College Physics (4th Edition)

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

Chapter 3 - Problems - Page 111: 5


Cindy must travel a distance of 4.92 miles at an angle of $24.0^{\circ}$ north of east.

Work Step by Step

The east component of Cindy's displacement is 1.50 mi + 3.00 mi = 4.50 mi east. The north component of Cindy's displacement is 2.00 mi north. We can find the distance Cindy must travel: $\sqrt{(4.50~mi)^2+(2.00~mi)^2} = 4.92~mi$ We can find the direction north of east: $tan~\theta = \frac{2.00~mi}{4.50~mi}$ $\theta = tan^{-1}(\frac{2.00~mi}{4.50~mi})$ $\theta = 24.0^{\circ}$
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