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Q 25P

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Found in: Page 73

### Physics For Scientists & Engineers

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?

$|\stackrel{\to }{R}|=9.48m,\theta =166°$

See the step by step solution

## Step 1: Define magnitude and direction

• 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).

## Step 2: Determine the resultant displacement vector

Let us consider the given data.

$\stackrel{\to }{{d}_{1}}=-3.5m\stackrel{^}{j}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{d}_{2}}=\left(8.2×cos45°m\right)\stackrel{^}{i}+\left(8.2×sin45°m\right)\stackrel{^}{j}=\left(5.8\stackrel{^}{i}+5.8\stackrel{^}{j}\right)m\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{d}_{3}}=-15m\stackrel{^}{i}$

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

$\stackrel{\to }{R}=\stackrel{\to }{{d}_{1}}+\stackrel{\to }{{d}_{2}}+\stackrel{\to }{{d}_{3}}\phantom{\rule{0ex}{0ex}}=\left(-15+5.8\right)m\stackrel{^}{i}+\left(5.8-3.5\right)m\stackrel{^}{j}\phantom{\rule{0ex}{0ex}}=\left(-9.2\right)m\stackrel{^}{i}+\left(2.3\right)m\stackrel{^}{j}$

## Step 3: Determine the magnitude and direction of the resultant vector

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

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

Hence, the resultant displacement is $|\stackrel{\to }{R}|=9.48m,\theta =166°$