IB PHYSICS SL. NORDEEN. Solutions to HW 21 Problems Giancoli pp. 64-65 #1
-19 Odd. 1. The resultant vector displacement of the car is given by. R west.
IB PHYSICS SL NORDEEN
Solutions to HW 21 Problems Giancoli pp. 64-65 #1-19 Odd 1.
The resultant vector displacement of the car is given by DR Dwest Dsouth- . The westward displacement is west
Dwest DR
Dsouth-
215 85cos 45o 275.1 km and the south displacement is
west
85sin 45o 60.1 km . The resultant displacement has a magnitude of
275.12 60.12 281.6 km
282 km . The direction is tan 1 60.1 275.1 12.3o 12o south of west .
3.
relationship V1 + X = V2 via the tail-to-tip method. Thus X V2 V1 .
5.
The vectors for the problem are drawn approximately to scale. The resultant has a length of 58 m and a direction 48o north of east. If calculations are done, the actual resultant should be 57.4 m at 47.5o north of east.
X
V2
Label the “INCORRECT” vector as vector X . Then Fig. 3-6 (c) illustrates the
V1
V3 VR V1 V2 V3
V2
V1 7.
(a)
See the accompanying diagram
(b) Vx 14.3cos 34.8o 11.7 units (c)
V V V 2 x
2 y
11.7
Vy 14.3sin 34.8o 8.16 units 2
8.16 14.3 units 2
V 34.8o
Vx
tan 1
9.
(a)
8.16 11.7
34.8o above the x axis
vnorth 735 km h cos 41.5o 550 km h
vwest 735 km h sin 41.5o 487 km h
(b) d north vnorth t 550 km h 3.00 h 1650 km
Vy
IB PHYSICS SL NORDEEN d west vwest t 487 km h 3.00 h 1460 km
11.
Ax 44.0 cos 28.0o 38.85 Cx 31.0 cos 270o 0.0
A C
x
AC
13.
C y 31.0sin 270o 31.0
38.85 0.0 38.85
38.85
2
Ay 44.0sin 28.0o 20.66
A C
51.66 64.6
20.66 31.0 51.66
tan 1
2
Ax 44.0 cos 28.0o 38.85
y
51.66
53.1o
38.85
Ay 44.0sin 28.0o 20.66
Bx 26.5 cos 56.0o 14.82
By 26.5sin 56.0 o 21.97
Cx 31.0 cos 270o 0.0
C y 31.0 sin 270 o 31.0
(a)
A B C
A B C
y
x
38.85 14.82 0.0 53.67
20.66 21.97 31.0 32.31
Note that since the x component is positive and the y component is negative, the vector is in the 4th quadrant. ABC
53.67
2
32.31 62.6
tan 1
2
32.31 53.67
31.0o below x axis
(b) A B C x 38.85 14.82 0.0 24.03
A B C
y
ABC
(c)
20.66 21.97 31.0 73.63
24.03
C A B
C A B
y
x
2
73.63 77.5 2
tan 1
73.63 24.03
71.9o
0.0 38.85 14.82 24.03
31.0 20.66 21.97 73.63
Note that since both components are negative, the vector is in the 3rd quadrant. CAB
24.03
2
73.63 77.5 2
tan 1
73.63 24.03
71.9o below x axis
IB PHYSICS SL NORDEEN Note that the answer to (c) is the exact opposite of the answer to (b). 15. The x component is negative and the y component is positive, since the summit is to the west of north. The angle measured counterclockwise from the positive x axis would be 122.4o. Thus the components are found to be x 4580 sin 32.4o 2454 m
y 4580 cos 32.4 o 3867 m
r 2450 m,3870 m,2450 m
r
2454
2
z 2450 m
4580 2450 5190 m 2
2
17. Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, vx 0 3.5 m s and ax 0 . In the vertical direction, v y 0 0 , a y 9.80 m s 2 , y0 0 , and the final location y 6.5 m . The time for the tiger to reach the ground is found from applying Eq. 2-11b to the vertical motion.
y y0 v y 0t 12 a y t 2
6.5m 0 0 12 9.8 m s 2 t 2
t
2 6.5m 9.8 m s 2
1.15 sec
The horizontal displacement is calculated from the constant horizontal velocity. x vx t 3.5 m s 1.15 sec 4.0 m
2.5
19.
Apply the range formula from Example 3-8. 2
R
v02 sin 2 0 g
sin 2 0
Rg 2 0
v
1.5
2.0 m 9.8 m 2 6.8 m s
2 0 sin 1 0.4239
s2
0.4239
0 13o , 77 o
1
0.5
0 0
0.5
1
1.5
2
-0.5
There are two angles because each angle gives the same range. If one angle is 45o , then
45o is also a solution. The two paths are shown in the graph.