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Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot Complete Solution Manual


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Mechanics of Materials: An Integrated Learning System 4th Edition Solution Manual by Engineering-Math.org

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Mechanics of Materials Eleventh Edition by R.C. Hibbeler Complete Solution Manual

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Strength of Materials Fourth Edition by Andrew Pytel and Ferdinand Singer Featured Image

Strength of Materials 4th Edition by Andrew Pytel and Ferdinand Singer



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Purchase Complete Solution Manual of University Physics with Modern Physics 15th Edition by Young and Freedman


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University Physics with Modern Physics 15th Edition Complete Solution Guides

University Physics with Modern Physics 15th Edition by Young and Freedman Complete Solution Guides

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Purchase Hydrology and Floodplain Analysis 5th Edition Solution Manual


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Hydrology and Floodplain Analysis Solution Manual

Hydrology and Floodplain Analysis 5th Edition Solution Guide

This is a PDF copy of the complete guide to the problems and exercises of the book Hydrology and Floodplain Analysis 5th Edition by Bedient, Huber, and Vieux. Expect the copy to be sent to your email address within 24 hours. If you have not heard from us within 24 hours, kindly send us a message to [email protected]

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College Physics by Openstax Chapter 3 Problem 2


Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side


Solution:

Part A

The total distance traveled is 

\begin{align*}

\text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\
\text{d} & = 1 200\ \text{m}  \ \qquad \  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}
\text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\
\text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\
\text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\
\text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The direction is

\begin{align*}
\theta & = \arctan \left( \frac{s_y}{s_x} \right) \\
\theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\
\theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

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