sach vat ly 03

We define the average velocity vSavg of a particle during the time interval Dt as the displacement of the particle divided by the time interval: v S avg; DSr Multiplying or dividing a v

3.4 components of a Vector and Unit Vectors 65

The graphical method of adding vectors is not recommended whenever high

accuracy is required or in three-dimensional problems In this section, we

describe a method of adding vectors that makes use of the projections of vectors

along coordinate axes These projections are called the components of the

vec-tor or its rectangular components Any vecvec-tor can be completely described by its

components

Consider a vector AS lying in the xy plane and making an arbitrary angle u

with the positive x axis as shown in Figure 3.12a This vector can be expressed as the

sum of two other component vectors ASx , which is parallel to the x axis, and ASy, which

is parallel to the y axis From Figure 3.12b, we see that the three vectors form a

right triangle and that AS 5 SAx1 ASy We shall often refer to the “components

of a vector AS,” written A x and A y (without the boldface notation) The

compo-nent A x represents the projection of AS along the x axis, and the component A y

represents the projection of AS along the y axis These components can be positive

or negative The component A x is positive if the component vector ASx points in

the positive x direction and is negative if ASx points in the negative x direction A

similar statement is made for the component A y

Use the law of sines (Appendix B.4) to find the direction

of RS measured from the northerly direction:

b 5 38.9°

The resultant displacement of the car is 48.2 km in a direction 38.9° west of north

Finalize Does the angle b that we calculated agree with an

estimate made by looking at Figure 3.11a or with an actual

angle measured from the diagram using the graphical

method? Is it reasonable that the magnitude of RS is larger

than that of both AS and BS? Are the units of RS correct?

Although the head to tail method of adding vectors

works well, it suffers from two disadvantages First, some

people find using the laws of cosines and sines to be ward Second, a triangle only results if you are adding two vectors If you are adding three or more vectors, the resulting geometric shape is usually not a triangle In Sec-tion 3.4, we explore a new method of adding vectors that will address both of these disadvantages

awk-Suppose the trip were taken with the two vectors in reverse order: 35.0 km at 60.0° west of north first and then 20.0 km due north How would the magnitude and the direction of the resultant vector change?

Answer They would not change The commutative law for vector addition tells us that the order of vectors in an addition is irrelevant Graphically, Figure 3.11b shows that the vectors added in the reverse order give us the same resultant vector

Wh At IF ?

Figure 3.12 (a) A vector AS

lying in the xy plane can be

rep-resented by its component vectors

A

S

x and ASy (b) The y component

vector ASy can be moved to the

right so that it adds to ASx The vector sum of the component

vectors is AS These three vectors form a right triangle.

y

x O

The magnitudes of these components are the lengths of the two sides of a right

tri-angle with a hypotenuse of length A Therefore, the magnitude and direction of AS

are related to its components through the expressions

u 5tan21aA A y

Notice that the signs of the components A x and A y depend on the angle u For

example, if u 5 120°, A x is negative and A y is positive If u 5 225°, both A x and A y are

negative Figure 3.13 summarizes the signs of the components when AS lies in the various quadrants

When solving problems, you can specify a vector AS either with its components

A x and A y or with its magnitude and direction A and u.

Suppose you are working a physics problem that requires resolving a vector into its components In many applications, it is convenient to express the components

in a coordinate system having axes that are not horizontal and vertical but that are still perpendicular to each other For example, we will consider the motion of objects sliding down inclined planes For these examples, it is often convenient to

orient the x axis parallel to the plane and the y axis perpendicular to the plane.

Q uick Quiz 3.4 Choose the correct response to make the sentence true: A ponent of a vector is (a) always, (b) never, or (c) sometimes larger than the mag-

com-nitude of the vector

Unit Vectors

Vector quantities often are expressed in terms of unit vectors A unit vector is a

dimensionless vector having a magnitude of exactly 1 Unit vectors are used to ify a given direction and have no other physical significance They are used solely

spec-as a bookkeeping convenience in describing a direction in space We shall use the

symbols i^, j^, and k^ to represent unit vectors pointing in the positive x, y, and z

directions, respectively (The “hats,” or circumflexes, on the symbols are a standard

notation for unit vectors.) The unit vectors i^, j^, and k^ form a set of mutually

perpen-dicular vectors in a right-handed coordinate system as shown in Figure 3.14a The magnitude of each unit vector equals 1; that is, 0 i^ 0 5 0 j^ 0 5 0 k^ 0 5 1.

Consider a vector AS lying in the xy plane as shown in Figure 3.14b The product

of the component A x and the unit vector i^ is the component vector ASx5A xi^,

Figure 3.13 The signs of the

components of a vector AS depend

on the quadrant in which the

vec-tor is located.

Figure 3.14 (a) The unit vectors

i^, j^, and k^ are directed along the x,

y, and z axes, respectively (b)

Vec-tor AS 5A x i^ 1 A y j^ lying in the xy

plane has components A x and A y.

x and y Components Equations 3.8

and 3.9 associate the cosine of

the angle with the x component

and the sine of the angle with the

y component This association is

true only because we measured the

angle u with respect to the x axis,

so do not memorize these

equa-tions If u is measured with respect

to the y axis (as in some problems),

these equations will be incorrect

Think about which side of the

tri-angle containing the components

is adjacent to the angle and which

side is opposite and then assign the

cosine and sine accordingly.

3.4 components of a Vector and Unit Vectors 67

which lies on the x axis and has magnitude 0 A x0 Likewise, ASy5A ySj is the

com-ponent vector of magnitude 0 A y 0 lying on the y axis Therefore, the unit-vector

notation for the vector AS is

A

S

For example, consider a point lying in the xy plane and having Cartesian

coordi-nates (x, y) as in Figure 3.15 The point can be specified by the position vector rS,

which in unit-vector form is given by

r

This notation tells us that the components of rS are the coordinates x and y.

Now let us see how to use components to add vectors when the graphical method

is not sufficiently accurate Suppose we wish to add vector BS to vector AS in

Equa-tion 3.12, where vector BS has components B x and B y Because of the bookkeeping

convenience of the unit vectors, all we do is add the x and y components separately

The resultant vector RS 5 SA 1 SB is

R

S

5 1A x i^ 1 A y j^2 1 1B x i^ 1 B y j^2or

R

S

5 1A x1B x2i^ 1 1A y1B y2j^ (3.14) Because RS 5R x i^ 1 R y j^, we see that the components of the resultant vector are

R x5A x1B x

R y5A y1B y

(3.15)

Therefore, we see that in the component method of adding vectors, we add all the

x components together to find the x component of the resultant vector and use the

same process for the y components We can check this addition by components with

a geometric construction as shown in Figure 3.16

The magnitude of RS and the angle it makes with the x axis are obtained from its

components using the relationships

At times, we need to consider situations involving motion in three component

directions The extension of our methods to three-dimensional vectors is

straight-forward If AS and BS both have x, y, and z components, they can be expressed in

the form

A

S

5A x i^ 1 A y j^ 1 A z k^ (3.18) B

S

5B x i^ 1 B y j^ 1 B z k^ (3.19) The sum of AS and BS is

R

S

5 1A x1B x2i^ 1 1A y1B y2j^ 1 1A z1B z2k^ (3.20)

Notice that Equation 3.20 differs from Equation 3.14: in Equation 3.20, the

resul-tant vector also has a z component R z 5 A z 1 B z If a vector RS has x, y, and z

com-ponents, the magnitude of the vector is R 5 !R x21R y21R z2 The angle ux

that RS makes with the x axis is found from the expression cos u x 5 R x /R, with

simi-lar expressions for the angles with respect to the y and z axes.

The extension of our method to adding more than two vectors is also

straight-forward For example, AS 1 BS 1 SC 5 1A x1B x1C x2i^ 1 1A y1B y1C y2j^ 1

1A z1B z1C z2k^ We have described adding displacement vectors in this section

because these types of vectors are easy to visualize We can also add other types of

y

x O

(x, y)

y x

ˆj ˆi

r

S

Figure 3.15 The point whose

Cartesian coordinates are (x, y)

can be represented by the position

resultant RS and the components

of the individual vectors.

Pitfall Prevention 3.3 tangents on Calculators Equa-

tion 3.17 involves the calculation

of an angle by means of a tangent function Generally, the inverse tangent function on calculators provides an angle between 290° and 190° As a consequence, if the vector you are studying lies in the second or third quadrant, the angle measured from the positive

x axis will be the angle your

calcu-lator returns plus 180°.

68 chapter 3 Vectors

Example 3.3 The Sum of Two Vectors

Find the sum of two displacement vectors AS and BS lying in the xy plane and given by

Categorize We categorize this example as a simple substitution problem Comparing this expression for AS with

the general expression AS 5A x i^ 1 A y j^ 1 A z k^, we see that A x 5 2.0 m, A y 5 2.0 m, and A z 5 0 Likewise, B x 5 2.0 m,

B y 5 24.0 m, and B z 5 0 We can use a two-dimensional approach because there are no z components.

S o l u t I o n

Use Equation 3.14 to obtain the resultant vector RS: RS 5 SA 1 SB 5 12.0 1 2.02i^ m 1 12.0 2 4.02j^ m

Evaluate the components of RS: R x54.0 m R y5 22.0 m

Use Equation 3.16 to find the magnitude of RS: R 5 "R x21R y25"14.0 m221122.0 m225"20 m 5 4.5 m

Find the direction of RS from Equation 3.17: tan u 5R y

R x

522.0 m4.0 m 5 20.50Your calculator likely gives the answer 227° for u 5 tan21(20.50) This answer is correct if we interpret it to mean 27°

clockwise from the x axis Our standard form has been to quote the angles measured counterclockwise from the 1x

axis, and that angle for this vector is u 5 333°

Conceptualize Although x is sufficient to locate a point

in one dimension, we need a vector rS to locate a point in

two or three dimensions The notation D rS is a

generaliza-tion of the one-dimensional displacement Dx in Equageneraliza-tion

2.1 Three-dimensional displacements are more difficult

to conceptualize than those in two dimensions because

they cannot be drawn on paper like the latter

For this problem, let us imagine that you start with your

pencil at the origin of a piece of graph paper on which

you have drawn x and y axes Move your pencil 15 cm

to the right along the x axis, then 30 cm upward along

the y axis, and then 12 cm perpendicularly toward you away

vectors, such as velocity, force, and electric field vectors, which we will do in later chapters

Q uick Quiz 3.5 For which of the following vectors is the magnitude of the vector equal to one of the components of the vector? (a) AS 52i^ 15j^

(b) BS 5 23j^ (c) CS 5 15k^

Example 3.4 The Resultant Displacement

A particle undergoes three consecutive displacements: D rS15 115i^ 130j^ 112k^ 2 cm, D rS

25 123i^ 214j^ 2 5.0k^2 cm,

and D rS351213i^ 115j^2 cm Find unit-vector notation for the resultant displacement and its magnitude

S o l u t I o n

from the graph paper This procedure provides the

dis-placement described by D rS1 From this point, move your

pencil 23 cm to the right parallel to the x axis, then 14 cm parallel to the graph paper in the 2y direction, and then

5.0 cm perpendicularly away from you toward the graph paper You are now at the displacement from the origin

described by D rS11 DSr2 From this point, move your

pencil 13 cm to the left in the 2x direction, and (finally!)

15 cm parallel to the graph paper along the y axis Your

final position is at a displacement D rS11 DSr21 DSr3

from the origin

3.4 components of a Vector and Unit Vectors 69

Example 3.5 Taking a Hike

A hiker begins a trip by first walking 25.0 km southeast from her car She stops

and sets up her tent for the night On the second day, she walks 40.0 km in a

direction 60.0° north of east, at which point she discovers a forest ranger’s tower

(A) Determine the components of the hiker’s displacement for each day

Conceptualize We conceptualize the problem by drawing a sketch as in Figure

3.17 If we denote the displacement vectors on the first and second days by AS and

B

S

, respectively, and use the car as the origin of coordinates, we obtain the

vec-tors shown in Figure 3.17 The sketch allows us to estimate the resultant vector as

shown

Categorize Having drawn the resultant RS, we can now categorize this problem

as one we’ve solved before: an addition of two vectors You should now have a

hint of the power of categorization in that many new problems are very similar to

problems we have already solved if we are careful to conceptualize them Once

we have drawn the displacement vectors and categorized the problem, this problem is no longer about a hiker, a walk,

a car, a tent, or a tower It is a problem about vector addition, one that we have already solved

Analyze Displacement AS has a magnitude of 25.0 km and is directed 45.0° below the positive x axis.

S o l u t I o n

Categorize Despite the difficulty in conceptualizing in three dimensions, we can categorize this problem as a tion problem because of the careful bookkeeping methods that we have developed for vectors The mathematical manip-ulation keeps track of this motion along the three perpendicular axes in an organized, compact way, as we see below

substitu-To find the resultant displacement,

S5 DSr11 DSr21 DSr3

5 115 1 23 2 132i^ cm 1 130 2 14 1 152j^ cm 1112 2 5.0 1 02k^ cm

5 125i^ 131j^ 17.0k^2 cmFind the magnitude of the resultant

210

220 Tent

E N

S W

The negative value of A y indicates that the hiker walks in the negative y direction on the first day The signs of A x and

A y also are evident from Figure 3.17

Find the components of BS using Equations 3.8 and 3.9: B x5B cos 60.08 5 140.0 km2 10.5002 5 20.0 km

B y5B sin 60.08 5 140.0 km2 10.8662 5 34.6 km

(B) Determine the components of the hiker’s resultant displacement RS for the trip Find an expression for RS in

terms of unit vectors

70 chapter 3 Vectors

▸ 3.5 c o n t i n u e d

Summary

Definitions

Scalar quantities are those that have only a

numerical value and no associated direction

Vector quantities have both magnitude and direction and

obey the laws of vector addition The magnitude of a vector is

always a positive number.

Concepts and Principles

When two or more vectors are added together, they

must all have the same units and they all must be the

same type of quantity We can add two vectors AS and

B

S

graphically In this method (Fig 3.6), the resultant

vector RS 5 AS 1 BS runs from the tail of AS to the

tip of BS

If a vector AS has an x component A x and a y compo-

nent A y, the vector can be expressed in unit-vector form

as AS 5A x i^ 1 A y j^ In this notation, i^ is a unit vector

pointing in the positive x direction and j^ is a unit

vec-tor pointing in the positive y direction Because i^ and j^

are unit vectors, 0 i^ 0 5 0 j^ 0 5 1.

A second method of adding vectors involves com­

ponents of the vectors The x component A x of the

vector AS is equal to the projection of AS along the

x axis of a coordinate system, where A x 5 A cos u The y component A y of AS is the projection of AS along

the y axis, where A y 5 A sin u.

We can find the resultant of two or more vectors

by resolving all vectors into their x and y components, adding their resultant x and y components, and then

using the Pythagorean theorem to find the magnitude

of the resultant vector We can find the angle that the

resultant vector makes with respect to the x axis by

using a suitable trigonometric function

Write the total displacement in unit-vector form: RS 5 137.7i^ 117.0j^2 km

Finalize Looking at the graphical representation in Figure 3.17, we estimate the position of the tower to be about

(38 km, 17 km), which is consistent with the components of RS in our result for the final position of the hiker Also,

both components of RS are positive, putting the final position in the first quadrant of the coordinate system, which is also consistent with Figure 3.17

After reaching the tower, the hiker wishes to return to her car along a single straight line What are the components of the vector representing this hike? What should the direction of the hike be?

Answer The desired vector RScar is the negative of vector RS:

R

S car5 2RS 5 1237.7i^ 217.0j^2 km

The direction is found by calculating the angle that the vector makes with the x axis:

tan u 5R car,y

R car,x

5 217.0 km237.7 km50.450which gives an angle of u 5 204.2°, or 24.2° south of west

Wh At IF ?

conceptual Questions 71

must be in which quadrant, (a) the first, (b) the ond, (c) the third, or (d) the fourth, or (e) is more than one answer possible?

7 Yes or no: Is each of the following quantities a vector?

(a)  force (b) temperature (c) the volume of water in

a can (d) the ratings of a TV show (e) the height of a building (f) the velocity of a sports car (g) the age of the Universe

8 What is the y component of the vector 13i^ 28k^2 m/s? (a) 3 m/s (b) 28 m/s (c) 0 (d) 8 m/s (e) none of those answers

9 What is the x component of the vector shown in Figure

OQ3.9? (a) 3 cm (b) 6 cm (c) 24 cm (d) 26 cm (e) none

24 0

Figure oQ3.9 Objective Questions 9 and 10.

10 What is the y component of the vector shown in Figure

OQ3.9? (a) 3 cm (b) 6 cm (c) 24 cm (d) 26 cm (e) none

of those answers

11 Vector AS lies in the xy plane Both of its components

will be negative if it points from the origin into which quadrant? (a) the first quadrant (b) the second quad-rant (c) the third quadrant (d) the fourth quadrant (e) the second or fourth quadrants

12 A submarine dives from the water surface at an angle of

30° below the horizontal, following a straight path 50 m long How far is the submarine then below the water surface? (a) 50 m (b) (50 m)/sin 30° (c) (50 m) sin 30° (d) (50 m) cos 30° (e) none of those answers

13 A vector points from the origin into the second

quad-rant of the xy plane What can you conclude about

its components? (a) Both components are positive

(b) The x component is positive, and the y component

is negative (c) The x component is negative, and the y

component is positive (d) Both components are tive (e) More than one answer is possible

1 What is the magnitude of the vector 110 i^ 210k^2 m/s?

(a) 0 (b) 10 m/s (c) 210 m/s (d) 10 (e) 14.1 m/s

2 A vector lying in the xy plane has components of

oppo-site sign The vector must lie in which quadrant? (a) the

first quadrant (b) the second quadrant (c) the third

quadrant (d) the fourth quadrant (e) either the second

or the fourth quadrant

3 Figure OQ3.3 shows two vectors DS1 and DS2 Which of the

possibilities (a) through (d) is the vector DS222 DS1,

or (e) is it none of them?

DS1

DS2

Figure oQ3.3

4 The cutting tool on a lathe is given two displacements,

one of magnitude 4 cm and one of magnitude 3 cm, in

each one of five situations (a) through (e) diagrammed

in Figure OQ3.4 Rank these situations according to

the magnitude of the total displacement of the tool,

putting the situation with the greatest resultant

magni-tude first If the total displacement is the same size in

two situations, give those letters equal ranks

Figure oQ3.4

5 The magnitude of vector AS is 8 km, and the magnitude

of BS is 6 km Which of the following are possible

val-ues for the magnitude of AS 1 BS? Choose all possible

answers (a) 10 km (b) 8 km (c) 2 km (d) 0 (e) 22 km

6 Let vector AS point from the origin into the second

quadrant of the xy plane and vector BS point from the

origin into the fourth quadrant The vector BS 2 AS

Objective Questions 1 denotes answer available in Student Solutions Manual/Study Guide

1 Is it possible to add a vector quantity to a scalar

quan-tity? Explain

2 Can the magnitude of a vector have a negative value?

Explain

3 A book is moved once around the perimeter of a

table-top with the dimensions 1.0 m by 2.0 m The book ends

up at its initial position (a) What is its displacement?

(b) What is the distance traveled?

4 If the component of vector AS along the direction of vector

B

S

is zero, what can you conclude about the two vectors?

5 On a certain calculator, the inverse tangent function

returns a value between 290° and 190° In what cases will this value correctly state the direction of a vector

in the xy plane, by giving its angle measured clockwise from the positive x axis? In what cases will it

counter-be incorrect?

Conceptual Questions 1 denotes answer available in Student Solutions Manual/Study Guide

72 chapter 3 Vectors

the resultant vector AS1BS points in the negative y

direction with a magnitude of 14 units Find the

mag-nitude and direction of BS

9 Why is the following situation impossible? A skater glides

along a circular path She defines a certain point on the circle as her origin Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin

10 A force FS1 of magnitude 6.00 units acts on an object at the ori-gin in a direction u 5 30.0° above

the positive x axis (Fig P3.10) A

second force FS2 of magnitude 5.00 units acts on the object in the direction of the positive

y axis Find graphically the

mag-nitude and direction of the

resul-tant force FS11 SF2

11 The displacement vectors AS

and BS shown in Figure P3.11 both have magnitudes of 3.00  m The direction of vec-

tor AS is u 5 30.0° Find

gra-phically (a) AS1BS, (b) AS2BS,

(c) BS 2 SA , and (d) AS 22 BS (Report all angles counterclock-

wise from the positive x axis.)

12 Three displacements are AS 5

13 A roller-coaster car moves 200 ft horizontally and then

rises 135 ft at an angle of 30.0° above the horizontal It next travels 135 ft at an angle of 40.0° downward What

is its displacement from its starting point? Use cal techniques

14 A plane flies from base camp to Lake A, 280 km away

in the direction 20.0° north of east After dropping off supplies, it flies to Lake B, which is 190 km at 30.0° west

of north from Lake A Graphically determine the tance and direction from Lake B to the base camp

Section 3.1 Coordinate Systems

1 The polar coordinates of a point are r 5 5.50 m and

u 5 240° What are the Cartesian coordinates of this

point?

2 The rectangular coordinates of a point are given by

(2, y), and its polar coordinates are (r, 30°) Determine

(a) the value of y and (b) the value of r.

3 Two points in the xy plane have Cartesian coordinates

(2.00, 24.00) m and (23.00, 3.00) m Determine (a) the

distance between these points and (b) their polar

coordinates

4 Two points in a plane have polar coordinates (2.50 m,

30.0°) and (3.80 m, 120.0°) Determine (a) the

Carte-sian coordinates of these points and (b) the distance

between them

5 The polar coordinates of a certain point are (r 5 4.30 cm,

u 5 214°) (a) Find its Cartesian coordinates x and y

Find the polar coordinates of the points with Cartesian

coordinates (b) (2x, y), (c) (22x, 22y), and (d) (3x, 23y).

6 Let the polar coordinates of the point (x, y) be (r, u)

Determine the polar coordinates for the points

(a) (2x, y), (b) (22x, 22y), and (c) (3x, 23y).

Section 3.2 Vector and Scalar Quantities

Section 3.3 Some Properties of Vectors

7 A surveyor measures the distance across a straight river

by the following method (Fig P3.7) Starting directly

across from a tree on the opposite bank, she walks

d 5 100 m along the riverbank to establish a baseline

Then she sights across to the tree The angle from

her baseline to the tree is u 5 35.0° How wide is the

river?

u

d

Figure P3.7

8 Vector AS has a magnitude of 29 units and points in

the positive y direction When vector BS is added to AS,

The problems found in this

chapter may be assigned

online in Enhanced WebAssign

1. straightforward; 2.intermediate;

3.challenging

1. full solution available in the Student

Solutions Manual/Study Guide

AMT Analysis Model tutorial available in

problems 73

Section 3.4 Components of a Vector and unit Vectors

15 A vector has an x component of 225.0 units and a y

component of 40.0 units Find the magnitude and

direction of this vector

16 Vector AS has a magnitude of 35.0 units and points in

the direction 325° counterclockwise from the positive

x axis Calculate the x and y components of this vector.

17 A minivan travels straight north in the right lane of a

divided highway at 28.0 m/s A camper passes the

mini-van and then changes from the left lane into the right

lane As it does so, the camper’s path on the road is a

straight displacement at 8.50° east of north To avoid

cutting off the minivan, the north–south distance

between the camper’s back bumper and the minivan’s

front bumper should not decrease (a) Can the camper

be driven to satisfy this requirement? (b) Explain your

answer

18 A person walks 25.0° north of east for 3.10 km How

far would she have to walk due north and due east to

arrive at the same location?

19 Obtain expressions in component form for the

posi-tion vectors having the polar coordinates (a) 12.8 m,

150°; (b) 3.30 cm, 60.0°; and (c) 22.0 in., 215°

20 A girl delivering newspapers covers her route by

travel-ing 3.00 blocks west, 4.00 blocks north, and then 6.00

blocks east (a) What is her resultant displacement?

(b) What is the total distance she travels?

21 While exploring a cave, a spelunker starts at the

entrance and moves the following distances in a

hori-zontal plane She goes 75.0 m north, 250 m east, 125 m

at an angle u 5 30.0° north of east, and 150 m south

Find her resultant displacement from the cave

entrance Figure P3.21 suggests the situation but is not

drawn to scale

Cave

entrance

Final position

u

E N

S W

Figure P3.21

22 Use the component method to add the vectors AS

and BS shown in Figure P3.11 Both vectors have

mag-nitudes of 3.00 m and vector AS makes an angle of

u 5 30.0° with the x axis Express the resultant AS 1 BS

24 A map suggests that Atlanta is 730 miles in a direction

of 5.00° north of east from Dallas The same map shows

that Chicago is 560 miles in a direction of 21.0° west of

north from Atlanta Figure P3.24 shows the locations

of these three cities Modeling the Earth as flat, use

5.00

730 mi

560 mi

Figure P3.24

25 Your dog is running around the grass in your back

yard He undergoes successive displacements 3.50 m south, 8.20 m northeast, and 15.0 m west What is the resultant displacement?

26 Given the vectors AS52.00i^ 16.00j^ and BS 53.00i^ 22.00j^ , (a) draw the vector sum CS5AS1BS and the vector dif ference DS5AS2BS (b) Calculate

in terms of polar coordinates, with angles measured

with respect to the positive x axis.

27 A novice golfer on the green

takes three strokes to sink the ball The successive dis-placements of the ball are 4.00 m to the north, 2.00 m northeast, and 1.00 m at 30.0° west of south (Fig

P3.27) Starting at the same initial point, an expert golfer could make the hole in what single displacement?

28 A snow-covered ski slope makes an angle of 35.0° with

the horizontal When a ski jumper plummets onto the hill, a parcel of splashed snow is thrown up to a maxi-mum displacement of 1.50 m at 16.0° from the verti-cal in the uphill direction as shown in Figure P3.28 Find the components of its maximum displacement (a) parallel to the surface and (b) perpendicular to the surface

35.0

16.0

Figure P3.28

29 The helicopter view in Fig P3.29 (page 74) shows two

people pulling on a stubborn mule The person on

the right pulls with a force FS1  of magnitude 120  N

M

W

1.00 m

30.0 2.00 m

4.00 m E N

S W

Figure P3.27

W

74 chapter 3 Vectors

student has learned that a single equation cannot be solved to determine values for more than one unknown

in it How would you explain to him that both a and b

can be determined from the single equation used in part (a)?

38 Three displacement vectors of a cro -

quet ball are shown in Figure P3.38, where 0 AS0 5 20.0 units,

0 BS0 5 40.0 units, and 0 CS0 5 30.0 units Find (a) the resultant in unit-vector notation and (b) the magni-tude and direction of the resultant displacement

39 A man pushing a mop across a floor causes it to undergo two displace-ments The first has a magnitude of

150 cm and makes an angle of 120° with the positive x

axis The resultant displacement has a magnitude of

140 cm and is directed at an angle of 35.0° to the

posi-tive x axis Find the magnitude and direction of the

second displacement

40 Figure P3.40 illustrates typical proportions of male (m)

and female (f) anatomies The displacements dS1m and

41 Express in unit-vector notation the following vectors,

each of which has magnitude 17.0 cm (a) Vector ES

is directed 27.0° counterclockwise from the positive x

axis (b) Vector FS is directed 27.0° counterclockwise

from the positive y axis (c) Vector GS is directed 27.0°

clockwise from the negative y axis.

42 A radar station locates a sinking ship at range 17.3 km and bearing 136° clockwise from north From the same station, a rescue plane is at horizontal range 19.6 km, 153° clockwise from north, with elevation 2.20 km (a) Write the position vector for the ship relative to the plane, letting i^ represent east, j^ north, and k^ up

(b) How far apart are the plane and ship?

43 Review As it passes over Grand Bahama Island, the

eye of a hurricane is moving in a direction 60.08 north

of west with a speed of 41.0 km/h (a) What is the vector expression for the velocity of the hurricane?

and direction of u1 5 60.0°

The person on the left pulls

with a force FS2 of tude 80.0 N and direction of

magni-u2 5 75.0° Find (a) the gle force that is equivalent

sin-to the two forces shown and (b) the force that a third per-son would have to exert on the mule to make the resul-tant force equal to zero The forces are measured in units

of newtons (symbolized N)

30 In a game of American

foot-ball, a quarterback takes the ball from the line of scrim-mage, runs backward a distance of 10.0 yards, and then runs sideways parallel to the line of scrimmage for 15.0 yards At this point, he throws a forward pass downfield 50.0 yards perpendicular to the line of scrimmage What is the magnitude of the football’s resultant displacement?

31 Consider the three displacement vectors AS 5

13i^ 23j^ 2 m, BS 51i^ 2 4 j^ 2 m, and CS 5 122i^ 15j^2 m

Use the component method to determine (a) the

magnitude and direction of DS 5 SA 1 SB 1 CS and

(b) the magnitude and direction of ES 5 2 AS 2

B

S

1 CS

32 Vector AS has x and y components of 28.70 cm and

15.0  cm, respectively; vector BS has x and y

com-ponents of 13.2  cm and 26.60 cm, respectively

If AS 2 SB 13 CS 50, what are the components of CS?

33 The vector AS has x, y, and z components of 8.00,

12.0, and 24.00 units, respectively (a) Write a vector

expression for AS in unit-vector notation (b) Obtain a

unit-vector expression for a vector BS one-fourth the

length of AS pointing in the same direction as AS

(c) Obtain a unit-vector expression for a vector CSthree times the length of AS pointing in the direction

opposite the direction of AS

34 Vector BS has x, y, and z components of 4.00, 6.00, and

3.00 units, respectively Calculate (a) the magnitude of

(a) Determine an expression for AS in unit-vector

nota-tion (b) Determine the magnitude and direction of AS

(c) What vector BS when added to AS gives a resultant

vector with no x component and a negative y

compo-nent 4.00 units in length?

36 Given the displacement vectors AS 513i^ 24j^ 14k^2 m

and  BS 512i^ 13j^ 27k^2 m, find the magnitudes of the following vectors and express each in terms of its

rectangular components (a) CS 5 AS 1 BS (b) DS 5 

2 AS 2 BS

37 (a) Taking AS 516.00i^ 28.00j^ 2 units, BS 5 128.00i^ 1

3.00j^ 2 units, and CS 5126.0i^ 119.0j^2 units,

deter-mine a and b such that aSA 1bBS 1 CS 50 (b) A

problems 75

circle of radius 3.70 cm that lies in a north–

south vertical plane Find (a) the magnitude of the total displacement of the object and (b)  the angle the total displacement makes with the vertical

Additional Problems

48 A fly lands on one wall

of a room The left corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate system If the fly is located at the point hav-ing coordinates (2.00, 1.00) m, (a) how far is it from the origin? (b) What is its location in polar coordinates?

49 As she picks up her riders, a bus driver traverses four

successive displacements represented by the expression126.30 b2i^ 214.00 b cos 4082i^ 214.00 b sin 4082j^

road-ney, American Journal of Physics 67(3) 252–256, March

1999

50 A jet airliner, moving initially at 300 mi/h to the east,

suddenly enters a region where the wind is blowing

at 100 mi/h toward the direction 30.0° north of east What are the new speed and direction of the aircraft relative to the ground?

51 A person going for a walk follows the path shown in

Figure P3.51 The total trip consists of four line paths At the end of the walk, what is the person’s resultant displacement measured from the starting point?

straight-End

x y

200 m 60.0 30.0150 m

300 m

100 m Start

Figure P3.51

52 Find the horizontal and vertical components of the

100-m displacement of a superhero who flies from the

W

M

East North

Figure P3.47

It maintains this velocity for 3.00 h, at which time the

course of the hurricane suddenly shifts due north,

and its speed slows to a constant 25.0 km/h This new

velocity is maintained for 1.50 h (b) What is the

unit-vector expression for the new velocity of the hurricane?

(c) What is the unit-vector expression for the

dis-placement of the hurricane during the first 3.00 h?

(d) What is the unit-vector expression for the

dis-placement of the hurricane during the latter 1.50 h?

(e) How far from Grand Bahama is the eye 4.50 h after

it passes over the island?

44 Why is the following situation impossible? A shopper

push-ing a cart through a market follows directions to the

canned goods and moves through a displacement

8.00 i^ m down one aisle He then makes a 90.0° turn

and moves 3.00 m along the y axis He then makes

another 90.0° turn and moves 4.00 m along the x axis

Every shopper who follows these directions correctly

ends up 5.00 m from the starting point

45 Review You are standing on the ground at the origin

of a coordinate system An airplane flies over you with

constant velocity parallel to the x axis and at a fixed

height of 7.60 3 103 m At time t 5 0, the airplane is

directly above you so that the vector leading from you

Figure P3.45 Determine the magnitude and

orienta-tion of the airplane’s posiorienta-tion vector at t 5 45.0 s.

46 In Figure P3.46, the line

seg-ment represents a path from

the point with position vector

15i^ 13j^2 m to the point with

location (16i^ 112j^) m Point

A is along this path, a fraction f

of the way to the destination

(a) Find the position vector of

point A in terms of f

(b) Evalu-ate the expression from part

(a) for f 5 0 (c) Explain whether

the result in part (b) is

reason-able (d) Evaluate the

expres-sion for f 5 1 (e) Explain whether the result in part (d)

is reasonable

47 In an assembly operation illustrated in Figure P3.47, a

robot moves an object first straight upward and then

also to the east, around an arc forming one-quarter

of a circle of radius 4.80 cm that lies in an east–west

vertical plane The robot then moves the object

upward and to the north, through one-quarter of a

AMT

(5, 3) (16, 12)

Q/C

76 Chapter 3 Vectors

57 A vector is given by RS 52i^ 1 j^ 13k^ Find (a) the

magnitudes of the x, y, and z components; (b) the

mag-nitude of RS; and (c) the angles between RS and

the x, y, and z axes.

58 A ferry transports tourists between three islands It sails from the first island to the second island, 4.76 km away, in a direction 37.0° north of east It then sails from the second island to the third island in a direc-tion 69.0° west of north Finally it returns to the first island, sailing in a direction 28.0° east of south Cal-culate the distance between (a) the second and third islands and (b) the first and third islands

59 Two vectors AS and BS have precisely equal mag-

nitudes For the magnitude of AS 1 SB to be 100 times

larger than the magnitude of AS 2 SB, what must be the angle between them?

60 Two vectors AS and BS have precisely equal

magni-tudes For the magnitude of AS 1 BS to be larger than

the magnitude of AS 2 SB by the factor n, what must

be the angle between them?

61 Let AS 5 60.0 cm at 270° measured from the

hori-zontal Let BS 5 80.0 cm at some angle u (a) Find the

magnitude of AS 1 BS as a function of u (b) From the answer to part (a), for what value of u does 0 AS 1 BS0 take on its maximum value? What is this maximum value? (c) From the answer to part (a), for what value

of u does 0 AS 1 BS0 take on its minimum value? What

is this minimum value? (d)  Without reference to the answer to part (a), argue that the answers to each of parts (b) and (c) do or do not make sense

62 After a ball rolls off the edge of a horizontal table at time

t 5 0, its velocity as a function of time is given by

v

S51.2i^ 2 9.8t j^

where vS is in meters per second and t is in seconds

The ball’s displacement away from the edge of the table, during the time interval of 0.380 s for which the ball is in flight, is given by

You can think of the units and unit vectors as

con-stants, represented by A and B Perform the

integra-tion to calculate the displacement of the ball from the edge of the table at 0.380 s

63 Review The instantaneous position of an object is

specified by its position vector leading from a fixed origin to the location of the object, modeled as a par-ticle Suppose for a certain object the position vector is

a function of time given by rS54i^ 13j^ 22t k^, where

r

S is in meters and t is in seconds (a) Evaluate dSr/dt

(b) What physical quantity does dSr/dt represent about

the object?

S

Q/C

Q/C W

top of a tall building

fol-lowing the path shown in

Figure P3.52

53 Review The biggest

stuffed animal in the

world is a snake 420 m

long, constructed by

Nor-wegian children

Sup-pose the snake is laid

out in a park as shown

in Figure P3.53,

form-ing two straight sides of a

105° angle, with one side

240 m long Olaf and Inge

run a race they invent

Inge runs directly from

the tail of the snake to

its head, and Olaf starts

from the same place at

the same moment but

runs along the snake

(a) If both children run

steadily at 12.0  km/h, Inge reaches the head of the

snake how much earlier than Olaf? (b) If Inge runs the

race again at a constant speed of 12.0 km/h, at what

constant speed must Olaf run to reach the end of the

snake at the same time as Inge?

54 An air-traffic controller observes two aircraft on his

radar screen The first is at altitude 800 m, horizontal

distance 19.2 km, and 25.0° south of west The second

aircraft is at altitude 1 100 m, horizontal distance

17.6 km, and 20.0° south of west What is the distance

between the two aircraft? (Place the x axis west, the

y axis south, and the z axis vertical.)

55 In Figure P3.55, a spider is

resting after starting to spin

its web The gravitational

force on the spider makes it

exert a downward force of

0.150 N on the junction of

the three strands of silk The

junction is supported by

dif-ferent tension forces in the

two strands above it so that

the resultant force on the junction is zero The two

sloping strands are perpendicular, and we have chosen

the x and y directions to be along them The tension

T x is 0.127 N Find (a) the tension T y, (b) the angle the

x axis makes with the horizontal, and (c) the angle the

y axis makes with the horizontal.

56 The rectangle shown in Figure

P3.56 has sides parallel to the x

and y axes The position vectors

of two corners are AS 5 10.0 m

at 50.0° and BS 5 12.0 m at 30.0°

(a) Find the perimeter of the

rect-angle (b) Find the magnitude

and direction of the vector from

the origin to the upper-right

cor-ner of the rectangle

30.0

100 m

problems 77

Challenge Problem

67 A pirate has buried his treasure on an island with five trees located at the points (30.0 m, 220.0 m), (60.0 m, 80.0 m), (210.0 m, 210.0 m), (40.0 m, 230.0 m), and (270.0 m, 60.0 m), all measured relative to some ori-gin, as shown in Figure P3.67 His ship’s log instructs

you to start at tree A and move toward tree B, but to cover only one-half the distance between A and B Then move toward tree C, covering one-third the distance between your current location and C Next move toward tree D, covering one-fourth the distance between where you are and D Finally move toward tree E, covering one-fifth the distance between you and E, stop, and dig (a) Assume you have correctly

determined the order in which the pirate labeled the

trees as A, B, C, D, and E as shown in the figure What

are the coordinates of the point where his treasure is

buried? (b) What If? What if you do not really know

the way the pirate labeled the trees? What would pen to the answer if you rearranged the order of the

hap-trees, for instance, to B (30 m, 220 m), A (60 m, 80 m),

E (210 m, 210 m), C (40  m, 230 m), and D (270 m,

60 m)? State reasoning to show that the answer does not depend on the order in which the trees are labeled

Q/C

64 Ecotourists use their global positioning system

indica-tor to determine their location inside a botanical

gar-den as latitude 0.002 43 degree south of the equator,

longitude 75.642 38 degrees west They wish to visit

a tree at latitude 0.001 62 degree north, longitude

75.644 26 degrees west (a) Determine the

straight-line distance and the direction in which they can walk

to reach the tree as follows First model the Earth

as a sphere of radius 6.37 3 106 m to determine the

westward and northward displacement components

required, in meters Then model the Earth as a flat

surface to complete the calculation (b) Explain why

it is possible to use these two geometrical models

together to solve the problem

65 A rectangular parallelepiped has dimensions a, b, and

c as shown in Figure P3.65 (a) Obtain a vector

expres-sion for the face diagonal vector RS1 (b) What is the

magnitude of this vector? (c) Notice that RS1, ck^, and

R

S

2 make a right triangle Obtain a vector expression

for the body diagonal vector RS2

y c

b

z a

66 Vectors AS and BS have equal magnitudes of 5.00

The sum of AS and BS is the vector 6.00j^ Determine

the angle between AS and BS

B

C

D

Figure P3.67

78

Fireworks erupt from the Sydney

Harbour Bridge in New South Wales,

Australia Notice the parabolic

paths of embers projected into

the air All projectiles follow a

parabolic path in the absence

4.4 Analysis Model: Particle in

Uniform Circular Motion

4.5 Tangential and Radial

In this chapter, we explore the kinematics of a particle moving in two dimensions

Knowing the basics of two-dimensional motion will allow us—in future chapters—to ine a variety of situations, ranging from the motion of satellites in orbit to the motion of electrons in a uniform electric field We begin by studying in greater detail the vector nature

exam-of position, velocity, and acceleration We then treat projectile motion and uniform circular motion as special cases of motion in two dimensions We also discuss the concept of relative motion, which shows why observers in different frames of reference may measure different positions and velocities for a given particle

In Chapter 2, we found that the motion of a particle along a straight line such as

the x axis is completely known if its position is known as a function of time Let

us now extend this idea to two-dimensional motion of a particle in the xy plane

We begin by describing the position of the particle In one dimension, a single numerical value describes a particle’s position, but in two dimensions, we indicate

its position by its position vector rS, drawn from the origin of some coordinate

sys-tem to the location of the particle in the xy plane as in Figure 4.1 At time t i, the

particle is at point A, described by position vector rSi At some later time t f, it is at

point B, described by position vector rSf The path followed by the particle from

4.1 the position, Velocity, and acceleration Vectors 79

A to B is not necessarily a straight line As the particle moves from A to B in the

time interval Dt 5 t f 2 t i, its position vector changes from rSi to rSf As we learned

in Chapter 2, displacement is a vector, and the displacement of the particle is the

difference between its final position and its initial position We now define the

dis-placement vector D rS for a particle such as the one in Figure 4.1 as being the

differ-ence between its final position vector and its initial position vector:

DSr ; rS

The direction of D rS is indicated in Figure 4.1 As we see from the figure, the

mag-nitude of D rS is less than the distance traveled along the curved path followed by the

particle

As we saw in Chapter 2, it is often useful to quantify motion by looking at the

displacement divided by the time interval during which that displacement occurs,

which gives the rate of change of position Two-dimensional (or three-dimensional)

kinematics is similar to one-dimensional kinematics, but we must now use full vector

notation rather than positive and negative signs to indicate the direction of motion

We define the average velocity vSavg of a particle during the time interval Dt as

the displacement of the particle divided by the time interval:

v

S avg; DSr

Multiplying or dividing a vector quantity by a positive scalar quantity such as Dt

changes only the magnitude of the vector, not its direction Because displacement

is a vector quantity and the time interval is a positive scalar quantity, we conclude

that the average velocity is a vector quantity directed along D rS Compare

Equa-tion 4.2 with its one-dimensional counterpart, Equation 2.2

The average velocity between points is independent of the path taken That is

because average velocity is proportional to displacement, which depends only

on the initial and final position vectors and not on the path taken As with one-

dimensional motion, we conclude that if a particle starts its motion at some point and

returns to this point via any path, its average velocity is zero for this trip because its

displacement is zero Consider again our basketball players on the court in Figure 2.2

(page 23) We previously considered only their one-dimensional motion back and

forth between the baskets In reality, however, they move over a two-dimensional

sur-face, running back and forth between the baskets as well as left and right across the

width of the court Starting from one basket, a given player may follow a very

compli-cated two-dimensional path Upon returning to the original basket, however, a

play-er’s average velocity is zero because the playplay-er’s displacement for the whole trip is zero

Consider again the motion of a particle between two points in the xy plane as

shown in Figure 4.2 (page 80) The dashed curve shows the path of the particle As

the time interval over which we observe the motion becomes smaller and smaller—

that is, as B is moved to B9 and then to B0 and so on—the direction of the

displace-ment approaches that of the line tangent to the path at A The instantaneous velocity

That is, the instantaneous velocity equals the derivative of the position vector with

respect to time The direction of the instantaneous velocity vector at any point in

a particle’s path is along a line tangent to the path at that point and in the

direc-tion of modirec-tion Compare Equadirec-tion 4.3 with the corresponding one-dimensional

version, Equation 2.5

The magnitude of the instantaneous velocity vector v 5 0 vS0 of a particle is called

the speed of the particle, which is a scalar quantity.

x

y

t i i



t f

f O

Figure 4.1 A particle moving

in the xy plane is located with

the position vector rS drawn from the origin to the particle The displacement of the particle as it moves from A to B in the time

interval Dt 5 t f 2 t i is equal to the

vector D rS5Srf2Sri.

80 chapter 4 Motion in two Dimensions

As a particle moves from one point to another along some path, its

instanta-neous velocity vector changes from vSi at time t i to vSf at time t f Knowing the velocity

at these points allows us to determine the average acceleration of the particle The

average acceleration aSavg of a particle is defined as the change in its instantaneous

velocity vector D vS divided by the time interval Dt during which that change occurs:

a

S avg; DSv

Because aSavg is the ratio of a vector quantity D vS and a positive scalar quantity Dt,

we conclude that average acceleration is a vector quantity directed along D vS As

indicated in Figure 4.3, the direction of D vS is found by adding the vector 2 vSi (the

negative of vSi) to the vector vSf because, by definition, D vS5Svf2Svi Compare Equation 4.4 with Equation 2.9

When the average acceleration of a particle changes during different time

inter-vals, it is useful to define its instantaneous acceleration The instantaneous eration aS is defined as the limiting value of the ratio D vS/Dt as Dt approaches zero:

Various changes can occur when a particle accelerates First, the magnitude

of the velocity vector (the speed) may change with time as in straight-line (one-

As the end point approaches , t

approaches zero and the direction

of approaches that of the green line tangent to the curve at

Sr

As the end point of the path is moved from to to , the respective displacements and corresponding time intervals become smaller and smaller.

 

AA

Figure 4.3 A particle moves from position A to

position B Its velocity vector changes from vSi to vSf The vector diagrams at the upper right show two

ways of determining the vector DvS from the initial and final velocities.

Figure 4.2 As a particle moves

between two points, its average

velocity is in the direction of the

displacement vector D rS By

defini-tion, the instantaneous velocity at

A is directed along the line

tan-gent to the curve at A.

Pitfall Prevention 4.1

Vector Addition Although the

vec-tor addition discussed in Chapter

3 involves displacement vectors,

vec-tor addition can be applied to any

type of vector quantity Figure 4.3,

for example, shows the addition of

velocity vectors using the graphical

approach.

4.2 two-Dimensional Motion with constant acceleration 81

dimensional) motion Second, the direction of the velocity vector may change with

time even if its magnitude (speed) remains constant as in two-dimensional motion

along a curved path Finally, both the magnitude and the direction of the velocity

vector may change simultaneously

Q uick Quiz 4.1 Consider the following controls in an automobile in motion: gas

pedal, brake, steering wheel What are the controls in this list that cause an

acceleration of the car? (a) all three controls (b) the gas pedal and the brake

(c) only the brake (d) only the gas pedal (e) only the steering wheel

with Constant Acceleration

In Section 2.5, we investigated one-dimensional motion of a particle under

con-stant acceleration and developed the particle under concon-stant acceleration model

Let us now consider two-dimensional motion during which the acceleration of a

particle remains constant in both magnitude and direction As we shall see, this

approach is useful for analyzing some common types of motion

Before embarking on this investigation, we need to emphasize an important

point regarding two-dimensional motion Imagine an air hockey puck moving in

a straight line along a perfectly level, friction-free surface of an air hockey table

Figure 4.4a shows a motion diagram from an overhead point of view of this puck

Recall that in Section 2.4 we related the acceleration of an object to a force on the

object Because there are no forces on the puck in the horizontal plane, it moves

with constant velocity in the x direction Now suppose you blow a puff of air on

the puck as it passes your position, with the force from your puff of air exactly in

the y direction Because the force from this puff of air has no component in the x

direction, it causes no acceleration in the x direction It only causes a momentary

acceleration in the y direction, causing the puck to have a constant y component

of velocity once the force from the puff of air is removed After your puff of air on

the puck, its velocity component in the x direction is unchanged as shown in Figure

4.4b The generalization of this simple experiment is that motion in two

dimen-sions can be modeled as two independent motions in each of the two perpendicular

directions associated with the x and y axes That is, any influence in the y

direc-tion does not affect the modirec-tion in the x direcdirec-tion and vice versa.

The position vector for a particle moving in the xy plane can be written

r

where x, y, and rS change with time as the particle moves while the unit vectors i^

and j^ remain constant If the position vector is known, the velocity of the particle

can be obtained from Equations 4.3 and 4.6, which give

component of the velocity,

are the same length in

both parts of the figure,

which demonstrates that

motion in two dimensions

can be modeled as two

independent motions in

perpendicular directions.

x y

x y

a

b

Figure 4.4 (a) A puck moves across a horizontal air hockey

table at constant velocity in the x

direction (b) After a puff of air

in the y direction is applied to the puck, the puck has gained a y com- ponent of velocity, but the x com-

ponent is unaffected by the force

in the perpendicular direction.

82 chapter 4 Motion in two Dimensions

Because the acceleration aS of the particle is assumed constant in this discussion,

its components a x and a y also are constants Therefore, we can model the particle as

a particle under constant acceleration independently in each of the two directions

and apply the equations of kinematics separately to the x and y components of the velocity vector Substituting, from Equation 2.13, v xf 5 v xi 1 a x t and v yf 5 v yi 1 a y t

into Equation 4.7 to determine the final velocity at any time t, we obtain

v

S

f5 1v xi1a x t 2i^ 1 1v yi1a y t 2j^ 5 1v xi i^ 1 v yij^2 1 1a x i^ 1 a yj^2t v

S

This result states that the velocity of a particle at some time t equals the vector

sum of its initial velocity vSi at time t 5 0 and the additional velocity aSt acquired

at time t as a result of constant acceleration Equation 4.8 is the vector version of

Equation 2.13

Similarly, from Equation 2.16 we know that the x and y coordinates of a particle

under constant acceleration are

x f5x i1v xi t 11

2a x t2 y f5y i1v yi t 11

2a y t2Substituting these expressions into Equation 4.6 (and labeling the final position

which is the vector version of Equation 2.16 Equation 4.9 tells us that the position

vector rSf of a particle is the vector sum of the original position rSi, a displacement

v

S

i t arising from the initial velocity of the particle, and a displacement 1

2Sat2 ing from the constant acceleration of the particle

We can consider Equations 4.8 and 4.9 to be the mathematical representation

of a two-dimensional version of the particle under constant acceleration model Graphical representations of Equations 4.8 and 4.9 are shown in Figure 4.5 The components of the position and velocity vectors are also illustrated in the figure

Notice from Figure 4.5a that vSf is generally not along the direction of either vSi or

a

S because the relationship between these quantities is a vector expression For the

same reason, from Figure 4.5b we see that rSf is generally not along the direction of

r

S

i, vSi, or aS Finally, notice that vSf and rSf are generally not in the same direction

Velocity vector as W

a function of time for a

particle under constant

acceleration in two

dimensions

Position vector as 

a function of time for a

particle under constant

acceleration in two

dimensions

Figure 4.5 Vector

representa-tions and components of (a) the

velocity and (b) the position of a

particle under constant

accelera-tion in two dimensions.

v yi t

t2 1

a x t2 1

4.2 two-Dimensional Motion with constant acceleration 83 Example 4.1 Motion in a Plane

A particle moves in the xy plane, starting from the origin at t 5 0 with an initial velocity having an x component of

20 m/s and a y component of 215 m/s The particle experiences an acceleration in the x direction, given by a x 5 4.0 m/s2

(A) Determine the total velocity vector at any time

Conceptualize The components of the initial velocity tell

us that the particle starts by moving toward the right and

downward The x component of velocity starts at 20 m/s and

increases by 4.0 m/s every second The y component of

veloc-ity never changes from its initial value of 215 m/s We sketch

a motion diagram of the situation in Figure 4.6 Because the

particle is accelerating in the 1x direction, its velocity

compo-nent in this direction increases and the path curves as shown

in the diagram Notice that the spacing between successive

images increases as time goes on because the speed is

increas-ing The placement of the acceleration and velocity vectors in

Figure 4.6 helps us further conceptualize the situation

Categorize Because the initial velocity has components in both the x and y directions, we categorize this problem

as one involving a particle moving in two dimensions Because the particle only has an x component of tion, we model it as a particle under constant acceleration in the x direction and a particle under constant velocity in the

Figure 4.6 (Example 4.1) Motion diagram for the particle.

Use Equation 4.8 for the velocity vector: Svf5Svi1Sat 5 1v xi1a x t2i^ 1 1v yi1a y t2j^

Substitute numerical values with the velocity in meters

S

f5 320 1 14.02t4 i^ 1 3215 1 102t4 j^

(1) vSf5 3 120 1 4.0t2i^ 2 15j^4

Finalize Notice that the x component of velocity increases in time while the y component remains constant; this result

is consistent with our prediction

(B) Calculate the velocity and speed of the particle at t 5 5.0 s and the angle the velocity vector makes with the x axis.

Analyze

S o l u t I o n

Evaluate the result from Equation (1) at t 5 5.0 s: Svf5 3 120 1 4.015.02 2i^ 2 15j^4 5 140i^ 2 15j^2 m/s

Determine the angle u that Svf makes with the x axis

at t 5 5.0 s:

u 5 tan21av v yf

xfb 5 tan21a215 m/s40 m/s b5 2218Evaluate the speed of the particle as the magnitude

of Svf:

v f5 0 vS

f 0 5 "v x f21v yf25"140221 121522 m/s 5 43 m/s

Finalize The negative sign for the angle u indicates that the velocity vector is directed at an angle of 21° below the

posi-tive x axis Notice that if we calculate v i from the x and y components of Svi , we find that v f v i Is that consistent with our prediction?

(C) Determine the x and y coordinates of the particle at any time t and its position vector at this time.

continued

84 chapter 4 Motion in two Dimensions

4.3 Projectile Motion

Anyone who has observed a baseball in motion has observed projectile motion

The ball moves in a curved path and returns to the ground Projectile motion of

an object is simple to analyze if we make two assumptions: (1) the free-fall tion is constant over the range of motion and is directed downward,1 and (2) the effect of air resistance is negligible.2 With these assumptions, we find that the path

accelera-of a projectile, which we call its trajectory, is always a parabola as shown in Figure 4.7

We use these assumptions throughout this chapter.

The expression for the position vector of the projectile as a function of time follows directly from Equation 4.9, with its acceleration being that due to gravity,

The expression in Equation 4.10 is plotted in Figure 4.8 for a projectile launched

from the origin, so that rSi50 The final position of a particle can be considered to

be the superposition of its initial position rSi ; the term vSi t, which is its displacement

if no acceleration were present; and the term 1

2Sgt2 that arises from its acceleration due to gravity In other words, if there were no gravitational acceleration, the par-

ticle would continue to move along a straight path in the direction of vSi Therefore, the vertical distance 1

2Sgt2 through which the particle “falls” off the straight-line path is the same distance that an object dropped from rest would fall during the same time interval

Finalize Let us now consider a limiting case for very large values of t.

What if we wait a very long time and then observe the motion of the particle? How would we describe the motion of the particle for large values of the time?

Answer Looking at Figure 4.6, we see the path of the particle curving toward the x axis There is no reason to assume this tendency will change, which suggests that the path will become more and more parallel to the x axis as time grows large Mathematically, Equation (1) shows that the y component of the velocity remains constant while the x compo- nent grows linearly with t Therefore, when t is very large, the x component of the velocity will be much larger than the y component, suggesting that the velocity vector becomes more and more parallel to the x axis The magnitudes of both x f and y f continue to grow with time, although x f grows much faster

Wh At IF ?

1 This assumption is reasonable as long as the range of motion is small compared with the radius of the Earth (6.4 3 10 6 m) In effect, this assumption is equivalent to assuming the Earth is flat over the range of motion considered.

2This assumption is often not justified, especially at high velocities In addition, any spin imparted to a projectile,

such as that applied when a pitcher throws a curve ball, can give rise to some very interesting effects associated with aerodynamic forces, which will be discussed in Chapter 14.

Use the components of Equation 4.9 with x i 5 y i 5 0 at

t 5 0 and with x and y in meters and t in seconds:

x f5v xi t 11a x t25 20t 1 2.0t2

y f5v yi t 5 215t Express the position vector of the particle at any time t: Srf5x f i^ 1 y fj^ 5 120t 1 2.0t22 i^ 2 15t j^

Analyze

S o l u t I o n

▸ 4.1 c o n t i n u e d

A welder cuts holes through a heavy

metal construction beam with a hot

torch The sparks generated in the

process follow parabolic paths.

Acceleration at the highest Point

As discussed in Pitfall Prevention

2.8, many people claim that the

acceleration of a projectile at the

topmost point of its trajectory is

zero This mistake arises from

confusion between zero vertical

velocity and zero acceleration If

the projectile were to experience

zero acceleration at the highest

point, its velocity at that point

would not change; rather, the

projectile would move horizontally

at constant speed from then on!

That does not happen, however,

because the acceleration is not zero

anywhere along the trajectory.

4.3 projectile Motion 85

Figure 4.7 The parabolic path

of a projectile that leaves the

ori-gin with a velocity vSi The velocity

vector vS changes with time in both magnitude and direction

This change is the result of

accel-eration aS5Sg in the negative

at the origin is vSi The vector vSi t

would be the displacement of the projectile if gravity were absent, and the vector 1Sgt2 is its vertical displacement from a straight-line path due to its downward gravita- tional acceleration.

R

x

y

h i

v y  0

i O

u

v

BA

Figure 4.9 A projectile launched over a flat surface from the origin

at t i 5 0 with an initial velocity

v

S

i The maximum height of the

projectile is h, and the horizontal range is R At A, the peak of the

trajectory, the particle has

coordi-nates (R/2, h).

In Section 4.2, we stated that two-dimensional motion with constant

accelera-tion can be analyzed as a combinaaccelera-tion of two independent moaccelera-tions in the x and y

directions, with accelerations a x and a y Projectile motion can also be handled in

this way, with acceleration a x 5 0 in the x direction and a constant acceleration a y 5

2g in the y direction Therefore, when solving projectile motion problems, use two

analysis models: (1) the particle under constant velocity in the horizontal direction

(Eq 2.7):

x f5x i1v xi t

and (2) the particle under constant acceleration in the vertical direction (Eqs

2.13–2.17 with x changed to y and a y = –g):

independent of each other and can be handled separately, with time t as the

com-mon variable for both components

Q uick Quiz 4.2 (i) As a projectile thrown upward moves in its parabolic path

(such as in Fig 4.8), at what point along its path are the velocity and

accelera-tion vectors for the projectile perpendicular to each other? (a) nowhere (b) the

highest point (c) the launch point (ii) From the same choices, at what point are

the velocity and acceleration vectors for the projectile parallel to each other?

Horizontal Range and Maximum Height of a Projectile

Before embarking on some examples, let us consider a special case of projectile

motion that occurs often Assume a projectile is launched from the origin at t i 5

0 with a positive v yi component as shown in Figure 4.9 and returns to the same

hori-zontal level This situation is common in sports, where baseballs, footballs, and golf

balls often land at the same level from which they were launched

Two points in this motion are especially interesting to analyze: the peak point A,

which has Cartesian coordinates (R/2, h), and the point B, which has coordinates

(R, 0) The distance R is called the horizontal range of the projectile, and the distance

h is its maximum height Let us find h and R mathematically in terms of v i, ui , and g.

acceleration in the x

direction.

v

S

The projectile is launched

with initial velocity vSi.

86 chapter 4 Motion in two Dimensions

We can determine h by noting that at the peak v y A 5 0 Therefore, from the

particle under constant acceleration model, we can use the y direction version of Equation 2.13 to determine the time tA at which the projectile reaches the peak:

Using the identity sin 2u 5 2 sin u cos u (see Appendix B.4), we can write R in the

more compact form

R 5 v i

2 sin 2ui

The maximum value of R from Equation 4.13 is Rmax5v i2/g This result makes

sense because the maximum value of sin 2ui is 1, which occurs when 2ui 5 90°

Therefore, R is a maximum when u i 5 45°

Figure 4.10 illustrates various trajectories for a projectile having a given initial speed but launched at different angles As you can see, the range is a maximum for ui 5 45° In addition, for any ui other than 45°, a point having Cartesian coordi-

nates (R, 0) can be reached by using either one of two complementary values of u i, such as 75° and 15° Of course, the maximum height and time of flight for one of these values of ui are different from the maximum height and time of flight for the complementary value

Q uick Quiz 4.3 Rank the launch angles for the five paths in Figure 4.10 with

respect to time of flight from the shortest time of flight to the longest

50 100

same value of R.

Figure 4.10 A projectile

launched over a flat surface from

the origin with an initial speed

of 50 m/s at various angles of

projection.

Pitfall Prevention 4.3

the Range Equation Equation

4.13 is useful for calculating R only

for a symmetric path as shown in

Figure 4.10 If the path is not

sym-metric, do not use this equation The

particle under constant velocity

and particle under constant

accel-eration models are the important

starting points because they give

the position and velocity

compo-nents of any projectile moving

with constant acceleration in two

dimensions at any time t.

4.3 Projectile Motion 87 Problem-Solving Strategy Projectile Motion

We suggest you use the following approach when solving projectile motion problems

1 Conceptualize Think about what is going on physically in the problem Establish

the mental representation by imagining the projectile moving along its trajectory

2 Categorize Confirm that the problem involves a particle in free fall and that air

resistance is neglected Select a coordinate system with x in the horizontal direction

and y in the vertical direction Use the particle under constant velocity model for the

x component of the motion Use the particle under constant acceleration model for

the y direction In the special case of the projectile returning to the same level from

which it was launched, use Equations 4.12 and 4.13

3 Analyze If the initial velocity vector is given, resolve it into x and y components

Select the appropriate equation(s) from the particle under constant acceleration

model for the vertical motion and use these along with Equation 2.7 for the horizontal

motion to solve for the unknown(s)

4 Finalize Once you have determined your result, check to see if your answers are

consistent with the mental and pictorial representations and your results are realistic

Example 4.2 The Long Jump

A long jumper (Fig 4.11) leaves the ground at an angle of 20.0° above the

hori-zontal and at a speed of 11.0 m/s

(A) How far does he jump in the horizontal direction?

Conceptualize The arms and legs of a long jumper move in a complicated way,

but we will ignore this motion We conceptualize the motion of the long jumper

as equivalent to that of a simple projectile

Categorize We categorize this example as a projectile motion problem

Because the initial speed and launch angle are given and because the final

height is the same as the initial height, we further categorize this problem as

satisfying the conditions for which Equations 4.12 and 4.13 can be used This

approach is the most direct way to analyze this problem, although the general methods that have been described will

always give the correct answer

(B) What is the maximum height reached?

Finalize Find the answers to parts (A) and (B) using the general method The results should agree Treating the

long jumper as a particle is an oversimplification Nevertheless, the values obtained are consistent with experience in

sports We can model a complicated system such as a long jumper as a particle and still obtain reasonable results

Figure 4.11 (Example 4.2) Romain Barras of France competes

in the men’s decathlon long jump at the 2008 Beijing Olympic Games.

88 chapter 4 Motion in two Dimensions

Figure 4.12 (Example 4.3) (a) Multiflash photograph of the projectile–target demonstration If the gun

is aimed directly at the target and is fired at the same instant the target begins to fall, the projectile will

hit the target (b) Schematic diagram of the projectile–target demonstration.

Example 4.3 A Bull’s-Eye Every Time

In a popular lecture demonstration, a projectile is fired at a target in such a way that the projectile leaves the gun at the same time the target is dropped from rest Show that if the gun is initially aimed at the stationary target, the pro-jectile hits the falling target as shown in Figure 4.12a

Conceptualize We conceptualize the problem by studying Figure 4.12a Notice that the problem does not ask for numerical values The expected result must involve an algebraic argument

AM

S o l u t I o n

Write an expression for the y coordinate

of the target at any moment after release,

noting that its initial velocity is zero:

(1) yT5y i T1102t 21

2gt25xT tan ui212gt2

Write an expression for the y coordinate

of the projectile at any moment:

(2) yP5y i P1v yi P t 21

2gt250 11v i P sinui 2t 21

2gt251v i P sinui 2t 21

2gt2

Write an expression for the x coordinate

of the projectile at any moment: xP

5x iP1v xi P t 5 0 1 1v i P cos ui 2t 5 1v iP cos ui 2t

Solve this expression for time as a function

of the horizontal position of the projectile:

t 5 xP

v i P cos uiSubstitute this expression into Equation (2): (3) yP5 1v iP sin ui2 a xP

v iP cos uib 21

2gt25xP tan ui212gt2

Finalize Compare Equations (1) and (3) We see that when the x coordinates of the projectile and target are the same—that is, when xT 5 xP—their y coordinates given by Equations (1) and (3) are the same and a collision results.

Categorize Because both objects are subject only to gravity, we categorize this problem as one involving two objects

in free fall, the target moving in one dimension and the projectile moving in two The target T is modeled as a particle under constant acceleration in one dimension The projectile P is modeled as a particle under constant acceleration in the

y direction and a particle under constant velocity in the x direction.

Analyze Figure 4.12b shows that the initial y coordinate y i T of the target is xT tan ui and its initial velocity is zero It falls

with acceleration a y 5 2g

4.3 projectile Motion 89 Example 4.4 That’s Quite an Arm!

A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal with an initial speed of

20.0 m/s as shown in Figure 4.13 The height from which the stone is thrown is 45.0 m above the ground

(A) How long does it take the stone to reach the ground?

Conceptualize Study Figure 4.13, in which we have

indi-cated the trajectory and various parameters of the motion

of the stone

Categorize We categorize this problem as a projectile

motion problem The stone is modeled as a particle under

con-stant acceleration in the y direction and a particle under concon-stant

velocity in the x direction.

Analyze We have the information x i 5 y i 5 0, y f 5 245.0 m,

a y 5 2g, and v i 5 20.0 m/s (the numerical value of y f is

negative because we have chosen the point of the throw as

under constant acceleration model:

y f5y i1v yi t 21

2gt2

2129.80 m/s22t2

Solve the quadratic equation for t: t 5 4.22 s

(B) What is the speed of the stone just before it strikes the ground?

S o l u t I o n

Analyze Use the velocity equation in the particle

under constant acceleration model to obtain the y

component of the velocity of the stone just before

it strikes the ground:

v y f5v yi2gt

Use this component with the horizontal

compo-nent v xf 5 v xi 5 17.3 m/s to find the speed of the

Substitute numerical values, using t 5 4.22 s:

Finalize Is it reasonable that the y component of the final velocity is negative? Is it reasonable that the final speed is

larger than the initial speed of 20.0 m/s?

What if a horizontal wind is blowing in the same direction as the stone is thrown and it causes the stone

to have a horizontal acceleration component a x 5 0.500 m/s2? Which part of this example, (A) or (B), will have a

dif-ferent answer?

Answer Recall that the motions in the x and y directions are independent Therefore, the horizontal wind cannot

affect the vertical motion The vertical motion determines the time of the projectile in the air, so the answer to part

(A) does not change The wind causes the horizontal velocity component to increase with time, so the final speed will

be larger in part (B) Taking a x 5 0.500 m/s2, we find v xf 5 19.4 m/s and v f 5 36.9 m/s

Figure 4.13

(Example 4.4) A stone is thrown from the top of a building.