Distance vs Displacement| Vector Addition and Subtraction: Graphical Method| Two-Dimensional Kinematics| College Physics Problem 3.1

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

3.1
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(120\:m\times 3\right)+\left(120\:m\times 1\right)

d=480\: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(1\times 120\:m\right)^2+\left(3\times 120\:m\right)^2}=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

 

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