College Physics by Openstax Chapter 3 Problem 1

Find the following for path A 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 

d=(3×120 m)+(1×120 m)d=480 m  (Answer)\begin{align*} \text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\ \text{d} & =480\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s =(sx)2 +(sy)2s =(1×120 m)2+(3×120 m)2s =379 m  (Answer)\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 }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=tan⁡−1(sxsy)θ=tan⁡−1(1×120 m3×120 m)θ=71.6∘, E of N  (Answer)\begin{align*} \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\ \theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\ \theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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