ch 03 vectors

7/28/2019 Ch 03 Vectors

1/27

2007 Pearson Prentice Hall, Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

57

Chapter 3: Vectors in Physics

Answers to Even-Numbered Conceptual Questions

2. Vectorsr

A ,r

G, andr

J are all equal to one another. In addition, vectorr

I is the same as vectorr

L .

4. No. The component and the magnitude can be equal if the vector has only a single component. If the vectorhas more than one nonzero component, however, its magnitude will be greater than either of its components.

6. No. If a vector has a nonzero component, the smallest magnitude it can have is the magnitude of thecomponent.

8. The vectorsr

A andr

B must point in the same direction.

10. The vectorr

A can point in the following directions: 45, 135, 225, and 315. In each of these directions

Ax = Ay .

12. Two vectors of unequal magnitude cannot add to zero, even if they point in opposite directions. Threevectors of unequal magnitude can add to zero if they can form a triangle.

14. When sailing upwind, your speed relative to the wind is greater than the speed of the wind itself. If you saildownwind, however, you move with the wind, and its speed relative to you is decreased.

Answers to Even-Numbered Conceptual Exercises

2. The vectors, in order of increasing magnitude (length), arer

B,r

C,r

A , andr

D .

4. The vectors, in order of increasing y component, arer

D ,r

B,r

A , andr

C. In arriving at this ranking, we note

that they component ofr

D is negative, they component ofr

B is zero, and they components ofr

A andr

C areincreasingly positive.

6. (a) The magnitude of1.4r

A is equal to the magnitude of 2.2r

B . These vectors point in different directions,

however. (b) The x component of 1.4r

A is less than the y component of 2.2r

B because the x component is

negative. The two components have the same magnitude, however.

8. (a) The magnitude ofr

A +r

D is less than the magnitude ofr

A +r

E . (b) The magnitude ofr

A +r

E is equal to

the magnitude ofr

A +r

F .

10. The vectorr

B points into the fourth quadrant, and hence its direction angle is between 270 and 360.Therefore, the answer is (d).

12. (a) Vectors 1 and 5 are position vectors. (b) Vectors 2, 3, 7, and 8 are velocity vectors. (c) Vectors 4 and 6are acceleration vectors.

7/28/2019 Ch 03 Vectors

2/27

Chapter 3: Vectors in Physics James S. Walker, Physics, 3rd Edition


58

Solutions to Problems

1. Picture the Problem: The press box is 38.0 ft above second baseand an unknown horizontal distance away.

Strategy: Use the tangent function to determine the horizontaldistance.

Solution: Use the tangentfunction to findx:

38.0 ft142 ft

tan tan15.0

yx

= = =

Insight: Dividing distances into right triangles in this manner is an important strategy for solving physics problems.

2. Picture the Problem: You drive 1.2 miles along an inclinedroadway, gaining 530 ft of altitude.

Strategy: Use the sine function to determine the angle and then theadditional distancex along the hypotenuse.

Solution:1. (a) Apply the sinefunction:

530 ftsin 1.2 mi 5280 ft/mi

y

s= =

2. Now solve for 1 530 ftsin 4.86300 ft/mi

= =

3. (b) Use the known angle together with the sine function to findx:150 ft

1800 ft 0.34 misin sin 4.8

yx

= = = =

Insight: It may be helpful for you to review the trigonometric functions sine, cosine, and tangent before tackling otherproblems in this chapter.

3. Picture the Problem: The road gains 6 ft in altitude for every 100 ft itspans in the horizontal direction..

Strategy: Use the tangent function to determine the angle.

Solution: Apply the tangent function: 16 ft

sin tan 3100 ft

y

x

= = =


4. Picture the Problem: The vector direction ismeasured counterclockwise from the +x axis.

Strategy: In each case find the vector

components.

Solution:1. (a) Find thex andy components: ( )

( )

cos 75 m cos35.0 61 m

sin 75 m sin 35.0 43 m

x

y

r r

r r

= = =

= = =

2. (b) Find thex andy components: ( )

( )

cos 75 m cos65.0 32 m

sin 75 m sin 65.0 68 m

x

y

r r

r r

= = =

= = =

Insight: Resolving vectors into their components is an important skill for solving physics problems.

x

y

35.0

rr

x

y

65.0

rr

(a) (b)

15.038.0 ft

x

530 ft

1.2 mi150 ftx

6 ft

100 ft

7/28/2019 Ch 03 Vectors

3/27



59

5. Picture the Problem: The base runner travels from C (home plate)to first base, then to A (second base), then to B (third base), andfinally back to C (home plate).

Strategy: The displacement vector rr

is the same as the position

vector rr

if we take home plate to be the origin of our coordinatesystem (as it is drawn). The displacement vector for a runner whohas just hit a double is drawn.

Solution:1. (a) Write the displacement vectorfrom C to A in terms of itsx andy components:

( ) ( ) 90 ft 90 ft= +r x yr

2. (b) Write the displacement vector from C to Bin terms of itsx andy components:

( ) ( ) ( ) 0 ft 90 ft 90 ft= + =r x y yr

3. (c) For a home run the displacement is zero: ( ) ( ) 0 ft 0 ft= +r x yr

Insight: The displacement is always zero when the object (or person) returns to its original position.

6. Picture the Problem: The ship approaches the rocks as depictedin the picture.

Strategy: The distance to the rocks can be determined from aright triangle that extends from the sailor to the top of thelighthouse to the base of the lighthouse and back to the sailor.Find the length of the bottom of that triangle and subtract 19 ft todetermine the distance to the rocks.

Solution:1. Use the tangent function to find the distanceL:( ) ( )49 14 ft 35 ft

tan 30 61 fttan30

LL

= = =

2. Subtract 19 ft fromL to find the distance to the rocks: 19 ft 61 19 ft 42 ftd L= = =

Insight: Identifying right triangles and manipulating the trigonometric functions are important skills to learn whensolving physics problems.

7. Picture the Problem: The water molecule forms a triangle with the positions of theoxygen and hydrogen nuclei as shown.

Strategy: Break the triangle up into two right triangles and use the sine function to findthe distance between the hydrogen nuclei. The angle is half of the 104.5 bond angle,or = 52.25.

Solution:1. Use the sine function to find thedistance d:

sin0.96

d=

2. The distance between hydrogen nuclei is 2d: ( ) ( )2 2 0.96 sin 52.25 1.5 d= =

Insight: Identifying right triangles and manipulating the trigonometric functions are important skills to learn whensolving physics problems.

rr

7/28/2019 Ch 03 Vectors

4/27



60

8. Picture the Problem: The given vector components correspond to the vector rr

as drawn atright.

Strategy: Use the inverse tangent function to determine the angle . Then use the PythagoreanTheorem to determine the magnitude of r

r

.

Solution:1. (a) Use the inverse tangent functionto find the distance angle :

1 9.5tan 3414

= =

or 34 below

the +x axis

2. (b) Use the Pythagorean Theorem todetermine the magnitude of r

r

:( ) ( )

2 22 2 14 m 9.5 m

17 m

x yr r r

r

= + = +

=

3. (c) If both xr and yr are doubled, the

direction will remain the same but the magnitudewill double: ( ) ( )

1

2 2

9.5 2tan 34

14 2

28 m 19 m 34 mr

= =

= + =

Insight: Any vector can be resolved into two components. The ability to convert a vector to and from its components isan essential skill for solving many physics problems.

9. Picture the Problem: The given vector components correspond to the vector rr

as drawn atright.

Strategy: Determine the angle from our knowledge of analog clocks. The given component

xr together with the angle will allow us to calculate the length ofrand the component yr .

Solution:1. (a) Find the angle : 1 360 3012

= =

2. Find the length ofr: 3.0 cmcos 3.5 cmcos cos30

x

x

rr r r

= = = =

o

3. (b) The componentsxr and

yr are only equal when =45. Since in this case =30, the component

yr will be

less thanxr or 3.0 cm.

4. (c) Findyr :

( )sin 3.5 cm sin 30 1.7 cmyr r = = =o


10. Picture the Problem: The trip takes you toward the east first and thentoward the north. The vector is depicted at right.

Strategy: Use the Pythagorean Theorem to determine the magnitude and

the inverse tangent function to determine the angle.

Solution:1. (a) Find the magnitude ofrr

: ( ) ( )2 2

660 ft 370 ft 760 ftr= + =

2. (b) I estimate an angle of close to 30 because such an angle would correspond to a north component that is half aslarge as the east component.

3. (c) Use the inverse tangent function to find : 1 370 fttan 29660 ft

= =


x

y

rr

II

III

IXII

x

y

rr

14 m

9.5 m

3.0 cm

370 ft

660 ft

rr

7/28/2019 Ch 03 Vectors

5/27



61

11. Picture the Problem: The two vectors Ar

(length 50 units) and Br

(length 120 units) are drawnat right.

Strategy: Resolve Br

into itsx andy components to answer the questions.

Solution:1. (a) FindBx: ( )120 units cos70 41 unitsxB = =

2. Since the vector Ar

points entirely in thex direction, we can see thatAx = 50 units and that

vector Ar

has the greaterx component.

3. (b) FindBy: ( )120 units sin70 113 unitsxB = =

4. The vector Ar

has noy component, so it is clear that vector Br

has the greatery component.


12. Picture the Problem: The four possible locations of the treasure arelabeledA,B, C, andD in the figure at right. The position vector for location

A is also drawn. North is up and east is to the right.Strategy: Use the vector components to find the magnitude and directionof each vector.

Solution:1. Find the magnitude of Ar

: ( ) ( )2 2

25.0 m 12.0 5.00 m 30.2 mA = + + =

2. Find the direction from north of Ar

: 125.0 m

tan 55.8 west of north12.0 5.00 mA

= = +

3. Find the magnitude of Br

: ( ) ( )2 2

25.0 m 12.0 5.00 m 26.0 mB = + =

4. Find the direction from north of Br

: 125.0 m

tan 74.4 west of north

12.0 5.00 m

B

= =

5. Find the magnitude of Cr

: ( ) ( )2 2

25.0 m 12.0 5.00 m 30.2 mC= + + =

6. Find the direction from north of Cr

: 125.0 m

tan 55.8 east of north12.0 5.00 mC

= = +

7. Find the magnitude of Dr

: ( ) ( )2 2

25.0 m 12.0 5.00 m 26.0 mD = + =

8. Find the direction from north of Dr

: 125.0 m

tan 74.4 east of north12.0 5.00 mD

= =

Insight: If you ever find a treasure map like this one, youll be glad you mastered vectors in physics!

13. Picture the Problem: The whale dives along a straight line tilted 20.0below horizontal for 150 m as shown in the figure.

Strategy: Resolve the whales displacement vector into horizontal andvertical components in order to find its depth ry and its horizontal traveldistance rx.

Solution:1. (a) The depth is given by ry: ( ) ( )sin 150 m sin 20.0 51 myr r = = =

2. (b) The horizontal travel distance is given by rx: ( ) ( )cos 150 m cos 20.0 140 m 0.14 kmxr r = = = =

Insight: Note that both answers are limited to two significant figures, because although 20.0 has three, 150 m hasonly two significant figures.

x

y

Br

Ar

70

12.0 m

25.0 m 25.0 m

5.00 m

A

B

C

D

palm tree

Ar A

7/28/2019 Ch 03 Vectors

6/27



62

14. Picture the Problem: The two vectors Ar

(length 50.0 m) and Br

(length 70.0 m) are drawnat right.

Strategy: Add vectors Ar

and Br

using the vector component method.

Solution:1. (a) A sketch of the vectors and their sum is shown at right.

2. (b) Add thex components: ( ) ( ) ( ) ( )50.0 m cos 20.0 70.0 m cos 50.0 92.0 mx x xC A B= + = + =

3. Add they components: ( ) ( ) ( ) ( )50.0 m sin 20.0 70.0 m sin 50.0 36.5 my y yC A B= + = + =

4. Find the magnitude ofC: ( ) ( )2 22 2 92.0 m 36.5 m 99.0 m

x yC C C= + = + =

5. Find the direction ofC: 1 1 36.5 mtan tan 21.792.0 m

y

C

x

C

C

= = =

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answersusing this approach than by using a ruler and protractor to add the vectors graphically.

15. Picture the Problem: The vectors involved in the problem are depicted at right.The control tower (CT) is at the origin and north is up in the diagram.

Strategy: Subtract vector Br

from Ar


Solution:1. (a) A sketch of the vectors and their difference is shown at right.

2. (b) Subtract thex components: ( ) ( ) ( ) ( )220 km cos 180 32 140 km cos 90 65 310 kmx x xD A B= = =

3. Subtract they components: ( ) ( ) ( ) ( )220 km sin 180 32 140 km sin 90 65 57 kmy y yD A B= = =

4. Find the magnitude ofD: ( ) ( )2 22 2 5310 km 57 km 320 km 3.2 10 mx yD D D= + = + = =

5. Find the direction ofD: 1 1 57 kmtan tan 10 180 170 or 10 north of west310 km

y

D

x

D

D

= = = + =

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answersusing this approach than by adding the vectors graphically. Notice, however, that when your calculator returns 10 asthe angle in step 5, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

16. Picture the Problem: The vectors involved in the problem are depicted atright.

Strategy: Subtract vector ivr

from fvr


Solution:1. (a) A sketch of the vectors and their difference is shown atright.

2. (b) Subtract thex components: ( ) ( ) ( ) ( )f , i, 66 km/h cos 75 45 km/h cos 0 28 km/hx x xv v v = = =

3. Subtract they components: ( ) ( ) ( ) ( )f , i,y 66 km/h sin 75 45 km/h sin 0 64 km/hy yv v v = = =

4. Find the magnitude of v : ( ) ( )2 22 2 28 km/h 64 km/h 70 km/hx yv v v = + = + =

Cr

x

yBr

Ar

20.0

50.0

x

y

fvr

ivr

75

ivr

vr

7/28/2019 Ch 03 Vectors

7/27



63

5. Find the direction of v : 1 164 km/h

tan tan 66 180 11428 km/h

y

v

x

v

v

= = = + =

where the angle is

measured counterclockwise from the positivex axis.

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answers

using this approach than by adding the vectors graphically. Notice, however, that when your calculator returns 66 asthe angle in step 5, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

17. Picture the Problem: The vectors involved in the problem are depicted at right.

Strategy: Deduce thex andy components of Br

from the information given about Ar

and Cr

. Use the known components to estimate the length and direction of Br

as wellas calculate them precisely.

Solution:1. (a) A sketch of the vectors is shown at right.

2.(b) The vector Br

must have anx component of 75 m so that when it is added toAr

thex components will cancel out. It must also have ay component of 95 m

because that is the length of Cr

and Ar

has noy component to contribute. Therefore

Br

must be longer than either Ar

or Cr

and it must have an angle of greater than 90.I estimate that its length is about 120 m and that it points at about 130.

3. (c) Using the known components of Br

we can find its magnitude:

2 2( 75 m) (95 m) 121 mB = + =

4. Find the direction of Br

: 195 m

tan 52 180 12875 mB

= = + =

Insight: Here the length and direction of Br

are determined by itsx andy components, which are determined from Ar

and Cr

. Learning to manipulate vector components will be a useful skill when tackling many physics problems.


Strategy: Since Ar

points entirely in thex direction, and Br

points entirely in they

direction, Ar

and Br

are thex andy components of their sum +A Br r

. Use the knownlengths of +A B

r r

and Ar

to findB.

Solution:1. Set the length of

+A Br r

equal to 37 units:

2 2

2 2 237

37

A B

A B

= +

= +

2. Solve forB: ( )22 2 237 37 22 30 unitsB A= = =

Insight: Here the length of Br

is determined by the lengths of the other two vectors because the directions of Ar

and Br

are stipulated. Learning to manipulate vector components will be a useful skill when tackling many physics problems.

x

y

Br

Ar

+A Br r

22 O

30

7/28/2019 Ch 03 Vectors

8/27



64


Strategy: Use the vector component method of addition and subtraction to

determine the components of each combination of Ar

and Br

. Once thecomponents are known, the length and direction of each combination can

be determined fairly easily.

Solution:1. (a) Determine the components of +A Br r

: ( ) ( ) 5 10 10 5+ = + =A B y x x yr r

2. Find the magnitude of +A Br r

: ( ) ( )2 2

10 5 11 units+ = + =A Br r

3. Determine the direction of +A Br r

, measuredcounterclockwise from the positivex axis.

1 5tan 27 or 33310

+

= =

A Br r

4. (b) Determine the components of A Br r

: ( ) ( ) 5 10 10 5 = = A B y x x yr r

5. Find the magnitude of A Br r

: ( ) ( )2 2

10 5 11 units = + =A Br r

6. Determine the direction of A Br r


1 5tan 27 180 20710

= = + =

A Br r

7. (c) Determine the components of B Arr

: ( ) ( ) 10 5 10 5 = = +B A x y x yrr

8. Find the magnitude of B Arr

: ( ) ( )2 2

10 5 11 units = + =B Arr

9. Determine the direction of B Arr


1 5tan 2710

= =

B A

rr

Insight: This problem is simplified by the fact that Ar

and Br

have only one component each, but a similar approachwill work even with more complicated vectors. Notice that you must have a picture of the vectors in your head (or onpaper) in order to correctly interpret the directions in steps 3, 6, and 9.

20. Picture the Problem: The vectors involved in the problem aredepicted at right.

Strategy: Add the vectors using the component method in order tofind the components of the vector sum. Use the components to findthe magnitude and the direction of the vector sum.

Solution:1. (a) Make estimates from the drawing: 20 m 1.5+ + A B Cr rr

2.(b) Add the vector components: ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

0 20.0 m cos 45 7.0 m cos 30

10.0 m 20.0 m sin 45 7.0 m sin 30

20.2 m 0.64 m

+ + = + + + + +

+ + = +

A B C x

y

A B C x y

r rr

r rr

+A Br r

+A Br r

A Br r

B Arr

x

y

Br

Ar

O

10

5

B Arr

A Br r

Cr

x

y

Br

Ar

+ +A B Cr rr

45

30

7/28/2019 Ch 03 Vectors

9/27



65

3. Use the components to find the magnitude: ( ) ( )2 220.2 m 0.64 m 20.2 m+ + = + =A B C

r rr

4. Use the components to find the angle: 1 0.64 mtan 1.820.2 m

= =

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more accurate answersusing this approach than by adding the vectors graphically. Notice, however, that when your calculator returns the angleof 1.8 in step 4, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

21. Picture the Problem: The vector involved in the problem is depicted at right.

Strategy: Determine thex andy components of and then express them in terms of the unitvectors.

Solution:1. Find thex andy components of rr

: ( ) ( )

( ) ( )

54 m cos 42 40 m

54 m sin 42 36 m

x

y

r

r

= =

= =

o

o

2. Now express rr

in terms of the unit vectors: ( ) ( ) 40 m 36 m = + r x yr

Insight: In general, an arbitrary two-dimensional vector Ar

can always be written as the sum of a vector component inthe x direction and a vector component in they direction.

22. Picture the Problem: The vector involved in the problem is depicted at right.

Strategy: Determine thex andy components of Ar

.

Solution:1. Find thex component: ( ) ( )2.5 m cos 140 1.9 mxA = =

2. Find they component: ( ) ( )2.5 m sin 140 1.6 myA = =


23. Picture the Problem: The vector Ar

has a length of 6.1 m and points in the negativex direction.

Strategy: In order to multiply a vector by a scalar, you need only multiply each component of the vector by the samescalar.

Solution:1. (a) Multiply each component of Ar

by 3.7: ( )

( )( ) ( )

6.1 m

3.7 3.7 6.1 m 23 m so 23 mxA

=

= = =

A x

A x x

r

r

2.(b) Since Ar

has only one component, its magnitude is simply 23 m.

Insight: Multiplying both components of a vector by a scalar will change the length of the vector but not its direction.

24. Picture the Problem: The vector 5.2 Ar

has a length of 44 m and points in the positivex direction.

Strategy: Divide the components of the vector 5.2 Ar

by 5.2 in order to find the components of Ar

. From there we

can easily find thex component and the magnitude of Ar

.

Solution:1. (a) Divide both sides by 5.2: 5.2 (44 m)( 8.5 m)

=

=

A x

A x

r

r

2. The vector Ar

has only anx component: 8.5 mxA =

3.(b) Since Ar

has only one component, its magnitude is simply 8.5 m.

Insight: Dividing both components of a vector by a scalar will change the length of the vector but not its direction.

Ar

x

y

140

rr

x

y

54 m42

7/28/2019 Ch 03 Vectors

10/27



66


Strategy: Determine the lengths and directions of the various vectors by using theirx andy components.

Solution:1. (a) Find the direction of Ar

from its components:

1 2.0 m

tan 225.0 m

= =

Ar

2. Find the magnitude of Ar

: ( ) ( )2 2

5.0 m 2.0 m 5.4 mA = + =

3. (b) Find the direction of Br

from itscomponents:

1 5.0 mtan 68 180 1102.0 m

= = + =

Br


: ( ) ( )2 2

2.0 m 5.0 m 5.4 mB = + =

5. (c) Find the components of +A Br r

: ( ) ( ) ( ) ( ) 5.0 2.0 m 2.0 5.0 m 3.0 m 3.0 m+ = + + = +A B x y x yr r

6. Find the direction of +A Br r

from its

components:

1 3.0 mtan 453.0 m

+

= =

A Br r


: ( ) ( )2 23.0 m 3.0 m 4.2 m+ = + =A B

r r

Insight: In the world of vectors 5.4+5.4 m can be anything between 0 m and 10.8 m, depending upon the directions thatthe vectors point. In this case their sum is 4.2 m.


Strategy: Determine the lengths and directions of the various vectors by using theirxandy components.

Solution:1. (a) Find the direction of

Ar

from its components:1 12 mtan 26

25 m

= =

Ar

2. Find the magnitude of Ar

: ( ) ( )2 225 m 12 m 28 mA = + =

3. (b) Find the direction of Br

from itscomponents:

1 15 mtan 822.0 m

= =

Br


: ( ) ( )2 22.0 m 15 m 15 mB = + =

5. (c) Find the components of +A Br r

:( ) ( ) ( ) ( )

25 2.0 m 12 15 m 27 m 3.0 m+ = + + + = +A B x y x yr r

6. Find the direction of +A Br r

from itscomponents:

1 3.0 mtan 6.327 m

+

= =

A Br r


: ( ) ( )2 227 m 3.0 m 27 m+ = + =A B

r r

Insight: In the world of vectors 28 + 15 m can be anything between 13 m and 43 m, depending upon the directions thatthe vectors point. In this case their sum is 27 m.

+A Br r

+A Br r

x

y

Br

Ar

2.0 m 5.0 mO

2.0 m

+A Br r

x

y

Br

Ar

O 25 m

12 m

+A Br r

7/28/2019 Ch 03 Vectors

11/27



67


Strategy: Since the components of the vectors are known, we need onlyadd or subtract the components separately as specified in the problemstatement.

Solution:1. (a) Subtract thecomponents:

( ) ( )

( ) ( )

25 m2 m 12 m15 m

23 m 27 m

= +

=

A B x y

x y

r r

2.(b) Multiply the answer topart (a) by 1:

( ) ( ) ( ) 23 m 27 m = = +B A A B x yr rr r

Insight: Adding and subtracting vectors in component form is often easierthan doing so graphically.


Strategy: Use the information given in the figure to determine the

components of each vector.Solution:1. ( ) ( ) ( ) ( )

( ) ( )

1.5 m cos 40 1.5 m sin 40

1.1 m 0.96 m

= +

= +

A x y

A x y

r

r

2. ( ) ( ) ( ) ( )

( ) ( )

2.0 m cos 19 2.0 m sin 19

1.9 m 0.65 m

=

= +

B

B x y

r

r

3. ( ) ( ) ( ) ( )

( ) ( )

1.0 m cos 180 25 1.0 m sin 180 25

0.91 m 0.42 m

= +

= +

C x y

C x y

r

r

4. ( ) 0 1.5 m= +D x y

r



Strategy: Use the information given in the figure to determine the

components of vectors , , andA B Cr rr

. Then add the components.

Solution:1. Add thexcomponent of each vector:

( ) ( )

( ) ( )

( ) ( )( )

1.5 m cos 40 1.1 m

2.0 m cos 19 1.9 m

1.0 m cos 180 25 0.91 m

2.1 m

x

x

x

x

A

B

C

= =

= =

= =

+ + =A B Cr rr

2. Add they component ofeach vector:

( ) ( )

( ) ( )

( ) ( )

( )

1.5 m sin 40 0.96 m

2.0 m sin 19 0.65 m

1.0 m sin 180 25 0.42 m

0.74 m

y

y

y

y

A

B

C

= =

= =

= =

+ + =A B Cr rr

3. Express the sum in unit vector notation: ( ) ( ) 2.1 m 0.74 m+ + = +A B C x yr rr

Insight: In this problem the vector component method of addition is much quicker than the graphical method.

25 mO

Ar

12 m

A Br r

y

x

B Arr

Ar

Br

Br

7/28/2019 Ch 03 Vectors

12/27



68

30. Picture the Problem: The two vectors involved in this problem aredepicted in the figure at right.

Strategy: Use the checkerboard squares as a coordinate grid to writevectors 1 and 2 in component form. Then use the components to determinethe magnitude and direction of each vector.

Solution:1. (a) Because each vector has components of length 3.5 cm and7.0 cm, the magnitude of displacement 1 will be equal to the magnitude ofdisplacement 2.

2. (b) Find the components of each vector: ( ) ( )

( ) ( )

1

2

7.0 cm 3.5 cm

3.5 cm 7.0 cm

= +

= +

r x y

r x y

r

r

3. Use the components to find the magnitude anddirection of displacement 1:

( ) ( )2 2

1

11

7.0 cm 3.5 cm 7.8 cm

3.5 cmtan 27 180 153

7.0 cm

r

= + =

= = + =

4. Use the components to find the magnitude anddirection of displacement 2:

( ) ( )2 2

2

12

3.5 cm 7.0 cm 7.8 cm

7.0 cmtan 63

3.5 cm

r

= + =

= =

Insight: Because the magnitude of a vector depends upon the squares of the components, it does not matter that thexcomponent of displacement 1 is negative; the displacements still have equal magnitudes.

31. Picture the Problem: The displacement vectors are depicted at right. North is in they direction and east is in thex direction.

Strategy: Sum the components of the vectors in order to determine +A Br r

.Multiply that vector by 1 in order to reverse its direction. Then find the magnitudeand direction of the reversed vector.

Solution:1. (a) Add the two displacementvectors:

( ) ( ) 72 m 120 m+ = +A B x yr r

2. Multiply by 1 in order to reverse thedirection of the net displacement and bring thecat back home:

( ) ( ) ( ) 72 m 120 m + = + A B x yr r

3. Find the magnitude of the desireddisplacement:

( ) ( ) ( )2 2

72 m 120 m 140 m + = + =A Br r

4. Find the direction of the desired displacement: 1 120 mtan 59 59 south of east72 m

= = =

5. (b) Vector addition is independent of the order in which the addition is accomplished. The initial displacement is thesame, so there is no change in the displacement for the homeward part of the trip.

Insight: In this problem we could claim the cats initial displacement is a single vector with the given components. Theanswers wouldnt change, but it would simplify the solution a little bit.

( ) +A Br r

x

y

Br

Ar

72 m O

120 m

7/28/2019 Ch 03 Vectors

13/27



69

32. Picture the Problem: The two legs of the cats path are indicated at right. North isin they direction and east is in thex direction.

Strategy: Determine the displacement from the known vectors that make up the twolegs of the cats journey. Divide the displacement by the total time of travel to findthe average velocity. Use thex andy components of the average velocity todetermine its magnitude and direction.Solution:1. (a) Determine thedisplacement:

( ) ( ) 120 m 72 m = + = + r A B y xr rr

2. Divide by the total time (45 min +17 min = 62 min) to find the averagevelocity: ( ) ( )

av72 m 1min 120m 1min

62 min 60s 62 m 60s

0.019 m/s 0.032 m/s

t

= = +

= +

rv x y

x y

r

r

3. Determine the magnitude of thevelocity:

2 2

avm m

0.019 0.032 0.037 m/ss s

v

= + =

4. Determine the direction of thevelocity:

1 120 mtan 59 180 12172 m

121 or 31 west of north

= = + =

=

Insight: The average speed would be calculated differently:( )

120 72 m0.052 m/s

45 17 min 60 s/min

ds

t

+= = =

+ . The

average speed is faster than the average velocity because the total distance traveled is larger than the displacement.

33. Picture the Problem: You travel due west for 130 s at 27 m/s then due south at 14 m/s for 62 s.

Strategy: Find the components of the displacement vector. Once the components are known the magnitude anddirection can be easily found. Let north be the positivey direction and east be the positivex direction.

Solution:1. Find the westward displacement: ( )( )27 m/s 130 s 3500 mx xr v t= = =

2. Find the southward displacement: ( ) ( )14 m/s 62 s 870 my yr v t= = =

3. Find the direction of the displacement: 1 1 870 mtan tan 14 180 1943500 m

or 14 south of west

y

x

r

r

= = = + =

4. Find the magnitude of the displacement: ( ) ( )2 2

3500 m 870 m 3600 m 3.6 kmr= + = =

Insight: The 14 refers to the angle below the negativex axis (west) because the argument of the inverse tangentfunction is

y xr r , or south divided by west.

= +r A B

r rr

x

y

Br

Ar

72 m O

120 m

7/28/2019 Ch 03 Vectors

14/27



70

34. Picture the Problem: You travel due east 1500 ft then due north 2500 ft.

Strategy: The components of the displacement are given, from which we can determine the magnitude and directionfairly easily. The direction of the average velocity will be the same as the direction of the displacement. The magnitudeof the average velocity is the magnitude of the displacement divided by the total time of travel. Let north be the positivey direction and east be the positivex direction.

Solution:1. Find the direction of the displacement: 1 1 2500 fttan tan 59 north of east1500 ft

y

x

r

r

= = =

2. Find the magnitude of the displacement: ( ) ( )2 2

1500 ft 2500 ft 2900 ft 0.305 m/ft 890 mr= + = =

3. Find the magnitude of the average velocity:av

890 m4.9 m/s

3.0 min 60 s/min

rv

t

= = =

Insight: The 59 refers to the angle above the positivex axis (east) because the argument of the inverse tangent functionis

y xr r , or north divided by east.

35. Picture the Problem: The jogger runs at 3.25 m/s in a direction 30.0 above the positivex axis.Strategy: Find the components of the velocity vector according to the method indicated in Figure 3-7(a).

Solution:1. (a) Find thex component of vr

: ( ) ( )3.25 m/s cos 30.0 2.81 m/sxv = =

2. Find they component of vr

: ( ) ( )3.25 m/s sin 30.0 1.63 m/syv = =

3. (b) If the joggers speed is halved, the direction will remain unchanged but thex andy components will be halved.

Insight: In this case the angle of 30.0 corresponds to the standard angle as indicated in Figure 3-7(a).

36. Picture the Problem: The ball rises straight upward, momentarily comes to rest, and then falls straight downward.

Strategy: After it leaves your hand the only acceleration of the ball is due to gravity, so we expect the answer to be

9.81 m/s2

. To calculate the acceleration we need only consider the initial and final velocities and the time elapsed.Because of the symmetry of the situation, the final velocity downward will have the same magnitude as the initialvelocity upward. Apply equation 3-5, taking upward to be the positive direction.

Solution: Apply equation 3-5:( ) ( ) 2

av

4.5 m/s 4.5 m/s9.8 m/s

0.92 sf i

t

= = =

v v y ya y

r r

r

Insight: We saw in Chapter 2 how a uniform acceleration will produce a symmetric trajectory, with the time to rise tothe peak of flight equaling the time to fall back down, and with equal initial and final speeds.

37. Picture the Problem: The skateboarder rolls down the ramp that is inclined 20.0 above the horizontal.

Strategy: To calculate the acceleration we need only consider the initial and final velocities and the time elapsed.Apply equation 3-5, taking the direction down the ramp to be the positive direction.

Solution: 1. Apply equation 3-5:( ) ( ) 2

av

10.0 m/s 0 m/s3.33 m/s

3.00 sf iv v

at

= = =

2. Compare with g sin : ( ) ( )2 2sin 9.81 m/s sin 20.0 3.36 m/sg = =

Insight: The two are equal to within rounding errors. Or perhaps there was a small amount of friction between theskateboard wheels and the ramp. In Chapter 5 well be able to rigorously prove the two are equal using a free bodydiagram. See, for instance, Example 5-9.

7/28/2019 Ch 03 Vectors

15/27



71

38. Picture the Problem: The skateboarder rolls down the ramp that is inclined 15.0 above the horizontal.

Strategy: The acceleration relates the change in velocity with the time elapsed. Solve equation 3-5 for the final speed,taking the direction down the ramp to be the positive direction.

Solution: Solve equation 3-5 forfv : ( ) ( ) ( )20 9.81 m/s sin 15.0 3.00 s 7.62 m/sf iv v a t = + = + =

Insight: Note that when the vector direction doesnt matter (this is essentially a one-dimensional problem) the equation3-5 looks exactly like equation 2-7. In Chapter 5 well be able to rigorously prove that sina g = using a free bodydiagram. See, for instance, Example 5-9.

39. Picture the Problem: The initial and final displacement vectors are depictedat right.

Strategy: Use the given formulae to determine the components of the initialand final positions. Then use those components to find the displacementvector. Divide the displacement vector by the elapsed time to find thevelocity vector, and then determine its magnitude and direction.

Solution:1. (a) Find the initialposition vector: ( )

[ ] [ ]{ }

( )

8

i8

3.84 10 m cos 0 sin 0

3.84 10 m

= +

=

r x y

x

r

2. Find the arguments of the sineand cosine functions for t= 7.38days. Let 6 radians/s2.46 10 = :

( )( )62.46 10 radians/s 7.38 d 86400 s/d

1.57 radians

t =

=

3. Find the final position vector: ( ) [ ] [ ]{ }

( ) [ ] [ ]{ }

( ) ( ) ( )

8f

8

5 8 8f

3.84 10 m cos sin

3.84 10 m cos 1.57 radians sin 1.57 radians

3.06 10 m 3.84 10 m 3.84 10 m

t t = +

= +

= +

r x y

x y

r x y y

r

r

4. Find the displacement vector: ( ) ( )8 8f i 3.84 10 m 3.84 10 m = = r r r y xr r r

5. Find the vector avvr

:( ) ( )

( ) ( )

8 8

av

av

3.84 10 m 3.84 10 m

7.38 d 86400 s/d 602 m/s 602 m/s

t

+ = =

= +

x yrv

v x y

r

r

r

6. Find the magnitude of avvr

: ( ) ( )2 2

av 602 m/s 602 m/s 852 m/sv = + =

7. Find the direction of avvr

: av,1 1

av,

602 m/stan tan 45 180 135

602 m/sy

x

v

v

= = = + =

8. (b) The instantaneous speed of the Moon is greater than the average velocity because the distance traveled is greaterthan the displacement in this case.

Insight: If the Moon had completed an entire orbit, instead of just one-quarter of an orbit, its displacement and itsaverage velocity would have been zero. Its speed remains constant, however, at about 947 m/s using the data given inthis problem. The given data correspond to a coordinate system where thex direction always points towards the centerof the Sun even as the Earth orbits the Sun.

rr

3.8410 m

3.84108 m

x

frr

irr

y

O

7/28/2019 Ch 03 Vectors

16/27



72

40. Picture the Problem: The initial and final velocity vectors are depicted at right.

Strategy: Use the given formulae to determine the components of the initial and finalvelocities. Then use those components to find the change in velocity vector. Divide thechange in velocity vector by the elapsed time to find the acceleration vector, and thendetermine its magnitude and direction.

Solution:1. (a) Find the initialvelocity vector:

( ) [ ] [ ]{ }

( )

i 945 m/s sin 0 cos 0

945 m/s

= +

=

v x y

y

r

2. Find the arguments of the sineand cosine functions for t= 0.100days. Let 6 radians/s2.46 10 = :

( )( )62.46 10 radians/s 0.100 d 86400 s/d

0.0213 radians

t

=

=


( ) [ ] [ ]{ }

( ) ( )

f

f

945 m/s sin cos

945 m/s sin 0.0213 radians cos 0.0213 radians

20.1 m/s 945 m/s

t t = +

= +

= +

v x y

x y

v x y

r

r

4. Find the acceleration vector: ( ) ( ) ( ) ( )2f i 20.1 m/s 945 m/s 945 m/s

0.00233 m/s0.100 d 86400 s/dt

= = =

x y yv va x

r r

r

5.(b) Find the arguments of thesine and cosine functions fort= 0.0100 days:

( )( )62.46 10 radians/s 0.0100 d 86400 s/d

0.00213 radians

t

=

=


( ) [ ] [ ]{ }

( ) ( )

f

f

945 m/s sin cos

945 m/s sin 0.00213 radians cos 0.00213 radians

2.01 m/s 945 m/s

t t = +

= +

= +

v x y

x y

v x y

r

r

7. Find the acceleration vector: ( ) ( ) ( ) ( )2f i 2.01 m/s 945 m/s 945 m/s

0.00233 m/s0.0100 d 86400 s/dt

= = =

x y yv va x

r r

r

Insight: The two answers ended up being the same because both time intervals are fairly small. If instead we hadexamined an interval of 1.00 days there would have been a y component of vr and a slightly different acceleration. InChapter 6 we will examine circular motion and find an even easier way to calculate the acceleration of the Moon. Thegiven data in this problem correspond to a coordinate system where thex direction always points towards the center ofthe Sun even as the Earth orbits the Sun.

41. Picture the Problem: The vectors involved in this problem are depicted at right.

Strategy: Let pg =vr

planes velocity with respect to the ground, ap =vr

attendantsvelocity with respect to the plane, and add the vectors according to equation 3-8 tofind ag =v

rattendants velocity with respect to the ground.

Solution:1. Apply equation 3-8: ( ) ( ) ( )ag ap pg

ag

1.22 m/s 16.5 m/s 15.3 m/s

15.3 m/sv

= + = + =

=

v v v x x xr r r

Insight: If the attendant were walking towards the front of the plane, her speed relative to the ground would be17.7 m/s, slightly faster than the airplanes speed.

vr

945 m/s

fvr

iv

r

y

Ox

apvr

pgvr

agvr

7/28/2019 Ch 03 Vectors

17/27



73


Strategy: Thex-component of the velocity was chosen perpendicular to themotion of the river. Therefore, the motion of the river will not affect the time ittakes to travel across it. Divide the width of the river by thex component of theboats velocity to find the time it takes to cross the river.

Solution:1. Find the velocity ofthe boat relative to the water:

( ) ( )

( ) ( )

bw 6.1 m/s cos 25 6.1 m/s sin 25

5.5 m/s 2.6 m/s

= +

= +

v x y

x y

r

2. Divide the width of the riverby thex component of bwv

r

bw,

25 m4.5 s

5.5 m/sx

xt

v= = =

Insight: In real life the velocity of the boat would not be constant during the trip across the river; the boat would have toaccelerate from rest at one side of the river and then decelerate as it approached the opposite shore, making the traveltime significantly longer than 4.5 s.


Strategy: Let yw =vr

your velocity with respect to the walkway, wg =vr

walkways

velocity with respect to the ground, and add the vectors according to equation 3-8 tofind yg =v

ryour velocity with respect to the ground. Then find the time it takes you

to travel the 85-m distance.

Solution:1. Find your velocitywith respect to the walkway:

( )yw85 m

1.25 m/s68 s

x

t

= = =

v x x xr

2. Apply equation 3-8 to find yourvelocity with respect to the ground: ( ) ( ) ( )yg yw wg

1.25 m/s 2.2 m/s 3.45 m/s= + = + =v v v x x xr r r

3. Now find the time of travel:yg

85 m25 s

3.45 m/s

xt

v

= = =

Insight: The moving walkway slashed your time of travel from 68 s to 25 s, a factor of 2.7! Note that we bent thesignificant figures rules a little bit by not rounding ywv

r to 1.3 m/s. This helped us avoid rounding error.


Strategy: Let yw =vr

your velocity with respect to the walkway, wg =vr

walkways

velocity with respect to the ground, and add the vectors according to equation 3-8 tofind yg =v

r

your velocity with respect to the ground. Then find the time it takes you

to travel the 85-m distance.

Solution:1. Find your velocity with respectto the walkway:

( )yw 85 m 1.3 m/s68 sxt

= = =

v x x xr

2. Apply equation 3-8 to find your velocitywith respect to the ground: ( ) ( ) ( )yg yw wg

1.3 m/s 2.2 m/s 0.9 m/s= + = + =v v v x x xr r r

3. Now find the time of travel:yg

85 m90 s

0.9 m/sx

tv

= = =

Insight: Going the wrong way on the moving walkway increases your time of travel from 68 s to about 90 s.

ywvr

wgvr

ygvr

wgv

r

ygvr

ywv

r

7/28/2019 Ch 03 Vectors

18/27



74

45. Picture the Problem: The vectors involved in this problem are depicted atright.

Strategy: Let pg =vr

velocity of the plane relative to the ground, pa =vr

velocity of the plane relative to the air, and ag=

v

r

velocity of the airrelative to the ground. The drawing at right depicts the vectors addedaccording to equation 3-8, pg pa ag= +v v v

r r r

. Determine the angle of the

triangle from the inverse sine function.

Solution:1. (a) Use the inverse sine function to find : ag1 1

pa

65 km/hsin sin 11 west of north

340 km/h

v

v

= = =

2.(b) The drawing above depicts the vectors.

3. (c) If the plane reduces its speed but the wind velocity remains the same, the angle found in part (a) should beincreased in order for the plane to continue flying due north.

Insight: If the planes speed were to be reduced to 240 km/h, the required angle would become 16.


Strategy: Let pf =vr

the passengers velocity relative to the ferry, pw =vr

the passengers

velocity relative to the water, and fw =vr

the ferrys velocity relative to the water. Apply

equation 3-8 and solve for fwvr

. Once the components of fwvr

are known, its magnitudeand direction can be determined.

Solution:1. Solve equation 3-8 for fwvr

: pw pf fw

fw pw pf

= +

=

v v v

v v v

r r r

r r r

2. Determine the components of fwvr

: ( ) ( ) ( )

( ) ( )

fw

fw

4.50 m/s sin 30 4.50 m/s cos30 1.50 m/s

2.25 m/s 2.40 m/s

= +

= +

v x y y

v x y

r

r

3. Find the direction of fwvr

: fw,1 1

fw,

2.25 m/stan tan 43 west of north

2.40 m/sx

y

v

v

= = =

4. Find the magnitude of fwvr

: ( ) ( )2 22 2

fw fw, fw, 2.25 m/s 2.40 m/s 3.29 m/sx yv v v= + = + =

Insight: If the person were to walk even faster with respect to the ferry, then fwvr

would have to be shorter and pointmore in the westerly direction.

47. Picture the Problem: The situation is similar to that depicted in the figure at right,except the boat is supposed to be a jet ski.

Strategy: Place thex-axis perpendicular to the flow of the river, such that the river isflowing in the negativey-direction. Let bw =v

rjet skis velocity relative to the water,

bg =vr

jet skis velocity relative to the ground, and wg =vr

waters velocity relative to

the ground. Use equation 3-8 to find the vector bwvr

, and then determine itsmagnitude.

Solution:1. Solve eq. 3-8 for bwvr

: bg bw wg bw bg wg= + = v v v v v vr r r r r r

pwvr

fwvr

pfvr

30

E

N

7/28/2019 Ch 03 Vectors

19/27



75

2. Find the components of bgvr

: ( ) ( )

( ) ( )

bg 9.5 m/s cos 20.0 9.5 m/s sin 20.0

8.9 m/s 3.2 m/s

= +

= +

v x y

x y

r

3. Subtract to find bwvr

: ( ) ( ) ( )

( ) ( )

bw bg wg 8.9 m/s 3.2 m/s 2.8 m/s

8.9 m/s 6.0 m/s

= = +

= +

v v v x y y

x y

r r r

4. Find the magnitude of bwvr

: ( ) ( )2 2

bw 8.9 m/s 6.0 m/s 11 m/sv = + =

Insight: Note that the 35 angle is extraneous information for this problem. If we work backwards to find the angle fromthe components of bwv

r

we get ( )1tan 6.0 8.9 34 = = , not exactly 35 due to rounding errors.

48. Picture the Problem: The situation is depicted in the figure at right, except the boatis supposed to be a jet ski.

Strategy: Place thex-axis perpendicular to the flow of the river, such that theriver is flowing in the negativey-direction. Let bw =v

r

jet skis velocity

relative to the water, bg =vr

jet skis velocity relative to the ground, and

wg =vr waters velocity relative to the ground. Set they component of bwv

r

equal to the magnitude of wgvr

so that they cancel, leaving only anx

component of bgvr

. Then determine the angle .

Solution:1. (a) Set bw, wg, 0y yv v+ = and solve for :

( )

bw, bw wg,

wg,1 1

bw

sin

2.8 m/ssin sin 13

12 m/s

y y

y

v v v

v

v

= =

= = =

2.(b) Increasing the jet skis speed relative to the water will increase bwv and therefore decrease the angle .

Insight: Airplanes must also make heading adjustments like the jet skis in order to fly in a certain direction when there

is a steady wind present.

49. Picture the Problem: The vectors for Jet Ski A and B are depictedat right. Note that bwv

r

has the same magnitude for each jet ski but if

you inspect the diagram for Jet Ski B you can see that bw bg,Bv v> .

Strategy: Usethex components of the velocities of each jet skirelative to the ground to determine the time required for each jet skito cross the river.

Solution:1. (a) The time required for each jet ski to cross the riverequals the width of the river divided by thex component of the jetskis velocity relative to the ground. From the diagrams you can see

that thex component of bg,Avr

equals bwvr

, but that thex componentof bg,Bv

r

is shorter than bwvr

. Therefore Jet Ski A has a higher velocity in thex direction relative to the ground, and will

cross the river first.

2.(b) Find the ratio of the times:

river

bg, A, bg, B, bw

river bg, A, bw

bg, B,

cos350.82x xA

B x

x

x

v v vt

xt v v

v

= = = =

Insight: The ratio is less than one, soA Bt t < and Jet Ski A reaches the opposite shore first. Remember this the next

time you race jet skis across a flowing river!

bg,Avr wg

vr

bwvr

bg,Bvr

wgvrbwv

r

Jet Ski A

Jet Ski B35

7/28/2019 Ch 03 Vectors

20/27



76

50. Picture the Problem: The ramp is depicted at right.

Strategy: Use the inverse sine function to find the angle using thepertinent sides of the triangle.

Solution:1. Write the definition of the sine function:

height

sin length

y

r= =

2. Use the inverse sine function to find the angle: 1 3.00 ftsin 17.510.0 ft

= =

Insight: Finding a right triangle in any physics problem allows you to use the arsenal of trigonometric tools to findvarious other quantities of interest. Learn to find them!

51. Picture the Problem: The vector components of Ar

and Br

are specified in the problem. Measure positive angles to becounterclockwise from the positivex axis.

Strategy: Multiply each component of Ar

by 2 and add them to Br

. Use the resulting components to determine thedirection and magnitude of the sum.

Solution:1. Multiply and add the components: ( ) ( ) ( ) ( ) 2 2 12.1 m 32.2 m 24.2 m 32.2 m+ = + = + A B x y x yr r

2. Find the angle: 1 32.2 mtan 53.1 or 30724.2 m

= =

3. Find the magnitude of the vector sum: ( ) ( )2 2

2 24.2 m 32.2 m 40.3 m+ = + =A Br r

Insight: Once the components of the vectors are known it is a fairly straightforward procedure to determine the scalarproduct, sum, magnitude, and direction.

52. Picture the Problem: The vector components of A Br r

, Cr

, and + +A B Cr rr

are specified in the problem.

Strategy: Use the given vector components to write three equations and solve them for Ar

and Br

.

Solution: 1. Add the three given equations to solve for Ar

: ( )

( )

( )

( ) ( )

13.8 m

51.4 m

62.2 m

Add: 2 99.8 m 49.9 m

+ + =

=

=

= =

A B C x

A B x

C x

A x A x

r rr

r r

r

r r

2. Now substitute the known vector Ar

into the secondequation:

( ) ( )( )

49.9 m 51.4 m1.5 m

=

=

x B x

B x

r

r

Insight: None of the vectors have anyy component. If they did, the problem would be a bit more difficult but stillsolvable as long as the number of unknowns is less than or equal to the number of equations.

3.00 ft

10.0 ft

7/28/2019 Ch 03 Vectors

21/27



77

53. Picture the Problem: The vectors involved in this problem are illustratedat right.

Strategy: Thex component of the balls velocity with respect to the trainmust be equal and opposite to the trains velocity in order for Gary to seethe ball rise straight upward. That fact, together with the angle of the

throw, can be used to find the speed btv of Michelles throw as well as thespeed bgv of the ball according to Gary.

Solution:1. (a) Michelle must have thrown the ball toward the rear of thetrain so that tgv

r

could cancel out thex component of btvr

and leave bgvr

completely vertical.

2.(b) Set the magnitudes ofthex components of btv

r

and tgvr

and solve for btv :

( )

( ) ( )

bt, bt tg,

tg,bt

cos 55.0

8.20 m/s14.3 m/s

cos 55.0 cos 55.0

x x

x

v v v

vv

= =

= = =

3. (c) The magnitude of bgv equals they component of btvr

: ( ) ( )bg 14.3 m/s sin 55.0 11.7 m/sv = =

Insight: Gary and Michelle disagree on the path taken by the ball, but each agree on the acceleration and time of flight.Well learn more about relative motion in Chapter 29.

54. Picture the Problem: The displacement vectors of the Hummer are depicted at right.

Strategy: Find the displacement vectors Ar

and Br

using the given speed and time

information. Use those vector components to find the final displacement vector Cr

.

Divide the displacement vector Cr

by the time in order to find the direction and speedof travel on the final leg. Let north point in the y direction and east in the x direction.

Solution:1. Find the vector Ar

: ( ) ( )

( )( )

( ) ( )

( ) ( )

sin 25 cos 25

6.5 km/h 15 min 1 h/60 min

sin 25 cos 25

0.69 km 1.5 km

A A A Av t v t = +

=

+

= +

A x y

x y

A x y

r

r

2. Find the vector Br

: ( )( ) ( ) 12 km/h 7.5 min 1 h/60 min 1.5 kmB Bv t= = =B x x xr

3. Find the vector Cr

: ( ) ( ) ( )

( ) ( )

( ) ( )

0.69 1.5 km 1.5 0 km

0.81 km 1.5 km

x x y yA B A B= + = + +

= + +

= +

C A B x y

x y

C x y

r r r

r

4. Find the direction angle : 1 1 1.5 kmtan tan 62 south of west0.81 km

y

x

C

C

= = =

5. Find the speed of travel:( ) ( )

2 20.81 km 1.5 km

4.6 km/h22 min 1 h/60 minC

vt

+= = =

Cr

Insight: Once the components of the vectors are known it is a fairly straightforward procedure to determine the scalarproduct, sum, magnitude, and direction.

x

y

Br

Ar

O

C

r25

7/28/2019 Ch 03 Vectors

22/27



78

55. Picture the Problem: The three-dimensional vector is depicted in the diagram atright.

Strategy: Determine thez component of Ar

by applying the cosine function to the

right triangle formed in thez direction. Then find the projection of Ar

onto thexy

plane (A sin 55) in order to find thex andy components of A

r

.Solution:1. Find thez component of A

r

: ( )65 m cos55 37 mzA = =

2. Find the projection onto thexy plane: ( )sin 55 65 m sin 55xyA A= =

3. Find thex component of Ar

: ( )65 m sin 55 cos 35 44 mxA = =

4. Find they component of Ar

: ( )65 m sin 55 sin 35 31 myA = =

Insight: A knowledge of right triangles can help you find the components of even a three dimensional vector. Once thecomponents are known, then addition and subtraction of vectors become straightforward procedures.

56. Picture the Problem: The football maintains its horizontal velocity but

increases its vertical velocity in the downward direction.Strategy: Find the vertical velocity

yv of the football after 1.75 s assuming

an initialyv of zero. The football maintains its horizontal velocity 0v , so the

two velocities form thex andy components of the balls velocity at 1.75 s.Use the components to find the magnitude and direction of the velocity.

Solution:1. (a) Use equation 3-6 tofind the components of

fvr

:( )

( ) ( )

0 2

m m 16.6 9.81 1.75 s

s s 16.6 m/s 17.2 m/s

f t

= + = +

=

v v a x y

x y

rr r

2.(b) Find the magnitude offv

r

: ( ) ( )2 2

16.6 m/s 17.2 m/s 23.9 m/sv = + =

3. Find the direction of fvr

: 1 1 17.2 m/stan tan 46.0 or 46.0 below horizontal16.6 m/s

y

x

vv

= = =

Insight: The motion of the football will be discussed in more detail in Chapter 4 when we consider projectile motion.

57. Picture the Problem: The path of the football traces out a parabola as the velocity increases at a constant rate in thedownward direction, but the velocity in the horizontal direction remains constant.

Strategy: Use equation 3-6 to find the average acceleration over a time interval t . Then substitute the various timeintervals into the formula to find ava

r

.

Solution:1. (a) Find the averageacceleration as a function of the timeinterval t . ( ) ( ){ } ( ) ( ){ }

( )( )

av

2 2

2

2av

16.6 m/s 9.81 m/s ( ) 16.6 m/s 9.81 m/s

9.81 m/s9.81 m/s

t t t

t t

t t t

t

t

t

+

= =

+ =

= =

v vva

x y x y

ya y

r rr

r

r

2. Determine avar

for t = 1.00 s: ( )2av 9.81 m/s= a yr

3. (b) Determine avar

for t = 2.50 s: ( )2av 9.81 m/s= a yr

4.(c) Determine avar

for t = 5.00 s: ( )2av 9.81 m/s= a yr

Insight: The acceleration due to gravity is constant, so the average acceleration is exactly the same for any time interval.

7/28/2019 Ch 03 Vectors

23/27



79

58. Picture the Problem: The vectors involved in this problem are illustrated at right.

Strategy: Let 1g =vr

velocity of plane 1 relative to the ground, 2g =vr

velocity of plane 2

relative to the ground, 12 =vr

velocity of plane 1 relative to plane 2, and 21 =vr

velocity ofplane 2 relative to plane 1 (not pictured). Let north be along the positivey-axis, east along

the positivex axis. Use equation 3-8 to find the components of 12vr

. Use the componentsto find the magnitude and direction.

Solution:1. (a) Find 12vr

byapplying equation 3-8:

1g 12 2g 12 1g 2g= + = v v v v v vr r r r r r

2. Use the given angles to find thecomponents of 12v

r

:( )

( ) ( )

( ) ( )

12 12 m/s

7.5 m/s cos 20 7.5 m/s sin 20

7.0 m/s 9.4 m/s

=

+

= +

v y

x y

x y

r

3. Find the direction of 12vr

: 12,1 1

12,

9.4 m/stan tan 53 north of east

7.0 m/sy

x

v

v

= = =

4. Find the magnitude of 12vr

: ( ) ( )2 2

12 7.0 m/s 9.4 m/s 12 m/sv = + =

5. (b) Since 21 12= v vr r

the vector 21vr

has the same magnitude as 12vr

but points in the opposite direction. Therefore,

21 12 m/s at 53 south of west.= vr

Insight: There are other ways to approach this problem. For instance, in step 1 we could say 12 1g g2= +v v vr r r

and then

use the fact that g2 2g= v vr r

to write 12 1g 2g= v v vr r r

. Its a little awkward to use g2vr

, which represents the velocity of

the ground relative to plane 2, but learning to think about velocity from both perspectives can help you solve difficultvector motion problems.

59. Picture the Problem: The vectors involved in this problem are illustrated at right.

Strategy: To use the graphical method you must make a scale drawing of thevectors and then measure the vector sum with a ruler. To use the component methodyou must independently add thex andy components of each vector.

Solution:1. (a) Using the scale drawing aboveyou can measure the length of the vector sum:

38 ft rr

2.(b) Independently add thex andycomponents of the vector sum: ( )

( )

( ) ( )

0 45 ft 35 ft 051ft 0 0 13 ft

10 ft 38 ft

= + + +

= + +

+ + +

= +

r A B C D

x

y

r x y

r rr rr

r

3. Find the magnitude of the sum: ( ) ( )2 2

10 ft 38 ft 39 ft = + =rr

4. Find the direction of the vector sum: 1 1 10 fttan tan 15 clockwise from38 ft

x

y

r

r

= = = Ar

Insight: When adding vectors graphically you must always ensure you are adding them head-to-tail. The vector sum isa vector that starts at the beginning of the first vector ( A

r

) and ends at the end of the last vector ( Dr

).

= + + +r A B C Dr rr rr

Dr

Br

Cr

Ar

12vr

1gvr

x

y

2gvr

O

20

7/28/2019 Ch 03 Vectors

24/27



80


Strategy: Use the component method of vector subtraction to find vr

. The averageacceleration is then v

r

divided by the time elapsed.

Solution:1. Find the

components of ivr

:

( ) ( )

( ) ( )

i 4.40 m/s cos 37.5 4.40 m/s sin 37.5

3.49 m/s 2.68 m/s

= +

= +

v x y

x y

r

2. Find the components of fvr

: ( ) ( )

( ) ( )

f 6.25 m/s cos 56.0 6.25 m/s sin 56.0

3.49 m/s 5.18 m/s

=

= +

v x y

x y

r

3. Subtract the vectors: ( ) ( )

( )

3.49 3.49 m/s 5.18 2.68 m/s

7.86 m/s

f i = =

=

v v v x y

y

r r r

4. Divide by the time elapsed:( )

( )av7.86 m/s

3.93 m/s2.00 st

= = =

yva y

r

r

Insight: Subtracting vectors A Br r

is the same as adding Br

to Ar

. Note that acceleration can change both themagnitude and the direction of the velocity vectors.


Strategy: Let bg =vr

velocity of the bus relative to the ground, rg =vr

velocity of the raindrops

relative to the ground, and rb =vr

velocity of the raindrops relative to the bus. Apply equation3-8 to form a right triangle of the velocity vectors as shown in the diagram. Use the righttriangle to find the ratio rg bgv v and the value of rgv .

Solution:1. (a) Write out equation 3-8:rg rb bg= +v v v

r r r

2. The vectors form a right triangle because the rain fallsvertically and the bus travels horizontally. Use thetriangle indicated in the diagram to find the ratio:

rg rb

bg rb

cos 1 1 3.7sin tan tan15

v vv v

= = = =

3. (b) Use the ratio to find rgv : ( )rg bg3.7 3.7 18 m/s 67 m/sv v= = =

Insight: The rain speed is a bit unrealistic; typical values for large raindrops are 10 m/s or about 20 mi/h.

62. Picture the Problem: The hands of the clock are depicted at right.

Strategy: Find the angles and that correspond to the time of 4:12 andthen use the angles to find the components of the position vectors for thetips of the hands. Then subtract the position vectors to find the tip-to-tipdistance.

Solution:1. Find the angle : 12 min 360 7260 min

= =

2. Find the angle :12604 360 126 90 36

12= = =

3. Find the components of Mr

:

( ) ( ) ( ) ( )

sin cos

14 ft sin 72 14 ft cos72 13 ft 4.3 ft

M M = +

= + = +

M x y

x y x y

r

= r M Hr rr

Mr

H

r

x

y

fvr

ivr

ivr

37.5

56.0

vr

bgvr

rbvr

rgvr

7/28/2019 Ch 03 Vectors

25/27



81

4. Find the components of Hr

:

( ) ( ) ( ) ( )

cos sin

9.0 ft cos36 9.0 ft sin 36 7.3 ft 5.3 ft

H H =

= = +

H x y

x y x y

r

5. Find the components of rr

: ( ) ( )( )

( ) ( )

13 7.3 ft 4.3 5.3 ft

6 ft 9.6 ft

= = +

= +

r M H x y

x y

r rr

6. Find the magnitude of rr

: ( ) ( )2 2

6 ft 9.6 ft 11 ftr = + =

Insight: The tip-to-tip distance changes from 5.0 ft (when the hands are aligned) to 23 ft (when the hands are oppositeeach other) during the course of a day.

63. Picture the Problem: The velocities of the surfer ssvr

and the waves wsvr

relative to the shore are shown in the diagram at right.

Strategy: Set they component of the surfers velocity equal to the velocityof the waves, and solve for the angle . Then apply equation 3-8 to find thesurfers velocity relative to the wave.

Solution:1. (a) Set ss, wsyv v= : ss ws

1 1ws

ss

sin1.3 m/s

sin sin 107.2 m/s

v v

v

v

=

= = =

2.(b) Apply equation 3-8: [ ]ss sw ws sw ss ws ss ss ws cos sinv v v = + = = + v v v v v v x y yr r r r r r

3. Since ss wssinv v= , they components cancel out:

( ) ( )sw ss cos 7.2 m/s cos10 7.1 m/sv = = =v x x xr

4.(c) If they component stays the same, but the vector increases in length, the angle it makes with thex-axis mustdecrease.

Insight: In this problem we assumed that the water is at rest relative to the shore, so that the surfers speed relative tothe water is the same as the surfers speed relative to the shore.

64. Picture the Problem: The diagram for Example 3-2 is shown at right. Thewidth of the river along thex direction is 25.0 m.

Strategy: Let bw =vr

boats velocity relative to the water, bg =vr

boats

velocity relative to the ground, and wg =vr

waters velocity relative to the

ground. Set they component of bwvr


so that

they cancel, leaving only anx component of bgvr



( )

bw, bw wg,

wg,1 1

bw

sin

1.4 m/ssin sin 136.1 m/s

y y

y

v v v

vv

= =

= = =

2.(b) Since they components cancel, it follows thatthex component of bwv

r

is the same as bgvr

: ( )bg bw cos 6.1 m/s cos13 5.9 m/sv v = = =

3. Now find the time to cross the river:bg

25.0 m4.2 s

5.9 m/s

xt

v

= = =

4.(c) If the speed of the boat is increased, it should make its heading more downstream in order that itsy componentremains 1.4 m/s and that it still lands directly across the river from its starting point.

Insight: Airplanes must also make heading adjustments like the jet skis in order to fly in a certain direction when thereis a steady wind present.

shore

ssvr

wsvr

x

y

7/28/2019 Ch 03 Vectors

26/27



82


Strategy: Because 0,+ + =A B Cr rr

, thex andy components of the vectors must

independently sum to zero. Since Ar

has noy component, we can use the known

y component of C

r

to find they component of B

r

. The known angle that B

r

makes with thex axis will yieldBx and give us a way to find the length of Ar

.

Solution:1. Set they componentsequal to zero:

( )

( )

0 sin 40.0 0

15 m sin 40.0 9.64 m

y y y y y y

y

A B C B C A C

B

+ + = = =

= =

2. DetermineBx using the tangentfunction:

9.64 mtan30.0 16.7 m

tan 30.0 tan 30.0y y

x

x

B BB

B = = = =

3. FindAx and then Ar

: ( ) ( )0 16.7 m 15 m cos 40.0 28 m

28 m

x x x x x xA B C A B C+ + = = = =

=Ar


: ( ) ( )2 22 2 16.7 m 9.64 m 19 mx yB B B= + = + =

Insight: We kept an extra significant figure when calculating the components of Br

in order to avoid rounding error.


Strategy: Let 1g =vr

velocity of boat 1 relative to the ground, 2g =vr

velocity of boat 2

relative to the ground, 12 =vr

velocity of boat 1 relative to boat 2. Let north be along thepositivey-axis, east along the positivex axis. Use equation 3-8 to find the components of

2gvr

, and use the components to find its magnitude and direction.

Solution:1. Apply equation

3-8 to find 2gvr :( )

( )

( )

( ) ( )

1g 12 2g 2g 1g 12

2g

2g

2.30 m/s sin 40.00.750 m/s

2.30 m/s cos 40.0

1.48 m/s 1.01 m/s

= + =

=

+

= +

v v v v v v

xv y

y

v x y

r r r r r r

r

r

2. Find the magnitude of 2gvr

: ( ) ( )2 2

2g 1.48 m/s 1.01 m/s 1.79 m/sv = + =

3. Find the direction of 2gvr

: 1 1.01 m/stan 34.3 south of west1.48 m/s

= =

Insight: If you pay attention carefully to the subscripts, equation 3-8 can help you solve complex relative motionproblems like this one. It is also helpful to draw a diagram as well. Then you can visually check your answer.

67. Picture the Problem: The diagram for Example 3-2 is shown at right.

Strategy: Let bw =vr

boats velocity relative to the water, bg =vr

boats

velocity relative to the ground, and wg =vr

waters velocity relative to the

ground. Set they component of bwvr


so that

they cancel, leaving only anx component of bgvr


12vr

1gvr

x

y

2gvr O

40.0

Cr

x

y

Br

Ar O

30.0

40.0

7/28/2019 Ch 03 Vectors

27/27



83


( )

bw, bw wg,

wg,1 1

bw

sin

1.4 m/ssin sin 12 upstream

7.0 m/s

y y

y

v v v

v

v

= =

= = =

2.(b) If the speed of the boat is increased, it should make its heading more downstream in order that itsy component

remains 1.4 m/s and that it still lands directly across the river from its starting point. Therefore the angle needed to godirectly across the river will decrease.

Insight: Airplanes must also make heading adjustments like the jet skis in order to fly in a certain direction when thereis a steady wind present.

68. Picture the Problem: The diagram for Example 3-2 is shown at right.

Strategy: Find the angle that bgvr

makes with the positivex axis by using

the coordinates of the dock (the displacement and velocity vectors areparallel for uniform motion). Apply equation 3-8 to determine therelationship between bgv

r

and bwvr

. Solve the two equations with two

unknowns to find the angle . Then the magnitude of bgvr

can be found

from its known components.Solution:1. (a) Because the displacementand velocity vectors are parallel, the ratiosof their components are equal:

bg,

bg,

28 m55 m

y y

x x

v r

v r

= =

2. Write equation 3-8 in component form:

[ ]

bg bw wg

bg, bg, bw bw wg cos sinx yv v v v v

= +

+ = +

v v v

x y x y y

r r r

3. Independently equate thex andycomponents:

bg, bw

bg, bw wg

cos

sinx

y

v v

v v v

=

=

4. Divide they equation by thex equation: bg, bw wg

bg, bw

sin

cosy

x

v v v

v v

=

5. Square both sides to get everything interms of sin . The units of m/s apply toeach of the velocities, but are omitted atright in order to save space:

( )

( ) ( ) ( ) ( )

( ) ( )

2 2 2 2 2 2 2bg, bw wg bw wg bw wg bw wg

2 2 2 2bg, bw bw

22 22

2 2 2

sin 2 sin sin 2 sin

cos 1 sin

6.7 sin 2 1.4 6.7 sin 1.428

55 6.7 6.7 sin

y

x

v v v v v v v v v

v v v

+ += =

+ =

6. Rearrange the equation into one that isquadratic in sin :

2 2

2

11.6 11.6sin 44.9sin 18.8sin 1.96

0 56.5sin 18.8sin 9.64

= +

=

7. Apply the quadratic formula.( ) ( )( )

( )( )

2

1

18.8 18.8 4 56.5 9.64sin 0.279, 0.612

2 56.5sin 0.612 38 upstream

= =

= =

8. (b) Now find the components of bgvr

: ( )

( )

bg, bw

bg, bw wg

cos 6.7 m/s cos38 5.3 m/s

sin 6.7 m/s sin 38 1.4 m/s 2.7 m/s

x

y

v v

v v v

= = =

= = =

9. Use the components to find bgv : ( ) ( )2 22 2

bg bg, bg, 5.3 m/s 2.7 m/s 5.9 m/sx yv v v= + = + =

Insight: While the boat is pointed 38 upstream, it is actually traveling ( )1tan 2.7 5.3 27 = upstream of the positivex

axis due to the water flow. Notice that we get the same angle for the displacement vector ( )1tan 28 m 55 m 27 = .

ch 03 vectors

Documents