## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Problems - Page 116: 84

#### Answer

(a) The plane must head at an angle of $76.4^{\circ}$ north of east. (b) The flight takes 2.72 hours.

#### Work Step by Step

(a) In order to travel directly north, the east component of the plane's airspeed must be equal in magnitude to the west component of the wind's speed. Let $\theta$ be the angle north of east at which the plane heads. We can find the angle $\theta$: $(300.0~km/h)~cos~\theta = (100.0~km/h)~cos~45^{\circ}$ $cos~\theta = \frac{(100.0~km/h)~cos~45^{\circ}}{300.0~km/h}$ $\theta = cos^{-1}(\frac{(100.0~km/h)~cos~45^{\circ}}{300.0~km/h})$ $\theta = 76.4^{\circ}$ To travel directly north, the plane must head at an angle of $76.4^{\circ}$ north of east. (b) We can find the north component of the plane's groundspeed: $v_y = (300.0~km/h)~sin~76.4^{\circ} - (100.0~km/h)~sin~45^{\circ}$ $v_y = 220.88~km/h$ We can find the time it takes to fly 600.0 km north: $t = \frac{d}{v_y} = \frac{600.0~km}{220.88~km/h} = 2.72~h$ The flight takes 2.72 hours.

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