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 - Problems - Page 58: 20b


$4.3^{\mathrm{o}}$ south of due west.

Work Step by Step

From part (a), $\vec{b}$=$(-5.01 \mathrm{k}\mathrm{m}) \hat{\mathrm{i}}-(0.38 \mathrm{k}\mathrm{m}) \hat{\mathrm{j}}$, (pointing to quadrant III). Using (3-6) for the angle $\displaystyle \tan^{-1}[\frac{-0.38 \mathrm{k}\mathrm{m}}{-5.01 \mathrm{k}\mathrm{m}}]=4.3^{\mathrm{o}} $(calc. gives angle in q. I) In q.III, $\theta=4.3^{\mathrm{o}}+180.0^{\mathrm{o}}=184.3^{\mathrm{o}},$ or $4.3^{\mathrm{o}}$ south of due west.
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