# College Physics 3.13 – Distance and Displacement

## Solution:

### Part A

The total distance traveled following Path C is

$d=\left(1\times 120\:m\right)+\left(5\times 120\:m\right)+\left(2\times 120\:m\right)+\left(1\times 120\:m\right)+\left(1\times 120\:m\right)+\left(3\times 120\:m\right)$

$d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is

$s_x=0+600+0-120+0-360=120\:m$

The displacement in the y-direction is

$s_y=120+0-240+0+120+0=0\:m$

The total displacement is

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

$s=\sqrt{\left(120\:m\right)^2+\left(0\:m\right)^2}$

$s=120\:m$

The direction angle from th x-axis is given by

$\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|$

$\theta _x=tan^{-1}\left|\frac{0\:m}{120\:m}\right|$

$\theta _x=0^{\circ}$

Therefore, the displacement is 120 m, east.