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 

d=(4×120m)+(3×120 m)+(3×120 m)d=1200 m  (Answer)\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 

s=(sx)2+(sy)2s=(1×120 m)2+(3×120 m)2s=(120 m)2+(360 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} & = \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

θ=arctan⁡(sysx)θ=arctan⁡(360 m120 m)θ=71.6∘, N of E (Answer)\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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