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Q1E

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Found in: Page 121

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

Find the following for path A in Figure,(a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

a. The total distance traveled is 480 m.

b. The magnitude and direction of the displacement from start to finish is 379.4 m & $71.{6}^{°}NE$

See the step by step solution

Step 1: Vector definition

A vector is a quantity that has both magnitude and direction. Displacement, velocity, force, torque, and other vector quantities are examples.

Step 2: Determine the total distance

a) Block = 120 m

For path A, the total distance traveled is:

$=4\mathrm{blocks}\phantom{\rule{0ex}{0ex}}=4×120\mathrm{m}\phantom{\rule{0ex}{0ex}}=480\mathrm{m}$

Therefore the total distance traveled by path A is 480 m.

Step 3: Determine the direction of displacement

b) The magnitude of the displacement from start to finish is:

$\left|\stackrel{⇀}{d}\right|{=}\sqrt{{{d}_{x}}^{2}+{{d}_{y}}^{2}}{=}\sqrt{{\left(1block\right)}^{2}+{\left(3block\right)}^{2}}$

$\begin{array}{rcl}\left|\stackrel{⇀}{\mathrm{d}}\right|& =& \sqrt{1+9}\mathrm{blocks}\\ & =& \sqrt{10}\mathrm{blocks}\\ & =& \sqrt{10}×120\mathrm{m}\\ & =& 3.162×120\mathrm{m}\\ & =& 379.4\mathrm{m}\end{array}$

The direction of the displacement:

$\begin{array}{rcl}\mathrm{\theta }& =& {\mathrm{tan}}^{-1}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\\ & =& {\mathrm{tan}}^{-1}\frac{3\mathrm{block}}{1\mathrm{block}}\\ & =& {\mathrm{tan}}^{-1}\left(3\right)\\ & =& 71.{6}^{°}\mathrm{NE}\end{array}$

Therefore the magnitude and direction of the displacement from start to finish is 379.4 m & $71.{6}^{°}NE$