Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 50

Answer

44.4$^o$ north of east.

Work Step by Step

Call east the positive x direction and north the positive y direction. Let P denote the Plane, A the Air, and G the Ground. The pair “PG”, for example, represents the plane’s motion relative to the ground. $$ \vec{v_{PG}} = \vec{v_{PA}} + \vec{v_{AG}} $$ See the diagram. The vectors form a triangle. Apply the Law of Sines. $$\frac{ v_{AG}}{sin \alpha} = \frac{ v_{PA}}{sin 128 ^{\circ} }$$ We find the small angle $\alpha$ to be 6.4$^o$. Therefore the heading of the plane should be 38 + 6.4 = 44.4$^o$ north of east.
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