Question

Find the following for path B 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 are m on a side.

Short answer

a. The total distance traveled is 1200 m .

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

Step-by-step solution

Determine total distance travel

A physical quantity is a material's physical property that can be quantified, such as mass, length, etc.

1 Block = 120 m.

For path B, the total distance traveled is:

= 4 blocks + 3 blocks + 3 blocks

= 10 blocks

$$\begin{gathered} =10\times 20 \\ =1200\mathrm{m} \end{gathered}$$

Therefore, the total distance traveled by path B is 1200 m.

Determine the direction of displacement

The magnitude of the displacement from start to finish is:

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

The direction of the displacement:

$$\begin{gathered} =\sqrt{1+9}\mathrm{blocks} \\ =\sqrt{10}\mathrm{blocks} \\ =\sqrt{10}\times 120\mathrm{m} \\ =3.162\times 120\mathrm{m} \\ =379.4\mathrm{m} \end{gathered}$$

The direction of the displacement:

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

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