Chapter 3 solutions
8.
y
Picture the Problem: The vector direction is
measured counterclockwise from the +x axis.
Strategy: In each case find the vector components.
y
35.0°
65.0°
x
(a)
Solution: 1. (a) Find the x and y components:
x
(b)
rx  r cos    75 m  cos 35.0  61 m
ry  r sin    75 m  sin 35.0  43 m
2. (b) Find the x and y components:
rx  r cos    75 m  cos 65.0  32 m
ry  r sin    75 m  sin 65.0  68 m
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 . Then use the Pythagorean
Theorem to determine the magnitude of r .
Solution: 1. (a) Use the inverse tangent function
to find the distance angle  :
 9.5 
  tan 
  34 or 34° below the
 14 
+x axis
2. (b) Use the Pythagorean Theorem to determine
the magnitude of r :
r  rx2  ry2 
3. (c) If both rx and ry are doubled, the direction
  tan 1 
will remain the same but the magnitude will
double:
r
1
14 m 
2
  9.5 m 
r  17 m
 9.5  2 
  34
 14  2 
 28 m    19 m 
2
2
 34 m
2
y
14 m
x
−9.5 m
20.
Picture the Problem: The two vectors A (length 40.0 m) and B (length 75.0 m) are drawn at
right.
y
50.0°
Strategy: Add vectors A and B using the vector component method.
20.0°
x
Solution: 1. (a) A sketch (not to scale) of the vectors and their sum is shown at right.
21.
2. (b) Add the x components:
Cx  Ax  Bx   40.0 m  cos  20.0    75.0 m  cos  50.0   85.8 m
3. Add the y components:
Cy  Ay  By   40.0 m  sin  20.0    75.0 m  sin  50.0   43.8 m
4. Find the magnitude of C :
C  Cx2  C y2 
5. Find the direction of C :
C  tan 1 
 Cy
 Cx
85.8 m 
2
  43.8 m   96.3 m
2

1  43.8 m 
  tan 
  27.0
 85.8 m 

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.
26.
2. (b) Subtract the x components:
Dx  Ax  Bx   220 km  cos 180  32   140 km  cos 90  65   310 km
3. Subtract the y components:
Dy  Ay  By   220 km  sin 180  32   140 km  sin 90  65   57 km
4. Find the magnitude of D:
D  Dx2  Dy2 
5. Find the direction of D:
 D  tan 1 
 310 km 
2
  57 km   320 km  3.2 105 m
2
 Dy 
1  57 km 
  tan 
  10  180  170 or 10 north of west
 310 km 
 Dx 
Picture the Problem: The vectors involved in the problem are depicted
at right.
y
30°
x
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.
45°
Solution: 1. (a) Make estimates
from the drawing:
A  B  C  20 m
2. (b) Add the vector components:
A  B  C  0   20.0 m  cos 45   7.0 m  cos  30   xˆ 
  1.5
 10.0 m    20.0 m  sin 45   7.0 m  sin  30   yˆ
A  B  C   20.2 m  xˆ   0.64 m  yˆ
 20.2 m
 0.64 m 
  1.8
 20.2 m 
  tan 1 
4. Use the components to find the angle:
32.
 20.2 m 2   0.64 m 2
ABC 
3. Use the components to find the magnitude:
Picture the Problem: The vectors involved in the problem are depicted at right.
y
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:
 12 m 
  –26
 25 m 
 A  tan 1 
O
41.
 25 m2   –12 m 2
2. Find the magnitude of A :
A
 28 m
3. (b) Find the direction of B from its
components:
B  tan 1 
4. Find the magnitude of B :
B
5. (c) Find the components of A  B :
A  B   25  2.0 m  xˆ   12  15 m  yˆ   27 m  xˆ  3.0 m  yˆ
6. Find the direction of A  B from its
components:
 AB  tan 1 
7. Find the magnitude of A  B :
AB 
25 m
x
12 m
 15 m 
  82
 2.0 m 
 2.0 m2  15 m2
 15 m
 3.0 m 
  6.3
 27 m 
 27 m 2  3.0 m 2
 27 m
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.
51.
 ry
 rx

1  2500 ft 
  tan 
  59 north of east
 1500 ft 

Solution: 1. Find the direction of the displacement:
  tan 1 
2. Find the magnitude of the displacement:
r
1500 ft 
3. Find the magnitude of the average velocity:
vav 
r
890 m

 4.9 m/s
t 3.0 min  60 s/min
2
  2500 ft   2900 ft  0.305 m/ft  890 m
2
Picture the Problem: The vectors involved in this problem are depicted at right.
Strategy: Let v yw  your velocity with respect to the walkway, v wg  walkway’s
velocity with respect to the ground, and add the vectors according to equation 3-8 to find
v yg  your velocity with respect to the ground. Then find the time it takes you to travel
the 85-m distance.
61.
Solution: 1. Find your velocity
with respect to the walkway:
 x 
 85 m 
v yw    xˆ  
 xˆ  1.25 m/s  xˆ
 t 
 68 s 
2. Apply equation 3-8 to find your
velocity with respect to the ground:
v yg  v yw  v wg  1.25 m/s  xˆ   2.2 m/s  xˆ   3.45 m/s  xˆ
3. Now find the time of travel:
t
x
85 m

 25 s
vyg 3.45 m/s
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:
2A  B  2 12.1 m  xˆ   32.2 m  yˆ   24.2 m  xˆ   32.2 m  yˆ
2. Find the angle:
  tan 1 
3. Find the magnitude of the vector sum:
2A  B 
 32.2 m 
  53.1 or 307
 24.2 m 
 24.2 m   32.2 m
2
2
 40.3 m