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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

Chapter 1: Simple Stress
Chapter 2: Simple Strain
Chapter 3:
Torsion
Chapter 4: Shear and Moment in Beams
Chapter 5: Stresses in Beams
Chapter 6: Beam Deflections
Chapter 7: Restrained Beams
Chapter 8: Continuous Beams
Chapter 9: Combined Stresses
Chapter 10: Reinforced Beams
Chapter 11: Columns
Chapter 12: Riveted, Bolted, and Welded Connections
Chapter 13: Special Topics
Chapter 14: Inelastic Action
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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

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

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Hydrology and Floodplain Analysis Solution Manual

Hydrology and Floodplain Analysis 5th Edition Solution Manual by Engineering-Math.Org

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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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