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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

AC 

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. ABC 

 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

ABC 

(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. CAB 

 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.