Question

A girl walks $4km$ towards west, then she walk $3km$ in a direction ${30}^o$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Answer 1

Let $O$ and $B$ be the initial and final positions of the girl respectively.

Then, the girl's position can be shown as:

Now, we have:

$\overrightarrow{OA}=-4\hat{i}$

$\overrightarrow{AB}=\hat{i}|\overrightarrow{AB}|\cos {60}^{\circ }+\hat{j}|\overrightarrow{AB}|\sin {60}^{\circ }$

$=\hat{i}3\times \frac{1}{2}+\hat{j}3\times \frac{\sqrt{3}}{2}$

$=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$

By the triangle law of vector addition, we have:

$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$

$=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j})$

$=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$

$=\left(\frac{-8+3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$

$=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$

Hence, the girl's displacement from her initial point of departure is

$\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$.