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: 30j

Answer

$260^{o}$

Work Step by Step

$\vec{w}=\vec{a}-\vec{b}$ $=[4.0m-6.0m]\hat{i}+[-3.0m-8.0m]\hat{j}$ $=(-2.0m)\hat{i}-(11m)\hat{j}$ Note that the vector points to the 3rd quadrant, where $180^{o} < \theta < 270^{o}.$ Using (3-6) $\vec{a}=a_{x}\hat{\mathrm{i}}+a_{y}\hat{\mathrm{j}} ,$ $a=\sqrt{a_{x}^{2}+a_{y}^{2}}$ and $\displaystyle \tan\theta=\frac{a_{y}}{a_{x}}$ (3-6) $\displaystyle \theta=\tan^{-1}\frac{-11m}{-2.0m}=80^{o}$, as given by the calculator, in quadrant 1, but we know that the angle is in quadrant 3, which is why we add (or subtract) $180^{o}$ to the obtained angle $\theta=80^{o}+180^{o}=260^{o}$ (counterclockwise to the +x axis)
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