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Q 25P

Expert-verifiedFound in: Page 73

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Your dog is running around the grass in your back** **yard. He undergoes successive displacements $3.50m$**** south, 8.20 m northeast, and 15.0 m west. What is the resultant displacement?**

$\left|\overrightarrow{R}\right|=9.48m,\theta =166\xb0$

- In simple terms, magnitude means 'distance or number.' It depicts the absolute or relative size or direction in which an object moves in the feeling of motion. It's a term for describing the size or scope of something.
- When we talk about direction, we're talking about a straight-line segment that connects two places in space. The direction is from the beginning (first) to the end (second).

Let us consider the given data.

$\overrightarrow{{d}_{1}}=-3.5m\hat{j}\phantom{\rule{0ex}{0ex}}\overrightarrow{{d}_{2}}=\left(8.2\times cos45\xb0m\right)\hat{i}+\left(8.2\times sin45\xb0m\right)\hat{j}=(5.8\hat{i}+5.8\hat{j})m\phantom{\rule{0ex}{0ex}}\overrightarrow{{d}_{3}}=-15m\hat{i}$

We sum the different displacements given in the problem to determine the resultant displacement.

$\overrightarrow{R}=\overrightarrow{{d}_{1}}+\overrightarrow{{d}_{2}}+\overrightarrow{{d}_{3}}\phantom{\rule{0ex}{0ex}}=(-15+5.8)m\hat{i}+(5.8-3.5)m\hat{j}\phantom{\rule{0ex}{0ex}}=(-9.2)m\hat{i}+(2.3)m\hat{j}$

The magnitude and direction of the resultant vector are then determined.

$\left|\overrightarrow{R}\right|=\sqrt{{R}_{x}^{2}+{R}_{y}^{2}}\phantom{\rule{0ex}{0ex}}\left|\overrightarrow{R}\right|=\sqrt{{(-9.2)}^{2}+{(2.3)}^{2}}\phantom{\rule{0ex}{0ex}}\left|\overrightarrow{R}\right|=9.48\text{m}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{{R}_{y}}{{R}_{x}}\right)+180\xb0\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{2.3}{-9.2}\right)+180\xb0\phantom{\rule{0ex}{0ex}}=166\xb0$

Hence, the resultant displacement is $\left|\overrightarrow{R}\right|=9.48m,\theta =166\xb0$

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