# College Physics 3.2 – Distance vs displacement

## Solution:

### Part A

So, the total distance traveled on path B as shown by the red line is just the sum of the distances.

$d=\left(4\times 120\:m\right)+\left(3\times 120\:m\right)+\left(3\times 120\:m\right)$

$d=1200\:m$

### Part B

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by

$s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}$

$s=\sqrt{\left(1\times 120\:m\right)^2+\left(3\times 120\:m\right)^2}$

$s=379\:m$

The direction of the displacement is given by

$\theta =tan^{-1}\left(\frac{1\times 120\:m}{3\times 120\:m}\right)=18.4^{\circ} ,\:East\:of\:North$