## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Problems - Page 118: 111

#### Answer

(a) The plane's velocity relative to the air is $160~km/h$ at an angle of $20.0^{\circ}$ north of east. (b) The magnitude of the plane's velocity relative to the ground is $150.54~km/h$ at an angle of $21.3^{\circ}$ north of east. (c) The wind speed is $10.0~km/h$ and it is blowing directly to the west.

#### Work Step by Step

(a) The plane flies at an airspeed of $160~km/h$ and heads at an angle of $20.0^{\circ}$ north of east. This is the plane's velocity relative to the air. (b) We can find the east component of the plane's velocity relative to the ground: $v_x = \frac{(320~km)~cos~20.0^{\circ}-20~km}{2.0~h} = 140.35~km/h$ We can find the north component of the plane's velocity relative to the ground: $v_y = \frac{(320~km)~sin~20.0^{\circ}}{2.0~h} = 54.72~km/h$ We can find the magnitude of the plane's velocity relative to the ground: $\sqrt{(140.35~km/h)^2+(54.72~km/h)^2} = 150.54~km/h$ We can find the angle $\theta$ north of east: $tan~\theta = \frac{54.72}{140.35}$ $\theta = tan^{-1}(\frac{54.72}{140.35})$ $\theta = 21.3^{\circ}$ The magnitude of the plane's velocity relative to the ground is $150.54~km/h$ at an angle of $21.3^{\circ}$ north of east. (c) We can find the x-component of the wind's velocity $v_{wx}$: $140.35~km/h = (160~km/h)~cos~20.0^{\circ}+v_{wx}$ $v_{wx} = 140.35~km/h - (160~km/h)~cos~20.0^{\circ}$ $v_{wx} = -10.0~km/h$ $v_{wx} = 10.0~km/h$ (west) We can find the y-component of the wind's velocity $v_{wy}$: $54.72~km/h = (160~km/h)~sin~20.0^{\circ}+v_{wy}$ $v_{wx} = 54.72~km/h - (160~km/h)~sin~20.0^{\circ}$ $v_{wx} = 0$ The wind speed is $10.0~km/h$ and it is blowing directly to the west.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.