PHYSICS 171
AQ 2009
Solutions to
Homework #3
(due Friday, October 2, 2009)
VR  V1  V2
 V3
1. Giancoli Chapter 3, Problem 4
The vectors for the problem are drawn approximately to scale. The
resultant has a length of 17.5 m and a direction 19 north of east.
If calculations are done, the actual resultant should be 17 m at 23o
north of east.
2. Giancoli Chapter 3, Problem 8
(a) V1  6.0ˆi  8.0ˆj
V1 
  tan 1
8.0
 127
6.0
2
2
1 5.0
 312
(b) V2  4.5ˆi  5.0ˆj V2  4.5  5.0  6.7   tan
4.5
(c) V1  V2  6.0ˆi  8.0ˆj  4.5ˆi  5.0ˆj  1.5ˆi  3.0ˆj

6.02  8.02  10.0
 

V1  V2  1.52  3.02  3.4
  tan 1
3.0
 117
1.5
(d) V2  V1  4.5ˆi  5.0ˆj  6.0ˆi  8.0ˆj  10.5ˆi  13.0ˆj

 

V2  V1  10.52  13.02  16.7
  tan 1
13.0
10.5
 309
3. Giancoli Chapter 3, Problem 10
Ax  44.0 cos 28.0  38.85
Ay  44.0sin 28.0  20.66
Bx  26.5cos 56.0  14.82
By  26.5sin 56.0  21.97
Cx  31.0cos 270  0.0
C y  31.0sin 270  31.0
(a)
 A + B + C
 A + B + C
y
x
 38.85   14.82   0.0  24.03  24.0
 20.66  21.97   31.0   11.63  11.6
V2
V1
VR
V3
(b)
 24.03
A+B+C 
  tan 1
11.63
24.03
2
 11.63  26.7
2
 25.8
4. Giancoli Chapter 3, Problem 16
Do this exercise with the numbers in the book, and again using halved numbers.
(a)
Use the Pythagorean theorem to find the possible x components.
90.02  x 2   55.0
(b)

2
x 2  5075
 x  71.2 units
Express each vector in component form, with V the vector to be determined.
71.2ˆi  55.0ˆj  V ˆi  V ˆj  80.0ˆi  0.0ˆj 

 
x
y

Vx   80.0  71.2   151.2
Vy  55.0
V  151.2ˆi  55.0ˆj
With halved numbers:
(a) 452= x2+(27.5)2 so x = ± 35.6 units
(b) V = (-40-35.6) i + 27.5 j = -75.6 i + 27.5 j  Halved results! All linear eqns.
5. Giancoli Chapter 3, Problem 22
Choose downward to be the positive y direction for this problem. Her acceleration is directed
along the slope.
(a) The vertical component of her acceleration is directed downward, and its magnitude will
be


given by a y  a sin   1.80 m s2 sin 30.0o  0.900 m s 2 .
(b) The time to reach the bottom of the hill is calculated from Eq. 2-12b, with a y
displacement of
325 m, v y 0  0, and a y  0.900 m s2 .
y  y0  v y 0t  12 a y t 2  325 m  0  0 
t
2  325 m 
 0.900 m s 
2
1
2
 0.900 m s   t 
2
2

 26.9 s
6. Giancoli Chapter 3, Problem 24
Since the acceleration vector is constant, Eqs. 3-13a and 3-13b are applicable. The particle
reaches its maximum x coordinate when the x velocity is 0. Note that v 0  5.0 m s ˆi and
r0  0.


v  v 0  at  5.0 ˆi m s  3.0t ˆi  4.5t ˆj m s
v x   5.0  3.0t  m s  v x  0   5.0  3.0t x  max  m s  t x  max 
5.0 m s
3.0 m s 2
 1.67 s
v  t x  max   5.0 ˆi m s   3.0 1.67  ˆi  4.5 1.67  t ˆj m s  7.5 m s ˆj
r  r0  v 0t  12 at 2  5.0 t ˆi m 
1
2
 3.0t

ˆi  4.5t 2 ˆj m
2
2
2
r  t x  max   5.0 1.67  ˆi m s  12  3.0 1.67  ˆi  4.5 1.67  ˆj m  4.2ˆi m  6.3ˆj m


7. Giancoli Chapter 3, Problem 42
Consider the downward vertical component of the motion, which will occur in half the total
time. Take the starting position to be y = 0, and the positive direction to be downward. Use
Eq. 2-12b with an initial vertical velocity of 0.
2
y  y0  v0 y t  a y t
1
2
2
 h  0  0  gt
1
2
2
down
 t  9.80 t 2  1.225t 2  1.2t 2
 g  
8
2
1
2
8. Giancoli Chapter 3, Problem 46
Choose the origin to be at ground level, under the place where the projectile is launched, and
upwards to be the positive y direction. For the projectile, v0  65.0 m s , 0  35.0,
a y   g , y0  115 m, and v y 0  v0 sin  0 .
(a)
The time taken to reach the ground is found from Eq. 2-12b, with a final height
of 0.
y  y0  v y 0t  12 a y t 2  0  y0  v0 sin  0t  12 gt 2 
 v0 sin  0  v02 sin 2  0  4   12 g  y0
t
(b)
 9.964 s ,  2.3655s  9.96s
2   12 g 
Choose the positive time since the projectile was launched at time t = 0.
The horizontal range is found from the horizontal motion at constant velocity.
x  v x t   v0 cos  0  t   65.0 m s   cos 35.0   9.964 s   531m
(c) At the instant just before the particle reaches the ground, the horizontal component of its
velocity is the constant v x  v0 cos  0   65.0 m s  cos 35.0  53.2 m s . The vertical
component is found from Eq. 2-12a.
v y  v y 0  at  v0 sin 0  gt   65.0 m s  sin 35.0   9.80 m s 2   9.964 s 
 60.4 m s
(d) The magnitude of the velocity is found from the x and y components calculated in part (c)
above.
v
v x2  v 2y 
 53.2 m s 2   60.4 m s 2
 80.5 m s
(e) The direction of the velocity is   tan 1
vy
vx
 tan 1
60.4
53.2
 48.6 , and so the object is
moving 48.6 below the horizon .
(f) The maximum height above the cliff top reached by the projectile will occur when the yvelocity is 0, and is found from Eq. 2-12c.
2
2
v y  v y 0  2a y  y  y0   0  v02 sin 2  0  2 gymax
ymax 
v02 sin 2  0
2g

 65.0 m s 2 sin 2 35.0

2 9.80 m s2

 70.9 m
9. Giancoli Chapter 3, Problem 62
Call the direction of the boat relative to the water the x direction, and upward the y direction.
Also see the diagram.
v passenger  v passenger  v boat rel.
rel. water
rel. boat
water

  2.12ˆi  0.42ˆj m s

 0.60 cos 45ˆi  0.60sin 45ˆj m s  1.70ˆi m s
v passenger
rel. water
v passenger

rel. boat
v boat rel.
water
10. Giancoli Chapter 3, Problem 70
Call the direction of the flow of the river the x direction (to the left in the diagram), and the
direction straight across the river the y direction (to the top in the diagram). From the
diagram,   tan 1 120 m 280 m  23. Equate the vertical components of the velocities to
find the speed of the boat relative to the shore.
vboat rel. cos   vboat rel. sin 45 
v water rel.
shore
120 m
water
vboat rel.   2.70 m s 
sin 45
 2.07 m s
cos 23
Equate the horizontal components of the velocities.
vboat rel. sin   vboat rel. cos 45  vwater

v boat rel.
shore
shore
shore
vwater
rel. shore
water
rel. shore
 vboat rel. cos 45  vboat rel. sin 
water
shore
  2.70 m s  cos 45   2.07 m s  sin 23  1.10 m s
shore
280 m
v boat rel.
water

45o
11. Giancoli Chapter 3, Problem 77
Choose upward to be the positive y direction. The origin is the point from which the pebbles
are released. In the vertical direction, a y  9.80 m s2 , the velocity at the window is
v y  0, and the vertical displacement is 8.0 m. The initial y velocity is found from Eq. 2-
12c.
v 2y  v 2y 0  2a y  y  y0 
vy0 

v 2y  2a y  y  y0  

0  2 9.80 m s 2
 8.0 m   12.5 m s
Find the time for the pebbles to travel to the window from Eq. 2-12a.
v y  v y 0 0  12.5m s
v y  v y 0  at  t 

 1.28s
a
9.80 m s2
Find the horizontal speed from the horizontal motion at constant velocity.
x  v x t  v x  x t  9.0 m 1.28 s  7.0 m s
This is the speed of the pebbles when they hit the window.