College Physics 3.2 – Distance vs displacement


Find the following for path B in Figure 3.52:

(a) the total distance traveled, and

(b) the magnitude and direction of the displacement from start to finish.

Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
Figure 3.52 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 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