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: 20a

Answer

5.0 km

Work Step by Step

Selecting $\hat{\mathrm{i}}$ as "east", $\hat{\mathrm{j}}$ as "north", the explorer wanted his displacement to be $\vec{r}=(5.6$ km $)\hat{\mathrm{j}}$. His real displacement is $\vec{a}=(7.8\mathrm{k}\mathrm{m})(\mathrm{c}\mathrm{o}\mathrm{s}50^{\mathrm{o}}\hat{\mathrm{i}}+\sin 50^{\mathrm{o}}\hat{\mathrm{j}})\\=(5.01\mathrm{k}\mathrm{m})\hat{\mathrm{i}}+(5.98\mathrm{k}\mathrm{m})\hat{\mathrm{j}}$ Now, we need a displacement $\vec{b}$ such that $\vec{r}=\vec{a}+ \vec{b}.$ $\vec{b}=\vec{r}-\vec{a} =(-5.01 \mathrm{k}\mathrm{m}) \hat{\mathrm{i}}-(0.38 \mathrm{k}\mathrm{m}) \hat{\mathrm{j}}$ ---- (a) Using (3-6) for the magnitude, $b= \sqrt{(-5.01\mathrm{k}\mathrm{m})^{2}+(-0.38\mathrm{k}\mathrm{m})^{2}}=5.0$ km.
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