when . . . when what? Because y is againzero at this point, the question then becomes: when

Your equations don’t give you a direct relationship between y and x, butknowing y you can find t, and then knowing t you can find x. The time t pro-vides the link between the horizontal and vertical situations (see Figure 3-40),because the two component or “shadow” motions are occurring together in time.

What you know/what you don’t.

y � ? when vy � 0 x � ? when y � 0 1t � ? when y � 02xo � yo � 0 ax � 0, ay � �g � �9.8 m/s2 vo � 19.21 m/s uo � 38.66°

y � 0.x � ?

vy � 0.y � ?

38.66°

Equations of motion in x direction

Equations of motion in y direction

Find t

Find t

Use t in…

Use t in…… to find x, vx, ax

… to find y, vy, ay

tFigure 3-40 A strategy forapproaching problems involvingmotion in both the x and ydirections.

Figure 3-39a, the cannonball in Figure 3-39b passes only while descending,even though the two trajectories are identical beyond point P. Both trajectories,in fact, are segments of a trajectory (the dotted curve) that has no beginning orend. The equations of motion give information about the whole dotted trajectoryfrom to but the information is meaningful only during the timeinterval when the projectile is actually in flight. You must reject solutions thatoccur outside this interval.

When—but only when—the trajectory is symmetrical, the velocity has thesame magnitude for any point on the upward path and the corresponding pointon the downward path. In that case, the upward and downward paths musttake the same amount of time. You could calculate the instant at which (the peak of this trajectory), and double it to obtain the time duration of the entiretrajectory. Symmetry arguments will simplify the solution in the next example.

vy � 0

t � ��,t � ��

y � h


Summary ◆ 77

The mathematical solution.• (Step 1 of Procedure 3-4) Obtain the component description. For velocity,

• (Step 2 of Procedure 3-4) Once you know Equation 3-20y permits youto solve for y when

becomes

so that the maximum height

Next, find the time t at which the ball returns to Finding t is simplifiedif you realize from symmetry that at this time. You can then sub-stitute for in Equation 3-17y

and solve for t, getting

Thus, the ball lands at

A second symmetry argument. Because the trajectory is symmetrical when theball starts and lands at the same height, you could also find the time t at whichthe ball reaches maximum height—that is, when half the trajectory iscompleted—and then double it. Using the fact that at this point, Equa-tion 3-17y gives you Solving for t, you get

at the halfway point, so that seconds when the ball lands,just as before.


t � 2.44t � 1.22 s0 � 12.00 m/s � 19.8 m/s2t.vy � 0

x � voxt � 115.00 m/s2 12.44/s2 � 36.6 m

t �2voy

g�

2112.00 m/s29.8 m/s2 � 2.44 s

�voy � voy � gt

vy�voy

vy � �voy

y � 0.

y �v2oy

2g�112.00 m/s22219.8 m/s22 � 7.35 m

02� voy

2� 2gy

vy2

� voy2

� 2gy

vy � 0:voy,

voy � vo sin 38.66° � 19.21 m/s 10.62472 � 12.00 m/s

vox � vo cos 38.66° � 19.21 m/s 10.78092 � 15.00 m/s

✦ S U M M A RY ✦

This chapter developed the conceptual and mathematical toolsfor treating motion in two dimensions. We began with the ideathat simple motions can be combined or superimposed toproduce more complex resultant motions. Reversing this, youcan break down or analyze or resolve a complex motion intoits component parts.

In particular, you can break a two-dimensional motiondown into two one-dimensional “shadow” motions, and do thesame things that we did in one dimension for the components

in each direction. We could solve problems (such as Examples3-6 and 3-7) using the component definitions of average veloc-ity and average acceleration:

(3-11x, 3-11y)

(3-14x, 3-14y) ax �¢vx

¢t and ay �

¢vy

¢t

vx �¢x

¢t and vy �

¢y

¢t

An equally valid approach to Example 3-10 is to use Equation 3-16y to findthe time t at which the ball lands, then use Equation 3-17x as before. The cal-culation is more difficult in this approach because you generally need to applythe quadratic formula to solve for t in Equation 3-16y. However, if the initial andfinal heights are different, it is necessary to do it this way because you cannotmake use of symmetry arguments. (To check your understanding of this secondapproach, try Problem 3-71.)


When the motion is uniformly accelerated (Section 3-6),we need to supplement these with the condition for uniformacceleration

(3-15x, 3-15y)

or use the derived equations of motion for uniform accelera-tion in each direction:

(3-16x) (3-16y)

(3-17x) (3-17y)

(3-18x) (3-18y)

Projectile motion (when air resistance is negligible) is aparticular instance of uniform acceleration. In this case, we take

and in the above equations. (In particular,3-16x becomes and 3-17x and 3-18x reduce to

)Vectors provide a framework for treating motion in two

dimensions mathematically. Vectors can be specified by amagnitude and direction description or by a componentdescription.

To convert from a magnitude and direction description toa component description, use

(3-6)

(3-7)

and

Vy � V sin u

Vx � V cos u

vx � vox.x � xo � voxt

ay � �gax � 0

v2y � v2

oy � 2ay 1 y � yo2v2

x � v2ox � 2ax 1x � xo2

vy � voy � aytvx � vox � axt

y � yo � voyt � 12 ayt2 x � xo � vox t � 1

2 ax t2

vx �vox � vx

2 and vy �

voy � vy

2


✦ Q UA L I TAT I V E A N D Q UA N T I TAT I V E P RO B L E M S ✦

H a n d s - O n A c t i v i t i e s a n d D i s c u s s i o n Q u e s t i o n sThe questions and activities in this group are particularly suitable forin-class use.

3-1. Hands-On Activity. Remove the labels from a large sodabottle so that it is transparent. Cut a circular hole in a stiffpiece of cardboard, just large enough to let the bottle slidefreely through it (Figure 3-41). Have a friend press down onthe bottle to hold it still. With the cardboard held stationaryat one height, make a circle on the bottle by tracing along theinner edge of the cardboard with a marking pen. Next, keep-ing the cardboard horizontal, lower it along the bottle at aslow, steady speed. As you do this, continue to mark the bot-tle by running your marker along the circular inner edge ofthe cardboard. As the figure shows, you should end up witha spiral trace on the bottle. Describe the two componentmotions that combine to produce the spiral motion for themarker tip. Figure 3-41 Problem 3-1

Marker circlesbottle

Marker circlesbottle

Movecardboard

(b)(a)

atthe

sameinstant

t as

to convert from component description to a magnitude anddirection description, use

(3-2)

and (3-5)

and then find from

Any vector equation is a summary of equations involvingcomponents in each direction. For example, sum-marizes the two equations and Itis useful to express vector equations in component formbecause the components add by ordinary arithmetic. But youshould also be able to form a picture (such as Figure 3-17) byProcedure 3-1 of what the sum of any two vectors looks like.

Subtracting vectors simply means adding the negative ofone vector to another: If has compo-nents and has components and and isopposite in direction to In particular, we use this when find-ing changes in a vector such as

(3-10) or (3-13)

Resolving vectors into components lets you solve two-dimensional motion problems using one-dimensional equationsof motion, each involving components of position, velocity,and acceleration vectors (see Procedure 3-4).

Because speed is defined as the magnitude of the veloc-ity vector,

(3-12)

an object changing direction can change velocity withoutchanging speed. Because the acceleration vector isnonzero whenever there is a change in the velocity, anobject is accelerating if its velocity vector is changing in anyway at all, even if only in direction (see Example 3-7).

¢vSaS

� ¢ vS

¢t

v or 0 vS 0 � 2v2x � v2

y

¢vS � vS2 � vS1¢ rS � rS2 � rS1

BS.

�By�BxBy, � BS

Bx

BS

AS

� BS

� AS

� 1�BS 2.

Ry � Ay � By.Rx � Ax � Bx

RS

� AS

� BS

tan u.u

tan u �Vy

Vx

V2� V 2

x � Vy2


3-2. Hands-On Activity.a. Draw a circle at least 0.2 m (about 8 in.) in diameter on a

large sheet of paper (a double page of newspaper will donicely). Move a drinking glass or beaker at steady speedalong this circle as though it were a circular track, makingcertain to keep the center of the glass’s bottom on the track.As the glass circles the track, trace around the bottom witha pencil so that a continuous path is marked out on thepaper (Figure 3-42).

b. How do the two component motions in this activity com-pare to the two component motions in Figure 3-4?

c. How does the resultant motion of your pencil point com-pare with the actual path of the Octopus rider in Figure3-4b?

3-3. Discussion Question. At the end of Section 3-1, a popfly in baseball was given as an example of projectile motion.A baseball player can also hit a line drive. How is this differ-ent than a pop fly? How is it the same? Is the motion of abullet fired from a gun projectile motion? In the example ofthe carnival game on the back of a truck (Case 3-2), how canyou vary the situation to make the trajectory of the weighteither more like a pop fly or more like a line drive?

••3-4. Discussion Question. A ball can go from point A topoint B by rolling down either of the tracks in Figure 3-43. Ifit is released from rest at A, will it get to B faster by rollingdown track I ortrack II? Explain.(Adapted from W. J.

Leonard and W. J.

Gerace, The Physics

Teacher, 34, 280–283

[1996].)

3-5. Discussion Question. The set-up described in Figure3-50 (see Problem 3-16) is altered so that now there is onlyone screen (Figure 3-44), but there are two trains traveling onseparate tracks between points X and Y. Suppose both trainstravel at constant speed, but not the same constant speed.However, both leave point X at the same time and both arriveat point Y at the same time. If you watch only the shadowsthat the two balls cast on the screen, how would the motionsof these two shadows compare?

Qualitative and Quantitative Problems ◆ 79

Figure 3-42 Problem 3-2


Motion ofmarker around

glass

Motion ofglass around circle

Track I

Track II


X

Screen

Y

Light from distant projector

Train A

Train B

Review and PracticeSection 3-1 Constructing Complex Motions from Simpler Motions

3-6. Analysis of complex motion into simpler motions: The“sawtooth” pattern in Figure 3-45 is traced on an xy recorder.The motion of the pen that left the trace can be analyzedinto two one-dimensional component motions. Describethese two motions. (Tocheck whether youridea works, move apiece of paper byhand and try it your-self.)

3-7. Synthesis of complex motions from simpler motions: A car-toon figure is drawn on a disk (Figure 3-46a). A pencil pokedthrough the disk’s center (Figure 3-46b) serves as an axis ofrotation. To animate the figure, you can rotate the disk aboutthe pencil at the same time you move the pencil in a straightline to the right, as shown. (You can trace or photocopy thedisk and cut it out to try this for yourself.) What does the


cartoon figure appear to be doing when both of these motionsare executed at once?

3-8. As the paper of an xy-recorder scrolls, the pen remains atheight for 1 s, drops abruptly to height (2 cm lower) whereit remains for the next second, returns to height for the thirdsecond, to for a fourth second, and continues alternating inthis way. Sketch the trace left by the pen on the paper.

h2

h1

h2h1


rotationof disk

motionof pencil

(b)(a)

SSM Solution is in the Student Solutions Manual WWW Solution is at http://www.wiley.com/college/touger

0540T_c03_49-90.qxd 08/25/2004 15:12 Page 79 EQA

3-9. Synthesis of com-plex motions from sim-pler motions: In Fig-ure 3-47, a heavy metalball hangs from a longrope over a sandbox onwheels. A slender metalrod attached to the bot-tom of the ball canleave a track in the sandwhen the ball swings.Suppose that the sand-box is pulled to theright at a slow constantspeed. Meanwhile, theball is given a gentlepush so that it swingstoward side A of thesandbox. Sketch anoverhead view of thetrack traced in the sand as the ball swings forward and backseveral times.

3-10. The pen of an xy-recorder moves between two heightsand (see Figure 3-48a) while the paper scrolls to the

right. Below are five descriptions of the pen’s motion. Fig-ure 3-48b shows five traces left by the pen on the paper. Foreach description of motion, write the number of the trace thatthat motion produces.a. Pen drops from to at constant speed.

b. Pen drops suddenly from to pauses for a second,goes up suddenly from to pauses for a second, thenrepeats the same sequence of steps.

h1,h2

h2,h1

h2h1

h2h1


3-13. SSM In Case 3-1, the weight goes up and down the slidewire of the carnival game while the truck is moving (see Fig-ure 3-7a). Assume friction in negligible. Indicate whether eachitem below is a true or false statement about the resulting two-dimensional motion of the weight.a. It has constant velocity in both the vertical and horizontal

directions.

b. It has the same acceleration as a ball thrown vertically intothe air.

c. It has a constant upward velocity and then a constant down-ward velocity.

d. It is accelerating in both the vertical and horizontal directions.

e. It has zero acceleration at the highest point in its path.

f. It has zero velocity at the highest point in its path.

g. Its horizontal velocity doesn’t change.Figure 3-48 Problem 3-10

c. Pen goes up and down between and at constant speed(except while quickly reversing direction).

d. Pen speeds up as it goes from to

e. Pen slows down as it goes from to

3-11. A principal step in manufacturing screws is to cut thethread that winds around the shaft. Describe two motions thatare necessary for the process of cutting the thread.

3-12. A straight track placed on a table top is seen from abovein Figure 3-49. It is shown at five equally spaced instants asit is slid across the table at constant speed As it beginssliding a cart at the left end of the track is givena velocity to the right. The cart rolls along the track as thetrack is slid, but the cart slows down at a constant rate becauseof friction. It just reaches the other end of the track at

a. The dot in the middle of the cart in the figure indicates itsinitial position. Draw dots on the figure showing roughlywhere the cart is when the track is at each of its laterpositions.

b. Now draw a smooth line connecting the five dots that rep-resent the cart’s position at the five depicted instants. Is thisline straight or curved?

c. The smooth line that you have drawn shows the path ofthe cart as seen by a stationary observer looking down atthe table. Can the motion along this path be understood asa superposition of straight line motions? Explain.

t4 � 1.00 s.

1at to � 02,vtrack.

h2.h1

h2.h1

h2h1


1

Trace left by pen

2

3 4

5

y

Paper scrolls atconstant speed

Pen moves up and down wire

y = h1

y = h2

vtrack

t0 = 0

t1 = 0.25 s

t2 = 0.50 s

t3 = 0.75 s

t4 = 1.00 s

Push

Pull

Ball

Rod


(a)

(b)

0540T_c03_49-90.qxd 08/24/2004 17:35 Page 80 EQA

Section 3-2 Breaking Down Two-Dimensional Motionsinto One-Dimensional Components: Projectile Motion

3-14. A basketball player shoots a foul shot in an outdoorplayground where the sun is directly overhead. As the balltravels from the player’s hands to the basket, the shadow thatthe ball casts on the ground ____. (accelerates; slows down;slows down then speeds up; speeds up then slows down; movesat constant speed)

3-15. Using the data in Figure 3-10 (Section 3-2) and makingany necessary measurements, calculate the speed of the golfball’s horizontal shadow motion.

3-16. A toy electric train moves along a curved track in theshadow box in Figure 3-50. A ball on a post is mounted onthe train. When light beams from distant slide projectors (orstrong narrow-beam flashlights) X and Y are directed at theball, the ball casts shadows on screens X and Y as shown.

g. As the train follows different possible paths from A to B butalways at constant speed, is it ever possible that one of thetwo shadows cast by the ball will be accelerating but notthe other? Briefly explain.

Section 3-3 Vectors

3-17. Which of the following are vector quantities and whichare scalars? Explain your reasoning.

speed length time

displacement velocity angle

acceleration

3-18.a. Turn to any page of your physics textbook. If the origin is

at the lower left-hand corner of this page (and the axes ori-ented in the usual way), determine by measurement themagnitude and direction of the position vector for the upperright-hand corner of the page.

b. If the origin is at the upper left-hand corner of this page,determine by measurement the magnitude and direction ofthe position vector for the lower right-hand corner of thepage.

3-19. SSM WWW Sketch the vector and find its componentsif its magnitude Aa. is 10 and its direction is above (counterclockwise from)

the axis.

b. is 10 and its direction is counterclockwise from the axis.

c. is 8 and its direction is below (clockwise from) the axis.

d. is 5 and its direction is below the negative x axis.

3-20. The aerial surveillance picture in Figure 3-51 show thelocation X of a proposed statue, but on each copy of the pic-ture, a different coordinate frame has been drawn for deter-mining positions. For each choice of coordinate frame, sketchthe position vector for X, then use a cm ruler to obtain thecomponents of this vector. Assume the scale is 1 cm � 50 m.

37°

�x60°

�x135°

�x37°

AS

vx

Figure 3-50 Problems 3-16, 3-80, and 3-81



Screen Y

Light fromdistantprojector X

Light fromdistantprojector Y

Screen X

A

B

X

X

0 50

X

scale100 m

X

y y

x

y

x

x

y

x

a. Suppose the train moves along the track at constant speedfrom A to B in the figure. Does the shadow that the ballcasts on screen X speed up, slow down, or move at con-stant speed? Explain your answer.

b. Does the shadow that the ball casts on screen Y speed up,slow down, or move at constant speed? Explain youranswer.

c. Can a constant speed motion in two dimensions have one-dimensional component motions that are accelerated?Explain.

d. Suppose the track were laid in a straight line from A to Binstead of being curved. In this case, would the shadow thatthe ball casts on screen X speed up, slow down, or moveat constant speed? Explain your answer.

e. In d, would the shadow that the ball casts on screen Yspeed up, slow down, or move at constant speed? Explainyour answer.

f. The answers to a and b are ____.

A. the same as the answers to d and e.

B. different from the answers to d and e because the train’sspeed is changing in a and b.

C. different from the answers to d and e because the train’sdirection is changing in a and b.

0540T_c03_49-90.qxd 08/24/2004 17:35 Page 81 EQA

3-21. Use Procedure 3-1 to draw the vector sum or resultantof each pair of one-dimensional vectors and shown inFigure 3-52. For each resultant you obtain, type in the valueof its magnitude and a single word (left, right, up, or down)to indicate its direction.

BS

AS

3-26. Vectors and are to be added to obtain a resultantVector has magnitude and is directed toward the

right. Vector has magnitude you are free to chooseits direction. Below are descriptions of six different possibleresultants For each description, sketch what the vector must look like, and sketch how you would add it to to pro-duce that resultant.a. Its magnitude is

b. Its magnitude is

c. Its magnitude is

d. Its magnitude is but its direction is different thanin c.

e. Its x component is

f. Its y component is

3-27. SSM Vectors and each have a magnitude of 10.Sketch the vector sum choosing the directions of and so that the resultant a. has a magnitude of 20.

b. is zero.

c. has a magnitude of 10.

d. has a magnitude of 2.

Section 3-4 Working with Vector Components

3-28. A jogger goes 300 m east, then 225 m north.a. Find the magnitude and direction of the jogger’s displace-

ment by drawing and measuring (to anticipate what a cal-culation should give you).

b. Calculate the magnitude and direction, and see if it agreeswith your measurements.

3-29. Using the same scale as in Example 3-1, use a protrac-tor and a centimeter ruler to finda. the magnitude and direction of the position vector of Hid-

den Harbor based on the coordinate frame in Figure 3-22.

b. the x and y components of this position vector.

3-30.a– e. Find the x and y components of each vector in Fig-

ure 3-55. Where possible, find the components both byinspection and by calculation to see if you can get thesame results both ways. You may wish to answer partsf and g first.

f. In each of parts a–e, what value of (measured counter-clockwise from the axis) could you use to calculate thecomponents?

g. In which parts could you find the components by inspection?

�xu

RS

BS

AS

AS

� BS,

BS

AS

Ry � �4.

Rx � �1.

R � 5,

R � 5.

R � 1.

R � 7.

AS

BS

RS

.

B � 4;BS

A � 3AS

RS.

BS

AS



Use thiscoordinateframe

8.05.0

37º

5.0120º

8.053º

6.0

y

x

(e)(d )

(c)(b)(a)

3-22. Use Procedure 3-1 to draw the vector sum or resultantof each pair of vectors and shown in Figure 3-53. Mea-sure the magnitude and directional angle for each resultant,and give their values.

BS

AS

3-23. SSM A hiking trail goes to Vernal Falls. The sign postedat the beginning of the trail indicates that it follows a river for1450 m in a direction S of E (south of east), and then cutsthrough woodland for 900 m in a direction N of E to reachthe falls. Taking the beginning of the trail as your origin, usea ruler and protractor to find the position vector (magnitudeand direction) of the falls.

3-24.a. The flight plan of a plane leaving O’Hare Airport first takes

it in a direction N of E (north of east), andthen in a direction N of W. Taking O’Hareas your origin, use a ruler and protractor to find the plane’sposition vector (magnitude and direction) at the end of theflight.

b. A second plane out of O’Hare follows the two stages of theflight plan in reverse order. Sketch the vector addition andthe final position vector for this plane’s flight. How do thetwo planes’ final position vectors compare?

3-25. Figure 3-54 showsfour different pairs ofvectors and All thevectors are equal inmagnitude. So are allthe vectors Rankthese four pairs in order of the magnitude of the resultant vec-tor starting with the smallest. Indicate which values, ifany, are equal. Briefly explain your reasoning.

AS

� BS,

BS.

AS

BS.A

S

70°4.8 � 104 m20°3.6 � 104 m

65°15°


Figure 3-53 Problems 3-22, 3-33, 3-36, 3-37, and 3-78

5

3

3

2

6

8

4

4

6

4A

B

(Numerical values are for A and B )

(e)(d )(c)(b)(a)

3

3 84

4 545°

5

10

45°5 3

45° 53°

345°

3

233°30°

A

B

(Numerical values are for A and B )

(e)(d )(c)(b)(a) ( f )

A AAA

BB

BB

(d )(c)(b)(a)


0540T_c03_49-90.qxd 08/24/2004 17:35 Page 82 EQA

3-31. Find the magnitude and direction of the velocity ifa.

b.

c.

d.

3-32. Connection with one-dimensional motion: SupposeEquation 3-6 is applied to the velocity vector of an object ona straight track directed along the x axis.a. What is the value of when the object is traveling to the

right? What is the value of

b. What is the value of when the object is traveling to theleft? What is now the value of

c. Are there any other possible values of for one-dimensionalmotion?

d. What does Equation 3-6 tell you about the relationshipbetween the velocity and the speed for one-dimensionalmotion in the x direction? Does this agree with the way weaccounted for direction in Chapter 2?

3-33. Use Procedure 3-3 to calculate the magnitude and direc-tion of the vector sum for each pair of vectors shownin Figure 3-53.

3-34. A primate research station attaches a tiny radio trans-mitter to a chimpanzee born in captivity before releasing itinto the wild. One day, the station picks up a signal indicat-ing the chimp is 4000 m from the station in a direction Sof W. Over the next day, the chimp wanders 2500 m in adirection N of E. At this point, how far from the stationand in what direction is the chimp?

3-35. SSM WWW An aircraft surveillance post reports anunidentified aircraft 800 km from them in a direction S ofE and traveling due north. If it continues in this direction for600 km, at what distance from the surveillance post and inwhat direction will the aircraft be observed?

3-36. By drawing and measuring, find the vector differencefor each pair of vectors in Figure 3-53.

3-37. By calculation, find the magnitude and direction of thevector difference for each pair of vectors in Figure 3-53.

3-38. Vector has components and

Vector has components and

Vector has components and Finda. a component description of

b. a magnitude and direction description of

c. a magnitude and direction description of the vector

d. a magnitude and direction description of the vector

3-39. Vectors and each have a magnitude of 5.a. What is the range of possible values for the magnitude of

b. On what does the actual value depend?

c. What is the range of possible values for the magnitude of

d. If the directions of and differ by what is the mag-nitude of A

S� B

S?

60°,BS

AS

AS

� BS

?

AS

� BS

?

BS

AS

�12 B

S.

3 AS.

AS

� BS

� CS

.

AS

� BS

� CS.

Cy � �4.Cx � �5CS

By � �2.Bx � 4BS

Ay � 3.Ax � �2AS

AS

� BS

AS

� BS

45°

40°

15°

AS

� BS

u

sin u?u

sin u?u

vx � �6 m/s and vy � �8 m/s.

vx � 6 m/s and vy � �8 m/s.

vx � �6 m/s and vy � 0.

vx � 6 m/s and vy � 8 m/s.vS

Figure 3-56 Problems 3-41, 3-42, and 3-43

3-40. A vector makes an angle of with the positive x axis.When you multiply the vector by 0.5, does the angle increase,decrease, or remain the same?

Section 3-5 Velocity and Acceleration Vectors

3-41. SSM WWW An ornithologist studying raptors releasesthree red-tailed hawks after fitting them with small radio trans-mitters. Figure 3-56a, shows hawk A’s position at a certaininstant, and its position at an instant 80.0 s later.a. To help picture what you will be calculating, first sketch

the change in hawk A’s position.

b. Calculate the components of

c. Find the components of hawk A’s average velocity overthe 80.0-s interval.

d. Find the magnitude and direction of

e. How does the direction of compare to the direction of¢ rS?

vS

vS

.

vS

¢ rS.

¢ rS,

rS2

rS1

60°

3-42. Repeat Problem 3-41 for Figure 3-56b.

3-43. Repeat Problem 3-41 for Figure 3-56c.

3-44. An object’s velocity vec-tor is shown at an instant and a later instant in Figure3-57.a. Give a component descrip-

tion of each vector.

b. Give a magnitude anddirection description of eachvector.

c. Find the magnitude and direction of

d. Has the object’s velocity changed?

e. Has the object’s speed changed?

3-45. A cheetah hasvelocity vector (Figure 3-58) at

and atFind the

magnitude and direc-tion of the cheetah’schange in velocity

over this interval.

3-46. Using the information from Problem 3-45, find thecheetah’s average acceleration (both the component descrip-tion and the magnitude and direction description) between

and t � 8 s.t � 3 s

¢vS

t � 8 s.vS2t � 3 s

vS1

¢vS.

t2

t1


60°r 2 =

1800

m

r1 = 500 m

r2 = 40 m

r1 = 40 m

r1 = 500 m

r2 = 750 m

Hawk A Hawk B Hawk C

(c)(b)(a)

v1 = 20 m/s

At t1

v2 = 20 m/s

At t2


v2 = 25 m/s30°v 1 = 20 m

/s

At t = 3s

127°

At t = 8s

Figure 3-58 Problems 3-45 and 3-46

0540T_c03_49-90.qxd 08/24/2004 17:35 Page 83 EQA

3-47. A motorcycle goes clockwise around a circular track ata constant speed of 20 m/s.a. Is the motorcycle accelerating?

b. How do the magni-tude and directionof its velocity vectorsat points A and B(figure) compare with

and in Prob-lem 3-44?

c. As the motorcyclegoes from A to B,how do the magni-tude and direction of

for the motor-cycle during thisperiod compare withthe solution in Prob-lem 3-44?

¢vS

vS2vS1

d. If it takes the motorcycle 4 s to get from A to B, use thesolution to Problem 3-44 to calculate the motorcycle’s aver-age acceleration (magnitude and direction) between A andB. Compare this with your answer to a.

Section 3-6 Solving Motion Problems in Two Dimensions:Projectile Motion Revisited

3-48. In Figure 3-60, a kicked soccer ball is shown in severalstop-action pictures taken at half-second intervals during its tra-jectory. The x and y shadows at each of these instants are alsoshown.a. Is the horizontal shadow motion uniform? Is the vertical

shadow motion uniform?

b. Using the coordinate frame in the figure to determine posi-tions, find the average horizontal and vertical velocities foreach half-second time interval (0 to 0.5 s, 0.5 s to 1.0 s,etc.). Do your calculations support your answer to part a?

3-49. Figure 3-61 shows a cannonball at five points during itsflight. The coordinate frame shown applies to all parts of thequestion. Without doing any calculations, answer each of thefollowing.a. Indicate whether the vertical acceleration is positive, nega-

tive, or zero at each of these points.

b. Indicate whether the speed is positive, negative, or zero ateach of these points.

c. At which of these positions is the speed a minimum?

d. At which of these positions is the speed zero?

Figure 3-61 Problem 3-49 Figure 3-62 Problem 3-51

e. Which of these quantities change sign during the cannon-ball’s flight?

3-50. When you shoot a free throw in basketball, does theball have zero velocity at its highest point? Briefly explain.

3-51. Figure 3-62 shows the trajectory of a ball thrown up inthe air at an angle from near the bottom of a pit. Fill in eachspace in the table below with or � to indicate whetherthe components of position, velocity, and acceleration are pos-itive, negative, or zero for the indicated parts of the trajectory.

�, 0,

x, y, vx, vy, v, ax, ay, a.


y (in m)

x (in m)0 10 20 30

2

4

6

8

10

01

23

01

23

01

23

01

23

01

23

01

23

01

23Figure 3-60 Problem 3-48

y

x

A

B

C

D

E

y

xB

O

D

A

C

E

A

B

Figure 3-59 Problem 3-47b


x y

While ball is going from A to B

While ball is going from B to C

When ball is at C

While ball is going from C to D

While ball is going from D to E

3-52. The theater at Einstein Community College has a slop-ing stage (Figure 3-63). Because of the slope, a ball releasedfrom rest on the stage would roll toward the edge of the stagewith a downhill acceleration of magnitude A groupof physics students map out coordinate axes on the stage, andthen give a ball an initial velocity as shown in the figure. Findthe x and y coordinates of the ball at each of the followinginstants: t � 0, t � 1 s, t � 2 s, t � 3 s, t � 4 s.

2.4 m/s2.

ayvyaxvx

3-53. Use the values obtained in Problem 3-52 to plot graphsof x versus t and y versus t. How do these graphs comparewith the graphs of x versus t and y versus t obtained inExample 3-8?

3-54. From Problem 3-52, find the components and ofthe ball’s velocity at each of the following instants:

3-55. Use the values obtained in Problem 3-54 to plot graphsof versus t and versus t. How do these graphs comparewith the graphs of versus t and versus t obtained inExample 3-8?

3-56. An artillery shell is fired horizontally from the top of acliff 78.4 m high. It leaves the gun with a velocity of 1200 m/s.a. How long does it take to reach the ground?

b. How far from the base of the cliff does it land?

c. Find the components of its velocity at the instant it strikesthe ground.

d. Find its speed at that instant.

3-57. Redo Example 3-9 taking the downward direction aspositive.

3-58. A golf ball is given an initial velocity of 49.0 m/s at anangle of above the horizontal. Ignoring air resistance,a. how long does it take the ball to reach its highest point?

b. how far has the ball gone horizontally when it is at its high-est point?

c. how long does the ball remain in flight?

3-59. SSM WWW A soccer ball leaves a player’s foot at anangle of to the ground. With negligible air resistance, ittakes 0.833 s to reach the highest point in its trajectory.a. With what speed did it leave the player’s foot?

b. What is its speed at its highest point?

c. How far away does it land?

22°

37°

vyvx

vyvx

t � 1 s, t � 2 s, t � 3 s, t � 4 s.t � 0vyvx



0 53°

v0 = 6.0 m/s

y

x

G o i n g F u r t h e rThe questions and problems in this group are not organized by sec-tion heading, so you must determine for yourself which ideas apply.Some of them will be more challenging than the Review and Practicequestions and problems (especially those marked with a • or ••).

3-60. In Figure 3-64, points P and Qare shown on a graph paper back-ground where each box representsone unit of distance. Copy the dia-gram four times, so that you can doeach part below on a separate dia-gram. Now sketch a coordinate frame(one for each part) in which the posi-tion vector fora. point P has components and

b. point P has components and

c. point P has components and

d. point Q has components and y � 0.x � 5

y � 3.x � �1

y � �2.x � 0

y � 0.x � 1


3-61.a. Figure 3-65 shows some unusually trained laboratory ani-

mals. Sketch the coordinate frame (superimpose it on thefigure) that would be best suited for describing the motionof animal A after the rope burns through. Give reasons foryour choice.

b. Repeat for animal B.

c. Repeat for animal C.

P

Q

Figure 3-64 Problem3-60

A

B

C

0540T_c03_49-90.qxd 08/24/2004 17:35 Page 85 EQA


37°

v0 = 150 m/s

altitude= 4000 m

1440 m

the road, the man’s apparent motion is the superposition ofthese two motions; that is, his apparent velocity is the vectorsum of the two given velocities. If the directions are as shownin Figure 3-66, find the magnitude and direction of the man’svelocity as it appears to an observer in the helicopter.

3-63. SSM Although a diver’s body rotates in space in a com-plicated way during a dive, the diver’s center of mass (we’lldefine this carefully in Chapter 5) moves like a projectile (seeFigure 3-67). A diver leaves a 10-m-high diving board with aspeed of 3.3 m/s at an angle above the horizontal. Findthe approximate speed with which she hits the water below.(Hint: The solution will be approximate when you make thefollowing simplifyingassumptions: Ignorethe difference in heightbetween the divingboard (where her feetare before the dive)and her center ofmass; ignore the timedifference between theinstant her hands makecontact with the waterand the instant hercenter of mass is at thewater’s surface; ignoreair resistance.)

3-64. A vector has amagnitude of 20 and isdirected vertically down-ward. Find its compo-nents and in eachof the coordinate framesshown in Figure 3-68.

3-65. SSM A radio signalis relayed from station A to station B to station C to station D.The distances and directions for these three stages are shownin Figure 3-69. How far and in what direction would a signalhave to go if it were to be sent directly from station A tostation D?

FyFx

FS

68°


3-66. Show that for any vector is the same for bothand

3-67. For each of the following situations, determine whetherthe velocity of the object cannot be constant, might be con-stant, or must be constant.a. A billiard ball rebounds off the cushion of a billiards table.

b. A car travels down a straight road at an average speed of20 m/s.

c. A car maintains a steady speed of 10 m/s around a trafficcircle.

3-68. The velocity com-ponents of a ball areplotted against time inFigure 3-70. What wouldyou observe the balldoing during the timeperiod graphed? Sketchits trajectory.

•3-69. A plane is in the situation depicted in Figure 3-71 whenit runs out of fuel. If it clears the mountain, it can glide to asafe landing. The mountain’s elevation is 4450 m. By a suitablecalculation, determine whether the plane will make it over.Neglect air resistance.

�VS

.VS

VS

, tan u

•3-70. Starting with Equations 3-17x and 3-17y, show by aseries of mathematical steps that

•3-71. A cannonball isfired from a hilltopbunker as shown in Fig-ure 3-72.a. If it is to land at the

foot of the hill (P),what must be its ini-tial speed?

b. If a truck at Prequires 11 s to movefar enough out of theway, will it escapedamage?

v2� v2

o � 2gy.


x

y

x

y

30° 30°

(b)(a)


30°

45°rA to B

= 300 km

rB to C = 100 kmrC to D = 120 km

vx

t

vy

t


40°

37°

500 m

P

v0


40°15°

vman on truck= 2 m/s

vtruck on road= 5 m/s



3-62. While a truck creeps along in traffic at a speed of 5 m/s,a man walks diagonally across the back of the truck at 2 m/s.To a traffic helicopter hovering in a stationary position above

0540T_c03_49-90.qxd 08/24/2004 17:36 Page 86 EQA

Figure 3-73 (for Problem 3-76)


•3-72. A projectile at ground level is given an initial velocityof 100 m/s at an angle of with the horizontal. At what twohorizontal distances does the projectile have a height of81.4 m?

•3-73. A kicked football leaves the kicker’s foot at a speed of20 m/s. At what angle above the horizontal was the ball kickedif it reaches a maximum height of 14.4 m?

•3-74. A model rocket carries no fuel; it is set in motionentirely by its ground-level launcher. 8.0 s after beinglaunched, the rocket lands on the ground 410 m from thelauncher.a. Find a component description of the rocket’s initial velocity.

b. Find the speed and directional angle at which the rocketwas launched.

c. If the launch occurs at at what two instants t doesthe rocket have a speed of 55 m/s?

••3-75. In 1918, the Paris Gun was state-of-the-art Germanartillery, with a muzzle velocity of (1 mile/s).It was set up at Laon, France, with the intent of shelling Paris,a horizontal distance of (108 km) away (slightlyfurther by land because of Earth’s curvature).a. To strike a target at this distance, at what angle did the gun

have to be fired? (Neglect air resistance.)

b. How long would the shell then take to reach its target?

3-76. In each of the following cases, what is the direction ofthe object’s average acceleration vector during the intervaldescribed? (Draw sketches if necessary to clarify your answer).a. A golf ball in a mini-golf game moves to the right, hits an

obstacle head on, and bounces straight back at the samespeed.

b. The golf ball hits the obstacle at a angle to the per-pendicular and bounces off it at the same angle and speed(see Figure 3-73).

c. A race car is driven halfway around a circular track at con-stant speed (see Figure 3-73).

d. The race car is driven two full laps around the circular trackat constant speed.

30°

1.08 � 105 m

1.61 � 103 m/s

t � 0,

30°3-78. Find the x and y compo-nents of each of the vectors in Fig-ure 3-53 if the coordinate frame isoriented as in Figure 3-74.

3-79. Give an example of twovectors that cannot be addedtogether.

3-80. Suppose the track in Figure 3-50 (described more fullyin Problem 3-16) is extremely flexible, so that it can bearranged to form a path of any shape from A to B. (Do notworry about difficulties the train may have in making turnsthat are too sharp.) Sketch a path from A to B such that if thetrain follows that path at constant speed from A to B, the ball’sshadow ona. screen X reverses direction once, but its shadow on screen Y

does not change directions at all;

b. screen Y reverses direction twice, but its shadow on screen Xdoes not change directions at all;

c. screen Y reverses direction twice, but its shadow on screen Xreverses direction only once.

3-81. Suppose the track in Figure 3-50 (see Problem 3-16) isextremely flexible. (Do not worry about difficulties the trainmay have in making turns that are too sharp.) Tell whether itis possible to bend the track to do each of the following. Ifit is possible, sketch one way of doing it. If it is not possible,tell why not.a. Arrange the flexible track so that when the ball moves at

constant speed, its shadow on screen X speeds up while itsshadow on screen Y slows down.

b. Arrange the flexible track so that when the ball moves atconstant speed, its shadow on screen X speeds up while itsshadow on screen Y also speeds up.

c. Arrange the flexible track so that when the ball moves atconstant speed, its shadow on screen X speeds up while itsshadow on screen Y moves at constant speed.

d. Arrange the flexible track so that when the ball speeds up,its shadow on screen X speeds up while its shadow onscreen Y slows down.

3-82. An object follows the path shown in red in Figure 3-75.Its position along the path is shown at 1-s intervals.a. Is the x component of the object’s velocity, changing as

the object travels? Explain.

b. Is the y component of the object’s velocity, changing asthe object travels? Explain.

vy,

vx,

•3-83. Find the position of the object in Figure 3-75 att � 5 s.

3-77. In Figure 3-37 in Section 3-6, the choice of origin isshown in the graph of y versus x. Suppose we move the originto the right so that is now directly under the highestpoint of the trajectory. Which of the other graphs in Figure 3-37would change?

x � 0


30°30°

v

v

Start

for 3-76b

Finish

for 3-76c

45°

y x


t = 0

t = 1 s

t = 2 s

0.10 m

t = 3 s

t = 4 s

y

x


a. Complete the following table.

Time Interval

from 0 to

from to

from to

from to

from to

from to

from to

from to

from to

b. Use the values from the table in a to find the average veloc-ity components and required to complete the follow-ing table.

Time Interval

from 0 to

from to

from to

from to

from to

from to

from to

from to

from to

c. Do either of the component motions have zero acceleration?Do either of the component motions have nonzero constantacceleration? Explain.

d. If either component motion has a nonzero constant accel-eration, calculate its value.

3-85.a. Can the magnitude of a vector remain constant while one

of its components is increasing? Explain.

t � 9 st � 8 s

t � 8 st � 7 s

t � 7 st � 6 s

t � 6 st � 5 s

t � 5 st � 4 s

t � 4 st � 3 s

t � 3 st � 2 s

t � 2 st � 1 s

t � 1 s

vyvx

vyvx

t � 9 st � 8 s

t � 8 st � 7 s

t � 7 st � 6 s

t � 6 st � 5 s

t � 5 st � 4 s

t � 4 st � 3 s

t � 3 st � 2 s

t � 2 st � 1 s

t � 1 s

¢y¢x

b. Can the magnitude of a vector remain constant while bothof its components are increasing? Explain.

3-86. A rock is thrown horizontally with an initial speed from a vertical cliff of height h. It lands on the flat plain belowat a distance R from the base of the cliff. This happens some-where on Earth. If the identical scenario is repeated on themoon,a. how is the horizontal component of the velocity with which

the rock hits the ground affected? Is it greater than, equalto, or less than it was on Earth? Explain. (Ignore differencesin air resistance.)

•b. how is the horizontal distance R traveled by the rockaffected? Is it greater than, equal to, or less than it was onEarth? Explain. (Again, ignore differences in air resistance.)

3-87. SSM A small craft pilot in A-ville wishes to get to C-town,which is located 480 km from A-ville in a direction N of E.But first she must make a stop in B-burg, 600 miles due eastof A-ville. How far and in what direction must she fly to getfrom B-burg to C-town?

3-88. Would you prefer to have cruise control keep your carat constant speed or constant velocity? Briefly explain why.

3-89. Vectors and make angles and with theis the vector sum of and The angle

that makes with the is (always, sometimes, or never)equal to Briefly explain your answer.

3-90.a. Sketch two vectors and that satisfy the following con-

dition: and are equal in magnitude, and their differ-ence is zero.

b. Now sketch two vectors and that satisfy this condi-tion: and are equal in magnitude, but their difference

is not zero. How do the vectors that you’vedrawn this time differ from the vectors you drew in part a?

c. Suppose an object’s speed is the same at instants and Is the change in velocity necessarily zero over the timeinterval Explain.

d. Is the average acceleration of the object in part c neces-sarily zero over the time interval Explain.

3-91.a. An object moves from position at instant to position

at instant so that its position changes byover the time interval If the

object has an average velocity over this same interval,how do the directions of the vectors and compare?Can you say anything about how their magnitudes compare?Explain.

b. An object’s velocity changes from at instant to atinstant so that over the time interval

If the object has an average acceleration over this same interval, how do the directions of the vec-tors and compare? Can you say anything about howtheir magnitudes compare? Explain.

3-92. A team of engineering students has designed a radio-controlled model car. On its first test run, a team memberstands with the control box while the car travels 7.8 m in adirection S of E, and then 5.2 m due north, and then60°

aSav¢vS

aSav¢t � t2 � t1.¢vS � vS2 � vS1t2,

vS2t1vS1

vS

¢ rSvS

¢t � t2 � t1.¢ rS � rS2 � rS1

t2,rS2

t1rS1

¢t � t2 � t1?

¢t � t2 � t1?¢vS

t2.t1

¢AS

� AS

2 � AS

1

AS

2AS

1

AS

2AS

1

¢AS

� AS

2 � AS

1

AS

2AS

1

AS

2AS

1

uA � uB.�x axisR

SBS

.AS

RS

� AS

� BS

�x axis.uBuAB

SAS

37°

vo



t = 0

y

x

t = 1 s

t = 2 s

t = 3 s

t = 4 s

t = 6 s

t = 5 s

t = 7 s

t = 8 s

0.20 m

t = 9 s

3-84. An object follows the path shown in red in Figure 3-76.Its position along the path is shown at 1-s intervals.

0540T_c03_49-90.qxd 08/24/2004 17:36 Page 88 EQA

15.0 m in a direction N of W. If the car sends a signalback to the control box at that point, how far and in whatdirection must the signal be sent?

3-93. SSM In Example 3-7, we calculated the magnitude andthe direction of the car’s average acceleration as it traveledaround a quarter of a circle at constant speed. If the car con-tinued for another quarter-circle at the same speed,a. would the magnitude of the average acceleration be the

same as for the first quarter-circle? Explain.

b. would the direction of the average acceleration be the sameas for the first quarter-circle? Explain.

c. are Equations 3-14x and 3-14y applicable to a car travelingin a circle at constant speed? Explain.

d. are Equations 3-16x through 3-18y applicable to a car trav-eling in a circle at constant speed? Explain.

3-94. A table tennis ball thrown from a window is stronglyaffected by air resistance, so it doesn’t keep speeding up indef-initely as it falls. Instead, its speed levels off to some finalvalue. Can Equations 3-16x through 3-18y be applied to themotion of this ball? Briefly explain.

3-95. Two students are arguing about playing soccer on themoon, where g has about of its value on Earth. Student Aclaims that with the same kick, the ball would rise higher andland farther down-field on the moon than on Earth. Student Bagrees that the ball will rise higher, but argues that becausethe gravitational acceleration only affects the vertical motion,the ball will not land farther downfield than on Earth. Withwhich student would you side? Briefly explain.

3-96. A runner doing acomplete lap aroundthe track in Figure 3-77at constant speed(always, sometimes, ornever) has a nonzeroacceleration. Choose thecorrect answer, andbriefly explain.

3-97. A car circles thetrack in Figure 3-78 atconstant speed. Con-sider its horizontalvelocity component Rank the labeled pointson the track in orderaccording to the valueof at each point,starting with the lowest(or most negative)value. Indicate where (ifat all) the values areequal.

3-98. Suppose you have a camcorder and viewer that let youproduce and view videos frame by frame. You decide to makea video of a child’s ball bouncing along a sidewalk in frontof a picket fence. How can you determine whether the ball’shorizontal velocity component is constant or not? What com-plications can you think of in this approach, and how wouldyou deal with them?

vx

vx.

16

15° 3-99. Describe a coordinate system (location of origin, direc-tion of axes) that you could use in Figure 3-16 if you wantedthe targets moving up the ramp to have positive velocities withno y components.

••3-100. A ball is thrown from the point withan initial velocity of 50 m/s at an angle of above the hor-izontal. Using the equations of motion for a projectile, obtainan equation that expresses the vertical position y in terms ofthe horizontal position x. It is possible to write this equationin the form a. Find the numerical values of a, b, and c.

b. Which of these constants is unitless?

3-101. Suppose that in Activity 3-2 you start with the pentoward the top of the ruler. You first move the ruler 10.0 cmto the right, and then move the pen 5.0 cm along theruler toward the bottom of the paper.a. What are the x- and y-components of the resultant vector?

b. Find the magnitude and direction of the resultant vector.

3-102. Both situations in Figure 3-79 take place on horizontalsurfaces. In situation 1, the ball is rolling around the circulartrack at a constant speed of 10 m/s. In situation 2, the ballstrikes a wall traveling at a constant speed of 10 m/s andbounces off in the direction shown at the same speed. Foreach situation, consider the ball’s average acceleration over theinterval that the ball takes to get from A to B.a. In which situation, if either, is the magnitude of this aver-

age acceleration greater?

b. In which of these situations (1, 2, both, or neither) is thedirection of this average acceleration towards the center ofthe circle?

y � ax2� bx � c.

30°x � 0, y � 0

•3-103. Consider the two situations described in Problem3-102.a. In situation 1, find the magnitude of the ball’s average

acceleration over the interval that the ball takes to get fromA to B.

b. Repeat for situation 2.

3-104. Suppose an object starts out traveling to the right(the direction) at a speed of 10 m/s. Suppose its velocitythen changes by and has a magnitude of 10 m/s. Inwhat two directions could be for the object to end upwith a speed of 3 m/s? Express each answer by giving theangle that makes with the axis.�x¢vS

¢vS¢vS¢vS,

�x




CB

A

D

x

y


r = 4.0 m

v = 10 m/s

v = 10 m/s

r = 4.0 m

Situation 1 Situation 2

v = 10 m/s

v = 10 m/s

B B

A A

0540T_c03_49-90.qxd 08/24/2004 17:36 Page 89 EQA


P ro b l e m s o n We b L i n k s3-105. Suppose that in Figure 3-80, points 1 and 2 are theendpoints of the path originally traced by the camper’s lanternin WebLink 3-1. Which of the paths in Figure 3-80 would havebeen traced by the lantern if the camper had not changed hismotion along the raft but the river’s flow had gained speed?

3-106. In WebLink 3-2, a three-dimensional motion is pro-duced by combining ____

a. a uniformly accelerated straight-line motion and a constant-speed straight-line motion.

b. a constant-speed straight-line motion and a circular motion.

c. a uniformly accelerated straight-line motion and a circularmotion.

d. two circular motions.

3-107. In WebLink 3-3, a motion that one might experienceon an amusement-park ride is produced by combining ____.a. a uniformly accelerated straight-line motion and a constant-

speed straight-line motion.

b. a constant-speed straight-line motion and a circular motion.

c. a uniformly accelerated straight-line motion and a circularmotion.

d. two circular motions.

3-108. In WebLink 3-4, the weight that moves up and downthe slide wire has a light on it that lets it be seen at night. Themotion observed at night that is actually shown in the WebLinkis a composite motion. Its two component motions are ____.a. uniformly accelerated upward motion and uniformly accel-

erated downward motion.

b. uniformly accelerated vertical motion and uniformly accel-erated horizontal motion.

c. uniformly accelerated vertical motion and constant speedhorizontal motion.


1

2

1

2

1

2

1

2

d. constant speed upward motion and uniformly accelerateddownward motion.

3-109. WebLink 3-5 shows that the sum of two vectors(always, sometimes, or never) depends on the order in whichyou add them.

3-110. In WebLink 3-6, suppose that instead of having equalmagnitudes, vector has a magnitude of 5 and vector hasa magnitude of 3. Then as vector rotates, the resultant can have a magnitude anywhere in the range from ____ to____.

3-111. In Problem 3-110, the resultant can have an x com-ponent anywhere in the range from ____ to ____ as vector rotates.

3-112. In Problem 3-110, the resultant can have a y com-ponent anywhere in the range from ____ to ____ as vector rotates.

3-113. According to WebLink 3-7, which (one or more) of thefollowing must you do when you add two vectors?a. Add their magnitudes to get the magnitude of the resultant.

b. Add their x components to get the x component of theresultant.

c. Add their y components to get the y component of theresultant.

d. Add their directional angles to get the directional angle ofthe resultant.

3-114. Figure 3-81 shows a portion of the circular motiontreated in WebLink 3-9. During the interval in which the objecttravels from A to B at constantspeed, ____.a. is positive and is

negative.

b. is negative and ispositive.

c. and are both positive.

d. and are both negative.

e. and are both zero.¢vy¢vx

¢vy¢vx

¢vy¢vx

¢vy¢vx

¢vy¢vx

BS

RS

BS

RS

RS

BS

BS

AS

A

B



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