Physics Chapter 3 Answers

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3 – 1

Chapter 3: Vectors in Physics

Answers to Even-Numbered Conceptual Questions 2. Vectors

A ,

G , and

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

I is the same as vector

L .

4. No. The component and the magnitude can be equal if the vector has only a single component. If the vector has 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 the component.

8. The vectors

A and

B must point in the same direction.

10. The vector

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. Three vectors 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 sail downwind, however, you move with the wind, and its speed relative to you is decreased.

Solutions to Problems and Conceptual Exercises

1. Picture the Problem: Each component of a vector is doubled in magnitude.

Strategy: Note the relationship between the components of a vector and its magnitude and direction to answer the conceptual question.

Solution: 1. (a) Doubling each of the components of a vector will double its magnitude, or increase its magnitude by a multiplicative factor of 2. You can picture this in your head or confirm it mathematically with a calculation like:

( ) ( ) ( )222 2 2 2 2 2 2 2 4 2 2x y x y x y x yA A A A A A A A A A= + ⇒ + = + = + =

2. (b) Doubling each of the components of a vector will not change its direction at all; the direction changes by a multiplicative factor of 1. You can picture this in your head or confirm it mathematically with a calculation like:

( ) ( ) ( )1 1 1tan tan 2 2 tany x y x y xA A A A A Aθ θ− − −= ⇒ = =

Insight: You can change a vector’s direction only by changing the relative magnitudes of its components. In this exercise each component was changed by the same multiplicative factor, so the relative magnitudes were unchanged.

2. Picture the Problem: Compare the magnitudes of the vectors depicted in the figure.

Strategy: Concentrate on the lengths of the vectors as drawn and ignore their direction.

Solution: By comparing the lengths of the vectors as drawn we can arrive at the ranking: B < C < A < D

Insight: Note that the symbol B refers to the magnitude of the vector and B refers to both magnitude and direction.

Chapter 3: Vectors in Physics James S. Walker, Physics, 4th Edition


3 – 2

3. Picture the Problem: Compare the magnitudes of the x components of the vectors depicted in the figure.

Strategy: Concentrate on the value of the x component of each vector. A vector that is oriented vertically in the diagram has an x component of zero, whereas horizontal vectors have x components with large magnitudes, either positive (to the right) or negative (to the left).

Solution: Note that the x component of D is large and negative. The value of its x component is therefore the smallest even though the magnitude of its x component is the largest. By comparing the values of the x components of the vectors as drawn we can arrive at the ranking: x x x xD C B A< < <

Insight: Note that the symbol xB refers to the value of the x component of vector .B 4. Picture the Problem: Compare the magnitudes of the y components of the vectors depicted in the figure.

Strategy: Concentrate on the value of the y component of each vector. A vector that is oriented horizontally in the diagram has a y component of zero, whereas vertical vectors have y components with large magnitudes, either positive (upward) or negative (downward).

Solution: Note that the y component of D is small and negative. The value of its y component is therefore even smaller than the zero y component of .B By comparing the values of the y components of the vectors as drawn we can

arrive at the ranking: y y y yD B A C< < <

Insight: Note that the symbol yB refers to the value of the y component of vector .B

5. Picture the Problem: The press box is 32.0 ft above second base

and an unknown horizontal distance away.

Strategy: Use the tangent function to determine the horizontal distance.

Solution: Use the tangent function to find x:

38.0 ft 119 fttan tan15.0

yxθ

= = =°

Insight: Dividing distances into right triangles in this manner is an important strategy for solving physics problems. 6. Picture the Problem: You drive 1.2 miles along an inclined

roadway, gaining 530 ft of altitude.

Strategy: Use the sine function to determine the angle and then the additional distance x along the hypotenuse.

Solution: 1. (a) Apply the sine function:

530 ftsin 1.2 mi 5280 ft/mi

ys

θ = =×

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

θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠

3. (b) Use the known angle together with the sine function to find x:

150 ft 1800 ft 0.34 misin sin 4.8

yxθ

Δ= = = =

°

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

15.0° 32.0 ft

x

θ 530 ft

1.2 mi 150 ftx



3 – 3

7. Picture the Problem: The road gains 6 ft in altitude for every 100 ft it

spans in the horizontal direction.

Strategy: Use the tangent function to determine the angle.

Solution: Apply the tangent function: 1 6 ftsin tan 3

100 ftyx

θ θ − ⎛ ⎞= ⇒ = = °⎜ ⎟⎝ ⎠


8. Picture the Problem: The vector direction is

measured counterclockwise from the +x axis.

Strategy: In each case find the vector components.

Solution: 1. (a) Find the x and y components: ( )( )

cos 75 m cos35.0 61 m

sin 75 m sin 35.0 43 mx

y

r r

r r

θ

θ

= = ° =

= = ° =

2. (b) Find the x and y components: ( )( )

cos 75 m cos 65.0 32 m

sin 75 m sin 65.0 68 mx

y

r r

r r

θ

θ

= = ° =

= = ° =

Insight: Resolving vectors into their components is an important skill for solving physics problems. 9. Picture the Problem: The base runner travels from C (home plate)

to first base, then to A (second base), then to B (third base), and finally back to C (home plate).

Strategy: The displacement vector Δr is the same as the position vector r if we take home plate to be the origin of our coordinate system (as it is drawn). The displacement vector for a runner who has just hit a double is drawn.

Solution: 1. (a) Write the displacement vector from C to A in terms of its x and y components:

( ) ( )ˆ ˆ90 ft 90 ft= +r x y

2. (b) Write the displacement vector from C to B in terms of its x and y components:

( ) ( ) ( )ˆ ˆ ˆ0 ft 90 ft 90 ft= + =r x y y

3. (c) For a home run the displacement is zero: ( ) ( )ˆ ˆ0 ft 0 ft= +r x y

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

x

y

35.0°

r

x

y

65.0°

r

(a) (b)

r

θ 6 ft

100 ft



3 – 4

10. Picture the Problem: The ship approaches the rocks as

depicted in the picture.

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

Solution: 1. Use the tangent function to find the distance L:

( ) ( )49 14 ft 35 fttan 30 61 ft

tan 30L

L−

° = ⇒ = =°

2. Subtract 19 ft from L 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 when solving physics problems.

11. Picture the Problem: The water molecule forms a triangle with the positions of the

oxygen and hydrogen nuclei as shown.

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

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

sin0.96 Å

dθ =

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

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

12. Picture the Problem: The given vector components correspond to the vector r as drawn at

right.

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

Solution: 1. (a) Use the inverse tangent function to find the distance angle θ :

1 9.5tan 3414

θ − −⎛ ⎞= = − °⎜ ⎟⎝ ⎠

or 34° below

the +x axis

2. (b) Use the Pythagorean Theorem to determine the magnitude of 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 magnitude will double: ( ) ( )

1

2 2

9.5 2tan 3414 2

28 m 19 m 34 mr

θ − − ×⎛ ⎞= = − °⎜ ⎟×⎝ ⎠

= + − =

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

x

y

r

14 m

−9.5 m



3 – 5

13. Picture the Problem: The given vector components correspond to the vector r as drawn at

right.

Strategy: Determine the angle θ from our knowledge of analog clocks. The given component xr together with the angle θ will allow us to calculate the length of r and the component yr .

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

θ = × ° = °

2. Find the length of r: 3.0 cmcos 3.5 cmcos cos30

xx

rr r rθ

θ= ⇒ = = =

3. (b) The components xr and yr are only equal when θ =45°. Since in this case θ =30°, the component yr will be less than xr or 3.0 cm.

4. (c) Find yr : ( )sin 3.5 cm sin 30 1.7 cmyr r θ= = =


14. Picture the Problem: The trip takes you toward the east first and then

toward 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 of r : ( ) ( )2 2680 ft 340 ft 760 ftr = + =

2. (b) I estimate an angle of close to 30° based on the sketch above.

3. (c) Use the inverse tangent function to find θ : 1 340 fttan 27 north of east

680 ftθ − ⎛ ⎞= = °⎜ ⎟

⎝ ⎠


15. Picture the Problem: The two vectors A (length 50 units) and B (length 120 units) are drawn

at right.

Strategy: Resolve B into its x and y components to answer the questions.

Solution: 1. (a) Find Bx: ( )120 units cos 70 41 unitsxB = ° =

2. Since the vector A points entirely in the x direction, we can see that Ax = 50 units and that vector A has the greater x component.

3. (b) Find By: ( )120 units sin 70 113 unitsxB = ° =

4. The vector A has no y component, so it is clear that vector B has the greater y component.


x

y

B

A70°

x

y

rθ

II

III

I XII

3.0 cm

θ 340 ft

680 ft

r



3 – 6

16. Picture the Problem: The four possible locations of the treasure are

labeled A, B, C, and D 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 direction of each vector.

Solution: 1. Find the magnitude of A : ( ) ( )2 222.0 m 15.0 5.00 m 29.7 mA = − + + =

2. Find the direction (from north) of A : 1 22.0 mtan 47.7 west of north

15.0 5.00 mAθ − ⎛ ⎞= = °⎜ ⎟+⎝ ⎠

3. Find the magnitude of B : ( ) ( )2 222.0 m 15.0 5.00 m 24.2 mB = − + − =

4. Find the direction (from north) of B : 1 22.0 mtan 65.6 west of north

15.0 5.00 mBθ − ⎛ ⎞= = °⎜ ⎟−⎝ ⎠

5. Find the magnitude of C : ( ) ( )2 222.0 m 15.0 5.00 m 29.7 mC = + + =

6. Find the direction (from north) of C : 1 22.0 mtan 47.7 east of north

15.0 5.00 mCθ − ⎛ ⎞= = °⎜ ⎟+⎝ ⎠

7. Find the magnitude of D : ( ) ( )2 222.0 m 15.0 5.00 m 24.2 mD = + − =

8. Find the direction (from north) of D : 1 22.0 mtan 65.6 east of north

15.0 5.00 mDθ − ⎛ ⎞= = °⎜ ⎟−⎝ ⎠

Insight: If you ever find a treasure map like this one, you’ll be glad you mastered vectors in physics! 17. Picture the Problem: The whale dives along a straight line tilted 20.0°

below horizontal for 150 m as shown in the figure.

Strategy: Resolve the whale’s displacement vector into horizontal and vertical components in order to find its depth ry and its horizontal travel distance rx.

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

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

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

18. Picture the Problem: Consider the two vectors and A B depicted in the figure.

Strategy: Remember the rules of adding and subtracting vectors. Vectors are always added head-to-tail. To subtract vectors, reverse the direction of the negative vector and add it head-to-tail to the positive vector.

Solution: 1. (a) To add +A B we must move B so that its tail is on the head of .A The resultant vector starts at the tail of A and ends at the head of ,B so it

points up and to the right like vector .E

2. (b) To subtract −A B we must reverse B (so that it points to the left) and add it head-to-tail to .A The resultant

vector points up and to the left like vector .F

15.0 m

22.0 m 22.0 m

5.00 m

A

B

C

D

palm tree

A θA



3 – 7

3. (c) To subtract −B A we must reverse A (so that it points down) and add it head-to-tail to .B The resultant vector

points down and to the right like vector .D Insight: In the figure the vector C points in a similar direction to .− −A B 19. Picture the Problem: Consider the six vectors through A F depicted in the

figure.

Strategy: Remember the rules of adding and subtracting vectors. Vectors are always added head-to-tail.

Solution: 1. (a) To add +A D we must move D so that its tail is on the head of .A The resultant vector starts at the tail of A and ends at the head of ,D so it

points up and to the right but is shorter than .A By the same analysis the vector ,+A E however, points up and to the right and is longer than .A We conclude that the magnitude of +A D is

less than the magnitude of .+A E 2. (b) As discussed in step 1, the vector +A E points up and to the right and is longer than .A Likewise the vector

+A F points up and to the left and is longer than .A We note the symmetry of the vectors and E F and conclude that

the magnitude of +A E is equal to the magnitude of .+A F Insight: In a similar fashion the vectors +A D and +A C would be equal in magnitude. 20. Picture the Problem: The two vectors A (length 40.0 m) and B (length 75.0 m) are

drawn at right.

Strategy: Add vectors A and B using the vector component method.

Solution: 1. (a) A sketch (not to scale) of the vectors and their sum is shown at right.

2. (b) Add the x components: ( ) ( ) ( ) ( )40.0 m cos 20.0 75.0 m cos 50.0 85.8 mx x xC A B= + = − ° + ° =

3. Add the y components: ( ) ( ) ( ) ( )40.0 m sin 20.0 75.0 m sin 50.0 43.8 my y yC A B= + = − ° + ° =

4. Find the magnitude of C : ( ) ( )2 22 2 85.8 m 43.8 m 96.3 mx yC C C= + = + =

5. Find the direction of C : 1 1 43.8 mtan tan 27.0

85.8 my

Cx

CC

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

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 using a ruler and protractor to add the vectors graphically.

Cx

y B A

20.0°

50.0°



3 – 8

21. 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 B from A using the vector component method.

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

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

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

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

5. Find the direction of D: 1 1 57 kmtan tan 10 180 170 or 10 north of west

310 kmy

Dx

DD

θ − −⎛ ⎞ ⎛ ⎞= = = − ° + ° = ° °⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

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 −10° as the angle in step 5, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.


right.

Strategy: Subtract vector iv from fv using the vector component method.

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

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

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

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

5. Find the direction of vΔ : 1 1 64 km/htan tan 66 180 114

28 km/hy

vx

vv

θ − −Δ

Δ⎛ ⎞ ⎛ ⎞= = = − ° + ° = °⎜ ⎟ ⎜ ⎟Δ −⎝ ⎠⎝ ⎠ where the angle is

measured counterclockwise from the positive x 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− ° as the angle in step 5, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

x

y

fv

iv

75°

i−v

Δv



3 – 9

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

Strategy: Deduce the x and y components of B from the information given about A and C . Use the known components to estimate the length and direction of B as well as calculate them precisely.

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

2. (b) The vector B must have an x component of −75 m so that when it is added to A the x components will cancel out. It must also have a y component of 95 m because that is the length of C and A has no y component to contribute. Therefore B must be longer than either A or C 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 B we can find its magnitude:

2 2( 75 m) (95 m) 121 mB = − + =

4. Find the direction of B : 1 95 mtan 52 180 128

–75 mBθ − ⎛ ⎞= = − ° + ° = °⎜ ⎟⎝ ⎠

Insight: Here the length and direction of B are determined by its x and y components, which are determined from A and C . Learning to manipulate vector components will be a useful skill when tackling many physics problems.


Strategy: Since A points entirely in the x direction, and B points entirely in the y direction, A and B are the x and y components of their sum +A B . Use the known lengths of +A B and A to find B.

Solution: 1. Set the length of +A B equal to 37 units:

2 2

2 2 237

37A B

A B= +

= +

2. Solve for B: ( )22 2 237 37 22 30 unitsB A= − = − − =

Insight: Here the length of B is determined by the lengths of the other two vectors because the directions of A and B are stipulated. Learning to manipulate vector components will be a useful skill when tackling many physics problems.

x

y

B

A

+A B

−22 O

30



3 – 10

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

at right.

Strategy: Use the vector component method of addition and subtraction to determine the components of each combination of A and B . Once the components are known, the length and direction of each combination can be determined fairly easily.

Solution: 1. (a) Determine the components of +A B : ( ) ( )ˆ ˆ ˆ ˆ–5 10 10 – 5+ = + =A B y x x y

2. Find the magnitude of +A B : ( ) ( )2 210 5 11 units+ = + − =A B

3. Determine the direction of +A B , measured counterclockwise from the positive x axis.

1 5tan 27 or 33310

θ −+

−⎛ ⎞= = − ° °⎜ ⎟⎝ ⎠A B

4. (b) Determine the components of −A B : ( ) ( )ˆ ˆ ˆ ˆ–5 10 10 – 5− = − = −A B y x x y

5. Find the magnitude of −A B : ( ) ( )2 210 5 11 units− = − + − =A B

6. Determine the direction of −A B , measured counterclockwise from the positive x axis.

1 5tan 27 180 20710

θ −−

−⎛ ⎞= = ° + ° = °⎜ ⎟−⎝ ⎠A B

7. (c) Determine the components of −B A : ( ) ( )ˆ ˆ ˆ ˆ10 –5 10 5− = − − = +B A x y x y

8. Find the magnitude of −B A : ( ) ( )2 210 5 11 units− = + =B A

9. Determine the direction of −B A , measured counterclockwise from the positive x axis.

1 5tan 2710

θ −−

⎛ ⎞= = °⎜ ⎟⎝ ⎠B A

Insight: This problem is simplified by the fact that A and B have only one component each, but a similar approach will work even with more complicated vectors. Notice that you must have a picture of the vectors in your head (or on paper) in order to correctly interpret the directions in steps 3, 6, and 9.

26. Picture the Problem: The vectors involved in the problem are

depicted at right.

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

Solution: 1. (a) Make estimates from the drawing:

20 m 1.5θ+ + ≈ ≈ °A B C

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

ˆ0 20.0 m cos 45 7.0 m cos 30ˆ10.0 m 20.0 m sin 45 7.0 m sin 30

ˆ ˆ20.2 m 0.64 m

+ + = + ° + − ° +⎡ ⎤⎣ ⎦− + ° + − °⎡ ⎤⎣ ⎦

+ + = +

A B C xy

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

C

x

yB

A

+ +A B C

θ

45°

30°

+A B

θ+A B

θ−A B

θ−B A x

y

B

A

O 10

5

−B A

−A B



3 – 11

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

θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠

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 the angle of 1.8° in step 4, you must have a picture of the vectors in your head (or on paper) to correctly determine the direction.

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

Strategy: Determine the x and y components of and then express them in terms of the unit vectors.

Solution: 1. Find the x and y components of Δr : ( ) ( )( ) ( )54 m cos 42 40 m

54 m sin 42 36 m

x

y

r

r

= − =

= − = −

2. Now express Δr in terms of the unit vectors: ( ) ( )ˆ ˆ40 m 36 mΔ = + −r x y

Insight: In general, an arbitrary two-dimensional vector A can always be written as the sum of a vector component in the x direction and a vector component in the y direction.

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

Strategy: Determine the x and y components of A .

Solution: 1. Find the x component: ( ) ( )3.50 m cos 145 2.87 mxA = ° = −

2. Find the y component: ( ) ( )3.50 m sin 145 2.01 myA = ° =


29. Picture the Problem: The vector A has a length of 6.1 m and points in the negative x direction.

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

Solution: 1. (a) Multiply each component of A by −3.7:

( )( )( ) ( )

ˆ6.1 m

ˆ ˆ3.7 3.7 6.1 m 23 m so 23 mxA

= −

− = − − = =⎡ ⎤⎣ ⎦

A x

A x x

2. (b) Since A 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.

30. Picture the Problem: The vector 5.2− A has a length of 34 m and points in the positive x direction.

Strategy: Divide the components of the vector 5.2− A by −5.2 in order to find the components of A . From there we can easily find the x component and the magnitude of A .

Solution: 1. (a) Divide both sides by −5.2: ( )( )

ˆ5.2 34 mˆ6.5 m

− == −

A xA x

2. The vector A has only an x component: 6.5 mxA = −

3. (b) Since A has only one component, its magnitude is simply 6.5 m.

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

A

x

y

145°

Δr

x

y

54 m 42°



3 – 12


Strategy: Determine the lengths and directions of the various vectors by using their x and y components.

Solution: 1. (a) Find the direction of A from its components:

1 2.0 mtan –225.0 m

θ − −⎛ ⎞= = °⎜ ⎟⎝ ⎠A

2. Find the magnitude of A : ( ) ( )2 25.0 m 2.0 m 5.4 mA = + − =

3. (b) Find the direction of B from its components:

1 5.0 mtan 68 180 110–2.0 m

θ − ⎛ ⎞= = − ° + ° = °⎜ ⎟⎝ ⎠B

4. Find the magnitude of B : ( ) ( )2 22.0 m 5.0 m 5.4 mB = − + =

5. (c) Find the components of +A B : ( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ5.0 2.0 m 2.0 5.0 m 3.0 m 3.0 m+ = − + − + = +A B x y x y

6. Find the direction of +A B from its components:

1 3.0 mtan 453.0 m

θ −+

⎛ ⎞= = °⎜ ⎟⎝ ⎠A B

7. Find the magnitude of +A B : ( ) ( )2 23.0 m 3.0 m 4.2 m+ = + =A B

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


Strategy: Determine the lengths and directions of the various vectors by using their x and y components.

Solution: 1. (a) Find the direction of A from its components:

1 12 mtan –2625 m

θ − −⎛ ⎞= = °⎜ ⎟⎝ ⎠A

2. Find the magnitude of A : ( ) ( )2 225 m –12 m 28 mA = + =

3. (b) Find the direction of B from its components:

1 15 mtan 822.0 m

θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠B

4. Find the magnitude of B : ( ) ( )2 22.0 m 15 m 15 mB = + =

5. (c) Find the components of +A B :

( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ25 2.0 m 12 15 m 27 m 3.0 m+ = + + − + = +A B x y x y

6. Find the direction of +A B from its components:

1 3.0 mtan 6.327 m

θ −+

⎛ ⎞= = °⎜ ⎟⎝ ⎠A B

7. Find the magnitude of +A B : ( ) ( )2 227 m 3.0 m 27 m+ = + =A B

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

+A B x

y

B

A

O 25 m

12 m

θ+A B

+A B

θ+A B x

y

B

A

2.0 m 5.0 m O

2.0 m



3 – 13


right. Strategy: Since the components of the vectors are known, we need only

add or subtract the components separately as specified in the problem statement.

Solution: 1. (a) Subtract the components:

( ) ( )( ) ( )

ˆ ˆ25 m 2 m –12 m 15 mˆ ˆ23 m 27 m

− = − + −

= + −

A B x yx y

2. (b) Multiply the answer to part (a) by −1: ( ) ( ) ( )ˆ ˆ23 m 27 m− = − − = − +B A A B x y

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

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

Strategy: Use the information given in the figure to determine the components of each vector.

Solution: 1. Find the components:

( ) ( ) ( ) ( )( ) ( )

ˆ ˆ1.5 m cos 40 1.5 m sin 40

ˆ ˆ1.1 m 0.96 m

= ° + °

= +

A x y

A x y

2. Repeat for :B ( ) ( ) ( ) ( )( ) ( )2.0 m cos 19 2.0 m sin 19

ˆ ˆ1.9 m 0.65 m

= − ° − − °

= + −

B

B x y

3. Repeat for :C ( ) ( ) ( ) ( )( ) ( )

ˆ ˆ1.0 m cos 180 25 1.0 m sin 180 25

ˆ ˆ0.91 m 0.42 m

= − ° + − °

= − +

C x y

C x y

4. Repeat for :D ( )ˆ ˆ0 1.5 m= +D x y

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.

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

Strategy: Use the information given in the figure to determine the components of vectors , , and A B C . Then add the components.

Solution: 1. Add the x component 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 C

2. Add the y component of each 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 C

3. Express the sum in unit vector notation: ( ) ( )ˆ ˆ2.1 m 0.74 m+ + = +A B C x y

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

25 m O

A12 m

−A B

y

x

−B A

−A

B

−B



3 – 14

36. Picture the Problem: The various vectors depicted in the diagram

represent position, velocity, or acceleration vectors.

Strategy: We can identify any position vectors because they do not originate on the particle’s path (unless the path goes through the origin) and they always terminate on the path. Velocity vectors must always originate on the path and point tangent to the path, while acceleration vectors always originate on the path and could point in any direction. For uniform circular motion the acceleration vector always points toward the center of the circle.

Solution: 1. (a) By applying the strategy outlined above we can identify vectors 1 and 5 as position vectors.

2. (b) By applying the strategy outlined above we can identify vectors 2, 3, 7, and 8 as velocity vectors. In principle they could also be acceleration vectors, but because vectors 2 and 3 have the same length, and vector 4 points toward the center of the circle, we can assume vectors 2 and 3 are velocity vectors and vector 4 is an acceleration vector.

3. (c) By applying the strategy outlined above we can identify vectors 4 and 6 as acceleration vectors.

Insight: Instantaneous velocity vectors always point tangent to the path, but average velocity vectors might not. 37. Picture the Problem: The two vectors involved in this problem are

depicted in the figure at right.

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

Solution: 1. (a) Because each vector has components of length 3.5 cm and 7.0 cm, the magnitude of displacement 1 will be equal to the magnitude of displacement 2.

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

1

2

ˆ ˆ7.0 cm 3.5 cmˆ ˆ3.5 cm 7.0 cm

Δ = − +

Δ = +

r x y

r x y

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

( ) ( )2 21

11

7.0 cm 3.5 cm 7.8 cm

3.5 cmtan 27 180 1537.0 cm

r

θ −

Δ = − + =

⎛ ⎞= = − ° + ° = °⎜ ⎟−⎝ ⎠

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

( ) ( )2 22

12

3.5 cm 7.0 cm 7.8 cm

7.0 cmtan 633.5 cm

r

θ −

Δ = + =

⎛ ⎞= = °⎜ ⎟⎝ ⎠

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

38. Picture the Problem: The displacement vectors are depicted at right. North is in

the y direction and east is in the x direction.

Strategy: Sum the components of the vectors in order to determine +A B . Multiply that vector by −1 in order to reverse its direction. Then find the magnitude and direction of the reversed vector.

Solution: 1. (a) Add the two displacement vectors:

( ) ( )ˆ ˆ72 m 120 m+ = − +A B x y

2. Multiply by −1 in order to reverse the direction of the net displacement and bring the cat back home:

( ) ( ) ( )ˆ ˆ72 m 120 m− + = + −A B x y

( )− +A B

x

yB

A

−72 m O

120 mθ



3 – 15

3. Find the magnitude of the desired displacement:

( ) ( ) ( )2 272 m 120 m 140 m− + = + =A B

4. Find the direction of the desired displacement:

1 120 mtan 59 59 south of east72 m

θ − −⎛ ⎞= = − ° = °⎜ ⎟⎝ ⎠

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

Insight: In this problem we could claim the cat’s initial displacement is a single vector with the given components. The answers wouldn’t change, but it would simplify the solution a little bit.

39. Picture the Problem: The two legs of the cat’s path are indicated at right. North is

in the y direction and east is in the x direction.

Strategy: Determine the displacement from the known vectors that make up the two legs of the cat’s journey. Divide the displacement by the total time of travel to find the average velocity. Use the x and y components of the average velocity to determine its magnitude and direction.

Solution: 1. Determine the displacement:

( ) ( )ˆ ˆ120 m 72 mΔ = + = + −r A B y x

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

av72 m 1min 120 m 1minˆ ˆ

62 min 60s 62 m 60sˆ ˆ0.019 m/s 0.032 m/s

t⎛ ⎞ ⎛ ⎞⎛ ⎞Δ −⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟Δ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

= − +

rv x y

x y

3. Determine the magnitude of the velocity: ( ) ( )2 2

av 0.019 m/s 0.032 m/s 0.037 m/sv = − + =

4. Determine the direction of the velocity:

1 120 mtan 59 180 12172 m

121 or 31 west of north

θ

θ

− ⎛ ⎞= = − ° + ° = °⎜ ⎟−⎝ ⎠

= ° °

Insight: The average speed would be calculated differently:

( )120 72 m 0.052 m/s.

45 17 min 60 s/mindst

+= = =

+ × The

average speed is faster than the average velocity because the total distance traveled is larger than the displacement. 40. Picture the Problem: You travel due west for 125 s at 27 m/s then due south at 14 m/s for 66 s.

Strategy: Find the components of the displacement vector. Once the components are known the magnitude and direction can be easily found. Let north be the positive y direction and east be the positive x direction.

Solution: 1. Find the westward displacement: ( )( )27 m/s 125 s 3375 mx xr v t= = − = −

2. Find the southward displacement: ( )( )14 m/s 66 s 924 my yr v t= = − = −

3. Find the direction of the displacement: 1 1 924 mtan tan 15 180 195 3375 m

or 15 south of west

y

x

rr

θ − −⎛ ⎞ −⎛ ⎞= = = ° + ° = °⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

°

4. Find the magnitude of the displacement: ( ) ( )2 23375 m 924 m 3500 m 3.5 kmr = − + − = =

Insight: The 15° refers to the angle below the negative x axis (west) because the argument of the inverse tangent function is y xr r , or south divided by west.

Δ = +r A B

x

yB

A

−72 m O

120 m

θ



3 – 16

41. 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 direction fairly easily. The direction of the average velocity will be the same as the direction of the displacement. The magnitude of the average velocity is the magnitude of the displacement divided by the total time of travel. Let north be the positive y direction and east be the positive x direction.

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

1500 fty

x

rr

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

2. Find the magnitude of the displacement: ( ) ( )2 21500 ft 2500 ft 2900 ft 0.305 m/ft 890 mr = + = × =

3. Find the magnitude of the average velocity: av

890 m 4.9 m/s3.0 min 60 s/min

rvt

Δ= = =

Δ ×

Insight: The 59° refers to the angle above the positive x axis (east) because the argument of the inverse tangent function is y xr r , or north divided by east.

42. Picture the Problem: The jogger runs at 3.25 m/s in a direction 30.0° above the positive x axis.

Strategy: Find the components of the velocity vector according to the method indicated in Figure 3-7(a).

Solution: 1. (a) Find the x component of v : ( ) ( )3.25 m/s cos 30.0 2.81 m/sxv = ° =

2. Find the y component of v : ( ) ( )3.25 m/s sin 30.0 1.63 m/syv = ° =

3. (b) If the jogger’s speed is halved, the direction will remain unchanged but the x and y components will be halved.

Insight: In this case the angle of 30.0° corresponds to the standard angle θ as indicated in Figure 3-7(a). 43. 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 initial velocity upward. Apply equation 3-5, taking upward to be the positive direction.

Solution: Apply equation 3-5:

( ) ( ) ( )2av

ˆ ˆ4.5 m/s 4.5 m/sˆ9.8 m/s

0.92 sf i

t− − −

= = = −Δ

v v y ya y

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

44. 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:

( ) ( ) 2av

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

3.00 sf iv v

at

− −= = =

Δ

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

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



3 – 17

45. Picture the Problem: The skateboarder rolls down the ramp that is inclined 17.5° 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 for fv : ( ) ( ) ( )20 9.81 m/s sin 17.5 3.25 s 9.59 m/sf iv v a t ⎡ ⎤= + Δ = + ° =⎣ ⎦

Insight: Note that when the vector direction doesn’t matter (this is essentially a one-dimensional problem) the equation 3-5 looks exactly like equation 2-7. In Chapter 5 we’ll be able to rigorously prove that sina g θ= using a free body diagram. See, for instance, Example 5-9.

46. Picture the Problem: The initial and final displacement vectors are depicted

at right.

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

Solution: 1. (a) Find the initial position vector:

( ) [ ] [ ]{ }( )

8i

8

ˆ ˆ3.84 10 m cos 0 sin 0

ˆ3.84 10 m

= × +

= ×

r x y

x

2. Find the arguments of the sine and cosine functions for t = 7.38 days. Let

6 radians/s2.46 10ω −= × :

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

1.57 radians

tω −= × ×

=

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

8f

8

5 8 8f

ˆ ˆ3.84 10 m cos sin

ˆ ˆ3.84 10 m cos 1.57 radians sin 1.57 radians

ˆ ˆ ˆ3.06 10 m 3.84 10 m 3.84 10 m

t tω ω= × +

= × +

= × + × ≅ ×

r x y

x y

r x y y

4. Find the displacement vector: ( ) ( )8 8f i ˆ ˆ3.84 10 m 3.84 10 mΔ = − = × − ×r r r y x

5. Find the vector avv :

( ) ( )

( ) ( )

8 8

av

av

ˆ ˆ3.84 10 m 3.84 10 m

7.38 d 86400 s/dˆ ˆ602 m/s 602 m/s

t

− × + ×Δ= =

Δ ×= − +

x yrv

v x y

6. Find the magnitude of avv : ( ) ( )2 2

av 602 m/s 602 m/s 852 m/sv = − + =

7. Find the direction of avv : av, 1 1

av,

602 m/stan tan 45 180 135602 m/s

y

x

vv

θ − −⎛ ⎞ ⎛ ⎞= = = − ° + ° = °⎜ ⎟ ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠

8. (b) The instantaneous speed of the Moon is greater than the average velocity because the distance traveled is greater than 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 its average velocity would have been zero. Its speed remains constant, however, at about 947 m/s using the data given in this problem. The given data correspond to a coordinate system where the x direction always points toward the center of the Sun even as the Earth orbits the Sun.

Δr

3.84×108 m

3.84×108 m

x

fr

ir

y

O

θ



3 – 18

47. 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 final velocities. Then use those components to find the change in velocity vector. Divide the change in velocity vector by the elapsed time to find the acceleration vector, and then determine its magnitude and direction.

Solution: 1. (a) Find the initial velocity vector:

( ) [ ] [ ]{ }( )

i ˆ ˆ945 m/s sin 0 cos 0

ˆ945 m/s

= − +

=

v x y

y

2. Find the arguments of the sine and cosine functions for t = 0.100 days. Let 6 radians/s2.46 10ω −= × :

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

0.0213 radians

tω −= × ×

=

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

f

f

ˆ ˆ945 m/s sin cos

ˆ ˆ945 m/s sin 0.0213 radians cos 0.0213 radians

ˆ ˆ20.1 m/s 945 m/s

t tω ω= − +

= − +

= − +

v x y

x y

v x y

4. Find the acceleration vector: ( ) ( ) ( ) ( )2f i

ˆ ˆ ˆ20.1 m/s 945 m/s 945 m/sˆ0.00233 m/s

0.100 d 86400 s/dt− − −⎡ ⎤− ⎣ ⎦= = = −

Δ ×

x y yv va x

5. (b) Find the arguments of the sine and cosine functions for t = 0.0100 days:

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

0.00213 radians

tω −= × ×

=

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

f

f

ˆ ˆ945 m/s sin cos

ˆ ˆ945 m/s sin 0.00213 radians cos 0.00213 radians

ˆ ˆ2.01 m/s 945 m/s

t tω ω= − +

= − +

= − +

v x y

x y

v x y

7. Find the acceleration vector:

( ) ( ) ( ) ( )2f iˆ ˆ ˆ2.01 m/s 945 m/s 945 m/s

ˆ0.00233 m/s0.0100 d 86400 s/dt

− − −⎡ ⎤− ⎣ ⎦= = = −Δ ×

x y yv va x

Insight: The two answers ended up being the same because both time intervals are fairly small. If instead we had examined an interval of 1.00 days there would have been a ˆ−y component of Δv and a slightly different acceleration. In Chapter 6 we will examine circular motion and find an even easier way to calculate the acceleration of the Moon. The given data in this problem correspond to a coordinate system where the x direction always points toward the center of the Sun even as the Earth orbits the Sun.

48. Picture the Problem: The photo shows two airplanes flying together during a midair refueling operation.

Strategy: Take note of the fact that if the two aircraft have different velocities they could not remain joined together for more than an instant.

Solution: 1. (a) The aircraft being refueled must have the same velocity as the KC-10A, or 125 m/s due east.

2. (b) The aircraft being refueled must have zero velocity relative to the KC-10A.

Insight: As the aircraft being refueled approaches the KC-10A it must have a slightly higher speed than the KC-10A, and then needs a slightly lower speed in order to pull away from the KC-10A after refueling.

Δv 945 m/s

fviv

y

O x



3 – 19

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

Strategy: Let pg =v plane’s velocity with respect to the ground, ap =v attendant’s velocity with respect to the plane, and add the vectors according to equation 3-8 to find ag =v attendant’s velocity with respect to the ground.

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

ag

ˆ ˆ ˆ1.22 m/s 16.5 m/s 15.3 m/s

15.3 m/sv

= + = − + =

=

v v v x x x

Insight: If the attendant were walking toward the front of the plane, her speed relative to the ground would be 17.7 m/s, slightly faster than the airplane’s speed.


Strategy: The x-component of the velocity was chosen perpendicular to the motion of the river. Therefore, the motion of the river will not affect the time it takes to travel across it. Divide the width of the river by the x component of the boat’s velocity to find the time it takes to cross the river.

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

( ) ( )( ) ( )

bw ˆ ˆ6.1 m/s cos 25 6.1 m/s sin 25ˆ ˆ5.5 m/s 2.6 m/s

= ° + °

= +

v x y

x y

2. Find the velocity of the boat relative to the ground:

( ) ( )( ) ( )

bg ˆ ˆ5.5 m/s 2.6 1.4 m/sˆ ˆ5.5 m/s 1.2 m/s

= + −

= +

v x y

x y

3. Divide the width of the river by the x component of

bgv : bg,

35 m 6.4 s5.5 m/sx

xtvΔ

= = =

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 travel time significantly longer than 6.4 s.


Strategy: Let yw =v your velocity with respect to the walkway, wg =v walkway’s velocity with respect to the ground, and add the vectors according to equation 3-8 to find yg =v 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 respect to the walkway: ( )yw

85 mˆ ˆ ˆ1.25 m/s68 s

xt

Δ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟Δ⎝ ⎠ ⎝ ⎠v x x x

2. Apply equation 3-8 to find your velocity with respect to the ground:

( ) ( ) ( )yg yw wg ˆ ˆ ˆ1.25 m/s 2.2 m/s 3.45 m/s= + = + =v v v x x x

3. Now find the time of travel:

yg

85 m 25 s3.45 m/s

xtvΔ

= = =

Insight: The moving walkway slashed your time of travel from 68 s to 25 s, a factor of 2.7! Note that we bent the significant figures rules a little bit by not rounding ywv to 1.3 m/s. This helped us avoid rounding error.

wgv

ygv

ywv

apv

pgv

agv



3 – 20


Strategy: Let yw =v your velocity with respect to the walkway, wg =v walkway’s velocity with respect to the ground, and add the vectors according to equation 3-8 to find yg =v 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 respect to the walkway: ( )yw

85 mˆ ˆ ˆ1.3 m/s68 s

xt

Δ⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟Δ⎝ ⎠ ⎝ ⎠v x x x

2. Apply equation 3-8 to find your velocity with respect to the ground: ( ) ( ) ( )yg yw wg ˆ ˆ ˆ1.3 m/s 2.2 m/s 0.9 m/s= + = − + =v v v x x x

3. Now find the time of travel:

yg

85 m 90 s0.9 m/s

xtvΔ

= = =

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


Strategy: Let pg =v velocity of the plane relative to the ground, pa =v

velocity of the plane relative to the air, and ag =v velocity of the air relative to the ground. The drawing at right depicts the vectors added according to equation 3-8, pg pa ag= +v v v . Determine the angle of the triangle from the inverse sine function.

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

pa

65 km/hsin sin 11 west of north340 km/h

vv

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

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 be increased in order for the plane to continue flying due north.

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


Strategy: Let pf =v the passenger’s velocity relative to the ferry, pw =v the passenger’s velocity relative to the water, and fw =v the ferry’s velocity relative to the water. Apply equation 3-8 and solve for fwv . Once the components of fwv are known, its magnitude and direction θ can be determined.

Solution: 1. Solve equation 3-8 for fwv : pw pf fw

fw pw pf

= +

= −

v v v

v v v

2. Determine the components of fwv : ( ) ( ) ( )( ) ( )

fw

fw

ˆ ˆ ˆ4.50 m/s sin 30 4.50 m/s cos30 1.50 m/s

ˆ ˆ2.25 m/s 2.40 m/s

= − ° + ° −⎡ ⎤⎣ ⎦= − +

v x y y

v x y

pwv

fwv

pfv

θ

30°

E

N

ywv

wgv

ygv



3 – 21

3. Find the direction of fwv : fw, 1 1

fw,

2.25 m/stan tan 43 west of north2.40 m/s

x

y

v

vθ − −

⎛ ⎞ ⎛ ⎞⎜ ⎟= = = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

4. Find the magnitude of fwv : ( ) ( )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 fwv would have to be shorter and point more in the westerly direction.

55. 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 the x-axis perpendicular to the flow of the river, such that the river is flowing in the negative y-direction. Let bw =v jet ski’s velocity relative to the water, bg =v jet ski’s velocity relative to the ground, and wg =v water’s velocity relative to the ground. Use equation 3-8 to find the vector bwv , and then determine its magnitude.

Solution: 1. Solve eq. 3-8 for bwv : bg bw wg bw bg wg = + ⇒ = −v v v v v v

2. Find the components of bgv : ( ) ( )( ) ( )

bg ˆ ˆ9.5 m/s cos 20.0 9.5 m/s sin 20.0

ˆ ˆ8.9 m/s 3.2 m/s

= ° + °

= +

v x y

x y

3. Subtract to find bwv : ( ) ( ) ( )( ) ( )

bw bg wg ˆ ˆ ˆ8.9 m/s 3.2 m/s 2.8 m/s

ˆ ˆ8.9 m/s 6.0 m/s

= − = + − −⎡ ⎤⎣ ⎦= +

v v v x y y

x y

4. Find the magnitude of bwv : ( ) ( )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 from the components of bwv we get ( )1tan 6.0 8.9 34θ −= = ° , not exactly 35° due to rounding errors.

56. Picture the Problem: The situation is depicted in the figure at right, except the

boat is supposed to be a jet ski.

Strategy: Place the x-axis perpendicular to the flow of the river, such that the river is flowing in the negative y-direction. Let bw =v jet ski’s velocity relative to the water, bg =v jet ski’s velocity relative to the ground, and

wg =v water’s velocity relative to the ground. Set the y component of bwv equal to the magnitude of wgv so that they cancel, leaving only an x component of bgv . Then determine the angle θ.

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

bw, bw wg,

wg, 1 1

bw

sin

2.8 m/ssin sin 13

12 m/s

y y

y

v v v

vv

θ

θ − −

= = −

− − −⎛ ⎞⎛ ⎞= = = °⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

2. (b) Increasing the jet ski’s speed relative to the water will increase bwv and therefore decrease the angle θ.

Insight: Airplanes must also make heading adjustments like the jet ski’s in order to fly in a certain direction when there is a steady wind present.



3 – 22

57. Picture the Problem: The vectors for Jet Ski A and B are depicted at

right. Note that bwv 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: Use the x components of the velocities of each jet ski relative to the ground to determine the time required for each jet ski to cross the river.

Solution: 1. (a) The time required for each jet ski to cross the river equals the width of the river divided by the x component of the jet ski’s velocity relative to the ground. From the diagrams you can see that the x compo-nent of bg,Av equals bwv , but that the x component of bg,Bv is shorter than

bwv . Therefore Jet Ski A has a higher velocity in the x direction relative to the ground, and will cross the river first.

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

river

bg, A, bg, B, bwA

riverB bg, A, bw

bg, B,

cos350.82x x

x

x

xv v vt

xt v vv

Δ

°Δ= = = =

ΔΔ

Insight: The ratio is less than one, so A Bt tΔ < Δ and Jet Ski A reaches the opposite shore first. Remember this the next time you race jet skis across a flowing river!

58. Picture the Problem: Vector A points in the positive x direction and vector B points in the negative x direction. Strategy: Use the definitions of vector magnitude and direction to answer the conceptual question.

Solution: 1. (a) The magnitude of a vector is determined by its length and is independent of its direction. Therefore we note that the magnitude of ( ) ˆ1.2 m=A x is less than the magnitude of ( ) ˆ3.4 m= −B x :

2. (b) The best explanation is I. The number 3.4 is greater than the number 1.2. Statement II is true, but not relevant, and statement III is true, but not relevant because the magnitude of a vector is independent of its direction.

Insight: Even if the signs were reversed the magnitude of ( ) ˆ1.2 m 1.2 m= − =A x would still be smaller than

( ) ˆ3.4 m 3.4 m.= =B x The magnitude of a vector is independent of its direction.

59. Picture the Problem: Vector A points in the negative x direction and vector B points in the positive y direction. Strategy: Use the procedure for multiplying a vector by a scalar to answer the conceptual question.

Solution: 1. (a) The magnitude of ( ) ˆ1.4 1.4 2.2 m 3.1 m= − =A x is equal to the magnitude of

( ) ˆ2.2 2.2 1.4 m 3.1 m.= =B y

2. (b) The best explanation is II. A number and its negative have the same magnitude. Statement I is true, but not relevant, and statement III is false (the vectors are perpendicular).

Insight: Even if the signs of each vector were reversed the products 1.4A and 2.2B would have the same magnitude.

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

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

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

lengthyr

θ = =

θ 3.00 ft

10.0 ft

bg,Avwgv

bwv

bg,Bv

wgvbwv

Jet Ski A

Jet Ski B 35°



3 – 23

2. Use the inverse sine function to find the angle: 1 3.00 ftsin 17.5

10.0 ftθ − ⎛ ⎞= = °⎜ ⎟

⎝ ⎠

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

61. Picture the Problem: The vector components of A and B are specified in the problem. Measure positive angles to be counterclockwise from the positive x axis.

Strategy: Multiply each component of A by 2 and add them to B . Use the resulting components to determine the direction and magnitude of the sum.

Solution: 1. Multiply and add the components: ( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ2 2 12.1 m 32.2 m 24.2 m 32.2 m+ = + − = + −A B x y x y

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

θ − −⎛ ⎞= = − ° °⎜ ⎟⎝ ⎠

3. Find the magnitude of the vector sum: ( ) ( )2 22 24.2 m 32.2 m 40.3 m+ = + − =A B

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

62. Picture the Problem: The x and y components of a vector are each negative.

Strategy: Use your knowledge of the quadrants of the coordinate system to answer the conceptual question.

Solution: The quadrant in which both vector components are negative is the third quadrant, where the direction angle is between 180° and 270°.

Insight: If the x component were negative but the y component were positive, the vector would lie in the second quadrant and its direction angle would be between 90° and 180°.

63. Picture the Problem: The x component of a vector is positive but the y component is negative.

Strategy: Use your knowledge of the quadrants of the coordinate system to answer the conceptual question.

Solution: The quadrant in which the x component of a vector is positive but the y component is negative is the fourth quadrant, where the direction angle is between 270° and 360°.

Insight: If the x and y components were each positive, the vector would lie in the first quadrant and its direction angle would be between 0° and 90°.

64. Picture the Problem: The vector components of −A B , C , and + +A B C are specified in the problem.

Strategy: Use the given vector components to write three equations and solve them for A and B .

Solution: 1. Add the three given equations to solve for A : ( )( )

( )( ) ( )

ˆ13.8 mˆ51.4 mˆ62.2 m

ˆ ˆAdd: 2 99.8 m 49.9 m

+ + =− = −

− = −

= − ⇒ = −

A B C xA B x

C x

A x A x

2. Now substitute the known vector A into the second equation:

( ) ( )( )

ˆ ˆ49.9 m 51.4 mˆ1.5 m

− − = −

=

x B xB x

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



3 – 24

65. Picture the Problem: The vectors involved in this problem are illustrated

at right.

Strategy: The x component of the ball’s velocity with respect to the train must be equal and opposite to the train’s velocity in order for Gary to see the ball rise straight upward. That fact, together with the angle of the throw, can be used to find the speed btv of Michelle’s throw as well as the speed bgv of the ball according to Gary.

Solution: 1. (a) Michelle must have thrown the ball toward the rear of the train so that tgv could cancel out the x component of btv and leave bgv completely vertical.

2. (b) Set the magnitudes of the x components of

btv and tgv and solve for btv :

( )

( ) ( )

bt, bt tg,

tg, bt

cos 65.0

8.35 m/s 19.8 m/scos 65.0 cos 65.0

x x

x

v v v

vv

= ° =

= = =° °

3. (c) The magnitude of bgv equals the y component of btv : ( ) ( )bg 19.8 m/s sin 65.0 17.9 m/sv = ° =

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

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

Strategy: Find the displacement vectors A and B using the given speed and time information. Use those vector components to find the final displacement vector C . Divide the displacement vector C by the time in order to find the direction and speed of travel on the final leg. Let north point in the y direction and east in the x direction.

Solution: 1. Find the vector A : ( ) ( )( )( )

( ) ( )( ) ( )

ˆ ˆsin 25 cos 25

6.5 km/h 15 min 1 h/60 min

ˆ ˆsin 25 cos 25

ˆ ˆ0.69 km 1.5 km

A A A Av t v t= − Δ ° + Δ °

= ×

× − ° + °⎡ ⎤⎣ ⎦= − +

A x y

x y

A x y

2. Find the vector B : ( )( ) ( )ˆ ˆ ˆ12 km/h 7.5 min 1 h/60 min 1.5 kmB Bv t= Δ = × =B x x x

3. Find the vector C : ( ) ( ) ( )( ) ( )

( ) ( )

ˆ ˆ

ˆ ˆ0.69 1.5 km 1.5 0 km

ˆ ˆ0.81 km 1.5 km

x x y yA B A B= − + = − + − +

= − − + − +

= − + −

C A B x y

x y

C x y

4. Find the direction angle θ: 1 1 1.5 kmtan tan 62 south of west

0.81 kmy

x

CC

θ − −−⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟− ⎝ ⎠⎝ ⎠

5. Find the speed of travel:

( ) ( )2 20.81 km 1.5 km4.6 km/h

22 min 1 h/60 minCvt

+= = =

Δ ×

C

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

x

y

B

A

O

θ

C25°

65.0°



3 – 25

67. Picture the Problem: The three-dimensional vector is depicted at right.

Strategy: Determine the z component of A by applying the cosine function to the right triangle formed in the z direction. Then find the projection of A onto the xy plane (A sin 55°) in order to find the x and y components of A .

Solution: 1. Find the z component of A : ( )65 m cos55 37 mzA = ° =

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

3. Find the x component of A : ( )65 m sin 55 cos 35 44 mxA = ° ° =⎡ ⎤⎣ ⎦

4. Find the y component of A : ( )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 the components are known, then addition and subtraction of vectors become straightforward procedures.

68. 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 initial yv of zero. The football maintains its horizontal velocity 0v , so the two velocities form the x and y components of the ball’s velocity at 1.75 s. Use the components to find the magnitude and direction of the velocity.

Solution: 1. (a) Use equation 3-6 to find the components of fv :

( )

( ) ( )

0 2

m mˆ ˆ16.6 9.81 1.75 ss s

ˆ ˆ16.6 m/s 17.2 m/s

f t ⎛ ⎞ ⎛ ⎞= + = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

v v a x y

x y

2. (b) Find the magnitude of fv : ( ) ( )2 216.6 m/s 17.2 m/s 23.9 m/sv = + − =

3. Find the direction of fv : 1 1 17.2 m/stan tan 46.0 or 46.0 below horizontal16.6 m/s

y

x

vv

θ − − −⎛ ⎞= = = − ° °⎜ ⎟⎝ ⎠

Insight: The motion of the football will be discussed in more detail in Chapter 4 when we consider projectile motion. 69. Picture the Problem: The path of the football traces out a parabola as the velocity increases at a constant rate in the

downward direction, but the velocity in the horizontal direction remains constant. Strategy: Use equation 3-6 to find the average acceleration over a time interval tΔ . Then substitute the various time

intervals into the formula to find ava . Solution: 1. (a) Find the average

acceleration as a function of the time interval tΔ . ( ) ( ){ } ( ) ( ){ }

( ) ( )

av

2 2

2

2av

ˆ ˆ ˆ ˆ16.6 m/s 9.81 m/s ( ) 16.6 m/s 9.81 m/s

ˆ9.81 m/sˆ9.81 m/s

t t t

t tt t t

tt

t

+Δ −Δ= =

Δ Δ⎡ ⎤ ⎡ ⎤− + Δ − −⎣ ⎦ ⎣ ⎦=

Δ⎡ ⎤− Δ⎣ ⎦= = −

Δ

v vva

x y x y

ya y

2. Determine ava for tΔ = 1.00 s: ( )2av ˆ9.81 m/s= −a y

3. (b) Repeat for tΔ = 2.50 s: ( )2av ˆ9.81 m/s= −a y

4. (c) Repeat for tΔ = 5.00 s: ( )2av ˆ9.81 m/s= −a y

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



3 – 26

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

Strategy: Let 1g =v velocity of plane 1 relative to the ground, 2g =v velocity of plane 2 relative to the ground, 12 =v velocity of plane 1 relative to plane 2, and 21 =v velocity of plane 2 relative to plane 1 (not pictured). Let north be along the positive y-axis, east along the positive x-axis. Use equation 3-8 to find the components of 12v . Use the components to find the magnitude and direction.

Solution: 1. (a) Find 12v by applying equation 3-8: 1g 12 2g 12 1g 2g = + ⇒ = −v v v v v v

2. Use the given angles to find the components of 12v :

( )( ) ( )

( ) ( )

12 ˆ12 m/s

ˆ ˆ 7.5 m/s cos 20 7.5 m/s sin 20

ˆ ˆ7.0 m/s 9.4 m/s

= −

− ° + °⎡ ⎤⎣ ⎦= +

v y

x y

x y

3. Find the direction of 12v : 12,1 1

12,

9.4 m/stan tan 53 north of east7.0 m/s

y

x

vv

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

4. Find the magnitude of 12v : ( ) ( )2 2

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

5. (b) Since 21 12= −v v the vector 21v has the same magnitude as 12v but points in the opposite direction. Therefore,

21 12 m/s at 53 south of west .= °v

Insight: There are other ways to approach this problem. For instance, in step 1 we could say 12 1g g2= +v v v and then use the fact that g2 2g= −v v to write 12 1g 2g= −v v v . It’s a little awkward to use g2v , which represents the velocity of the ground relative to plane 2, but learning to think about velocity from both perspectives can help you solve difficult vector motion problems.

71. 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 the vectors and then measure the vector sum with a ruler. To use the component method you must independently add the x and y components of each vector.

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

38 ftΔ ≈r

2. (b) Independently add the x and y components of the vector sum: ( )

( )( ) ( )

ˆ0 45 ft 35 ft 0ˆ 51ft 0 0 –13 ft

ˆ ˆ10 ft 38 ft

Δ = + + += + − +

+ + +Δ = +

r A B C Dx

yr x y

3. Find the magnitude of the sum: ( ) ( )2 210 ft 38 ft 39 ftΔ = + =r

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

38 ftx

y

rr

θ − −⎛ ⎞Δ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎜ ⎟Δ ⎝ ⎠⎝ ⎠

A

Insight: When adding vectors graphically you must always ensure you are adding them head-to-tail. The vector sum is a vector that starts at the beginning of the first vector ( A ) and ends at the end of the last vector ( D ).

Δ = + + +r A B C D

D

BC

A

θ

12v

1gv

x

y

2gv

O

θ

20°

θ



3 – 27


Strategy: Use the component method of vector subtraction to find Δv . The average acceleration is then Δv divided by the time elapsed.

Solution: 1. Find the components of iv :

( ) ( )( ) ( )

i ˆ ˆ4.10 m/s cos33.5 4.10 m/s sin 33.5ˆ ˆ3.42 m/s 2.26 m/s

= ° + °

= +

v x y

x y

2. Find the components of fv : ( ) ( )( ) ( )

f ˆ ˆ6.05 m/s cos59.0 6.05 m/s sin 59.0ˆ ˆ3.12 m/s 5.19 m/s

= ° − °

= + −

v x y

x y

3. Subtract the vectors: ( ) ( )( ) ( )

ˆ ˆ3.12 3.42 m/s 5.19 2.26 m/s

ˆ ˆ0.30 m/s 7.45 m/sf iΔ = − = − + − −

= − + −

v v v x y

x y

4. Divide by the time elapsed:

( ) ( ) ( ) ( )2 2av

ˆ ˆ0.30 m/s 7.45 m/sˆ ˆ0.15 m/s 3.73 m/s

2.00 st− + −Δ

= = = − + −Δ

x yva x y

Insight: Subtracting vectors −A B is the same as adding −B to A . Note that acceleration can change both the magnitude and the direction of the velocity vectors.


Strategy: Let bg =v velocity of the bus relative to the ground, rg =v velocity of the raindrops relative to the ground, and rb =v velocity of the raindrops relative to the bus. Apply equation 3-8 to form a right triangle of the velocity vectors as shown in the diagram. Use the right triangle 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

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

rg rb

bg rb

cos 1 1 3.7sin tan tan15

v vv v

θθ θ

= = = =°

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.

bgv

rbv rgvθ

fv

iv

i−v

33.5°

59.0°

Δv



3 – 28

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

Strategy: Find the angles θ and α that correspond to the time of 4:12 and then use the angles to find the components of the position vectors for the tips of the hands. Then subtract the position vectors to find the tip-to-tip distance.

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

θ = × ° = °

2. Find the angle α:

12604

360 126 90 3612

α = × ° = ° − ° = °

3. Find the components of M : ( ) ( ) ( ) ( )

ˆ ˆsin cosˆ ˆ ˆ ˆ14 ft sin 72 14 ft cos 72 13 ft 4.3 ft

M Mθ θ= +

= ° + ° = +

M x yx y x y

4. Find the components of H : ( ) ( ) ( ) ( )

ˆ ˆcos sinˆ ˆ ˆ ˆ9.0 ft cos36 9.0 ft sin 36 7.3 ft 5.3 ft

H Hα α= −

= ° − ° = + −

H x yx y x y

5. Find the components of Δr : ( ) ( )( )( ) ( )

ˆ ˆ13 7.3 ft 4.3 5.3 ft

ˆ ˆ6 ft 9.6 ft

Δ = − = − + − −

= +

r M H x y

x y

6. Find the magnitude of Δr : ( ) ( )2 26 ft 9.6 ft 11 ftrΔ = + =

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

75. Picture the Problem: The velocities of the surfer ssv and the waves wsv

relative to the shore are shown in the diagram at right. Strategy: Set the y component of the surfer’s velocity equal to the

velocity of the waves, and solve for the angle θ. Then apply equation 3-8 to find the surfer’s velocity relative to the wave.

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

ss ws

1 1ws

ss

sin

1.3 m/ssin sin 107.2 m/s

v v

vv

θ

θ − −

=

⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

2. (b) Apply equation 3-8: [ ]ss sw ws sw ss ws ss ss wsˆ ˆ ˆ cos sinv v vθ θ= + ⇒ = − = + −v v v v v v x y y 3. Since ss wssinv vθ = , the

y components cancel out: ( ) ( )sw ss ˆ ˆ ˆcos 7.2 m/s cos10 7.1 m/sv θ= = ° =v x x x

4. (c) If the y component stays the same, but the vector increases in length, the angle it makes with the x-axis must decrease.

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

shore

ssvwsv

x

y

θ

Δ = −r M HM

H

θ

α

x

y



3 – 29

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

width of the river along the x direction is 25.0 m.

Strategy: Let bw =v boat’s velocity relative to the water, bg =v boat’s velocity relative to the ground, and wg =v water’s velocity relative to the ground. Set the y component of bwv equal to the magnitude of wgv so that they cancel, leaving only an x component of bgv . Then determine the angle θ.

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

( )bw, bw wg,

wg, 1 1

bw

sin

1.4 m/ssin sin 13

6.1 m/s

y y

y

v v v

vv

θ

θ − −

= = −

− − −⎛ ⎞⎛ ⎞= = = °⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

2. (b) Since the y components cancel, it follows that the x component of bwv is the same as bgv : ( )bg bw cos 6.1 m/s cos13 5.9 m/sv v θ= = ° =

3. Now find the time to cross the river:

bg

25.0 m 4.2 s5.9 m/s

xtvΔ

Δ = = =

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


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

right.

Strategy: Because 0,+ + =A B C the x and y components of the vectors must

independently sum to zero. Since A has no y component, we can use the known y component of C to find the y component of B . The known angle that B makes with the x axis will yield Bx and give us a way to find the length of A .

Solution: 1. Set the y components equal to zero:

( )( )

0 sin 40.0 0

15 m sin 40.0 9.64 my y y y y y

y

A B C B C A C

B

+ + = ⇒ = − − = − − ° −

= ° =

2. Determine Bx using the tangent function:

9.64 mtan 30.0 16.7 mtan 30.0 tan 30.0

y yx

x

B BB

B° = ⇒ = = =

° °

3. Find Ax and then A : ( ) ( )0 16.7 m 15 m cos 40.0 28 m

28 mx x x x x xA B C A B C+ + = ⇒ = − − = − − ° = −

=A

4. Find the magnitude of B : ( ) ( )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 B in order to avoid rounding error.

C

x

y

B

A O

30.0°

40.0°



3 – 30


Strategy: Let 1g =v velocity of boat 1 relative to the ground, 2g =v velocity of boat 2 relative to the ground, 12 =v velocity of boat 1 relative to boat 2. Let north be along the positive y-axis, east along the positive x-axis. Use equation 3-8 to find the components of 2gv , and use the components to find its magnitude and direction.

Solution: 1. Apply equation 3-8 to find 2gv :

( )( )

( )( ) ( )

1g 12 2g 2g 1g 12

2g

2g

ˆ2.15 m/s sin 47.0ˆ0.775 m/s

ˆ2.15 m/s cos 47.0

ˆ ˆ1.57 m/s 0.691 m/s

= + ⇒ = −

°⎡ ⎤= − ⎢ ⎥

+ °⎢ ⎥⎣ ⎦= − + −

v v v v v v

xv y

y

v x y

2. Find the magnitude of 2gv : ( ) ( )2 22g 1.57 m/s 0.691 m/s 1.72 m/sv = − + − =

3. Find the direction of 2gv : 1 0.691 m/stan 23.8 south of west1.57 m/s

θ − −⎛ ⎞= = °⎜ ⎟−⎝ ⎠

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

79. Picture the Problem: A dragonfly approaches its prey along a path that

makes it appear motionless to the prey.

Strategy: Use vector components to make the y component of dv exactly the same as the y component of p .v

Solution: Set d, p, y yv v=

and solve for dv :

d p

pd

sin

0.750 m/s 1.00 m/ssin sin 48.5°

v v

vv

θ

θ

=

= = =

Insight: In the extreme case of θ = 0° the dragonfly would have to fly infinitely fast in order to match the y component of the prey’s velocity. If the maximum speed of the dragonfly were 2.00 m/s, then it must approach the prey at an angle greater than ( )1sin 0.750 m/s 2.00 m/s 22.0°.θ −= =

80. Picture the Problem: A dragonfly approaches its prey along a path that makes it appear motionless to the prey.

Strategy: Consider the effect of the angle θ on the magnitudes of the components of dv to answer the conceptual question.

Solution: If the angle θ were made larger, a larger fraction of the dragonfly’s velocity would be in the y direction. It would not need to fly as fast in order to match the y component of the prey’s velocity. That means its speed would be less than the 1.00 m/s determined in Problem 79.

Insight: In the extreme case of θ = 90° the dragonfly’s speed would only need to match the prey’s speed of 0.750 m/s.

12v

1gv

x

y

2gvOθ

47.0°



3 – 31

81. Picture the Problem: A dragonfly approaches its prey along a path that




and solve for dv :

d p

pd

sin

0.750 m/s 0.806 m/ssin sin 68.5°

v v

vv

θ

θ

=

= = =

Insight: Our answer to Problem 80 is confirmed; the dragonfly’s speed is less than the 1.00 m/s required in Problem 79. 82. Picture the Problem: A dragonfly approaches its prey along a path that




and solve for θ:

d p

p1 1

d

sin

0.750 m/ssin sin 52.1°0.950 m/s

v v

vv

θ

θ − −

=

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Insight: The dragonfly must adjust its angle constantly to correct for wind and the possibility that its prey might not continue flying in a straight line at constant speed.

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

Strategy: Let bw =v boat’s velocity relative to the water, bg =v boat’s velocity relative to the ground, and wg =v water’s velocity relative to the ground. Set the y component of bwv equal to the magnitude of wgv so that they cancel, leaving only an x component of bgv . Then determine the angle θ.

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

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

vv

θ

θ − −

= = −

− − −⎛ ⎞⎛ ⎞= = = °⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

2. (b) If the speed of the boat is increased, it should make its heading more downstream in order that its y component remains 1.4 m/s and that it still lands directly across the river from its starting point. Therefore the angle needed to go directly across the river will decrease.




3 – 32

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

Strategy: Find the angle that bgv makes with the positive x axis by using the coordinates of the dock (the displacement and velocity vectors are parallel for uniform motion). Apply equation 3-8 to determine the relationship between bgv and bwv . Solve the two equations with two unknowns to find the angle θ. Then the magnitude of bgv can be found from its known components.

Solution: 1. (a) Because the displace ment and velocity vectors are parallel, the ratios of their components are equal:

bg,

bg,

28 m55 m

y y

x x

v rv r

Δ= =

Δ

2. Write equation 3-8 in component form:

[ ]bg bw wg

bg, bg, bw bw wgˆ ˆ ˆ ˆ ˆcos sinx yv v v v vθ θ

= +

+ = + −

v v v

x y x y y

3. Independently equate the x and y components: bg, bw

bg, bw wg

cos

sinx

y

v v

v v v

θ

θ

=

= −

4. Divide the y equation by the x equation: bg, bw wg

bg, bw

sincos

y

x

v v vv v

θθ

−=

5. Square both sides to get everything in terms of sin θ. The units of m/s apply to each of the velocities, but are omitted at right in order to save space:

( )( ) ( )( ) ( )

( ) ( )

2 2 2 2bg, bw wg bw wg

2 2bg, bw

2 2 2bw wg bw wg

2 2bw

22 22

2 2 2

sin 2 sincos

sin 2 sin

1 sin

6.7 sin 2 1.4 6.7 sin 1.42855 6.7 6.7 sin

y

x

v v v v vv v

v v v v

v

θ θθ

θ θ

θ

θ θ

θ

⎛ ⎞ − +=⎜ ⎟⎜ ⎟

⎝ ⎠− +

=−

− +⎛ ⎞ =⎜ ⎟ −⎝ ⎠

6. Rearrange the equation into one that is quadratic in sin θ:

2 2

2

11.6 11.6sin 44.9sin 18.8sin 1.960 56.5sin 18.8sin 9.64θ θ θ

θ θ

− = − +

= − −

7. Apply the quadratic formula.

( ) ( )( )( )

( )

2

1

18.8 18.8 4 56.5 9.64sin 0.279, 0.612

2 56.5

sin 0.612 38 upstream

θ

θ −

± − − −= = −

= = °

8. (b) Now find the components of bgv : ( )( )

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/sx

y

v v

v v v

θ

θ

= = ° =

= − = ° − =

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

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

Physics Chapter 3 Answers

Documents

Transcript of Physics Chapter 3 Answers