Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 3 - Vectors - Questions - Page 56: 12

Answer

Magnitude=81 km angle=$40^{\circ}$ north of east

Work Step by Step

Here x component is given as : $X=50+25 cos 60^{\circ}$ =62.5 km y component, $Y=30+25 sin 60^{\circ}$ =51.7 km So magnitude is given by: $|S|=\sqrt (x^2+y^2)$ =81 km $\theta=y/x=tan ^- 51.7/62.5$=$40^{\circ}$ north of east.
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