Question

A person going for a walk follows the path shown in FigureP3.51. The total trip consists of four straight line paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point?

Short answer

The person’s resultant displacement measured from the starting point is $240m$

Step-by-step solution

Finding the displacements of four individuals

The resultant displacement $\overrightarrow{R}$is equal to the four individual displacements,

$\overrightarrow{R}=\overrightarrow{d1}+\overrightarrow{d2}+\overrightarrow{d3}+\overrightarrow{d4}$

The individual displacement in unit vector notations:

$$\begin{gathered} 1=100\hat{i}m \\ \overrightarrow{d2}=-300\hat{j}m \\ \overrightarrow{d3}=(-150\cos 30°)\hat{i}m+(-150\sin 30°)\hat{j}m=-130\hat{i}m-75\hat{j}m \\ \overrightarrow{d4}=(-200\cos 60°)\hat{i}m+(200\sin 60°)\hat{j}m=-100\hat{i}m+173\hat{j}m \end{gathered}$$

Summing the components

The sum of all four components, we have

$$\begin{gathered} Rx=d_{1x}+d_{2x}+d_{3x}+d_{4x}=(100+0-130-100)m=-130m \\ Ry=d_{1y}+d_{2y}+d_{3y}+d_{4y}=(0-300-75+173)m=-220m \end{gathered}$$

Therefore, $\overrightarrow{R}=\overrightarrow{d1}+\overrightarrow{d2}+\overrightarrow{d3}+\overrightarrow{d4}=(-130\hat{i}-202\hat{j})m$

Its magnitude is, $\sqrt{{\left(103\right)}^2+{\left(202\right)}^2}=240m$