College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 3 - Problems - Page 111: 12


(a) The pilot must fly 54.2 miles at an angle of $26.4^{\circ}$ north of east. (b) The pilot must fly an extra distance of 134.2 miles.

Work Step by Step

(a) We can find the total west component of the two extra parts of the flight: $55~mi - (25~mi)~sin~15^{\circ} = 48.5~mi$ We can find the total south component of the two extra parts of the flight: $0 +(25~mi)~cos~15^{\circ} = 24.1~mi$ To fly back to the original destination, the pilot needs to fly 48.5 miles east and 24.1 miles north. We can find the distance $d$: $d = \sqrt{(48.5~mi)^2+(24.1~mi)^2} = 54.2~mi$ We can find the direction north of east: $tan~\theta = \frac{24.1~mi}{48.5~mi}$ $\theta = tan^{-1}(\frac{24.1~mi}{48.5~mi})$ $\theta = 26.4^{\circ}$ (b) We can find the extra distance the pilot must fly: $55~mi+25~mi+54.2~mi = 134.2~mi$
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