Question

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 are120 mon a side.

Short answer

a. The total distance traveled is 480 m.

b. The magnitude and direction of the displacement from start to finish is 379.4 m & $71.6^{°}NE$

Step-by-step solution

Vector definition

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

Determine the total distance

a) Block = 120 m

For path A, the total distance traveled is:

$$\begin{gathered} =4\mathrm{blocks} \\ =4\times 120\mathrm{m} \\ =480\mathrm{m} \end{gathered}$$

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

Determine the direction of displacement

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

$\left|\stackrel{\rightharpoonup }{d}\right|=\sqrt{{d_x}^2+{d_y}^2}=\sqrt{{\left(1block\right)}^2+{\left(3block\right)}^2}$

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

The direction of the displacement:

$\begin{array}{rcl}\mathrm{\theta }& =& {\tan }^{-1}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\\ & =& {\tan }^{-1}\frac{3\mathrm{block}}{1\mathrm{block}}\\ & =& {\tan }^{-1}\left(3\right)\\ & =& 71.6^{°}\mathrm{NE}\end{array}$

Therefore the magnitude and direction of the displacement from start to finish is 379.4 m &$71.6^{°}NE$