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: 26b

Answer

$\text{Using Eq. 3-6, the magnitude is }$ $$ |\vec{R}|=\sqrt{(-3.18 \mathrm{m})^{2}+(4.72 \mathrm{m})^{2}}=5.69 \mathrm{m} $$

Work Step by Step

The vector equation is $\vec{R}=\vec{A}+\vec{B}+\vec{C}+\vec{D}$ . Expressing $\vec{B}$ and $\vec{D}$ in unit-vector notation, we have $(1.69 \hat{\mathrm{i}}+3.63 \hat{\mathrm{j}}) \mathrm{m}$ and $(-2.87 \hat{\mathrm{i}}+4.10 \hat{\mathrm{j}}) \mathrm{m},$ respectively. Where the length unit is not displayed in the solution below, the unit meter should be understood. we have $\text{(Eq. 3-6)}$ $$\vec{R}=(-3.18 \mathrm{m}) \hat{\mathrm{i}}+(4.72 \mathrm{m}) \hat{\mathrm{j}}$$ Using Eq. $\text{3-6}$, the magnitude is $$ |\vec{R}|=\sqrt{(-3.18 \mathrm{m})^{2}+(4.72 \mathrm{m})^{2}}=5.69 \mathrm{m} $$
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