# Distance vs Displacement| Vector Addition and Subtraction: Graphical Method| Two-Dimensional Kinematics| College Physics Problem 3.2

### Find the following for path B in Figure 3.54: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.


Figure 3.54 The various lines represent paths taken by different people
walking in a city. All blocks are 120 m on a side.

SOLUTION:

PART A

So, the total distance traveled on path A as shown by the green 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

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