# Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.14

### Find the following for path D in Figure 3.58: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

SOLUTION:

Part A

The total distance traveled following Path D is

$d=\left(2\times 120\:m\right)+\left(6\times 120\:m\right)+\left(4\times 120\:m\right)+\left(1\times 120\:m\right)$

$d=1560\:m$

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

$s_x=0+720\:m+0-120\:m=600\:m$

The displacement in the y-direction is

$s_y=-240\:m+0+480\:m+0=240\:m$

The total displacement is

$s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}=\sqrt{\left(600\:m\right)^2+\left(240\:m\right)^2}=646.22\:m$

The direction angle from th x-axis is given by

$\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|=tan^{-1}\left|\frac{240}{600}\right|=21.8^{\circ} \:North\:of\:East$

Therefore, the displacement is 646.22 m, directed at 21.8 degrees north of east.