Question

A girl walks $4km$ towards west,then she walk $3km$ in a direction $30°$east of north and stops.Determine the girls displacement from her initial point to 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
$\overrightarrow{OA}=-4\hat{i}$
$\overrightarrow{AB}=\hat{i}|\overrightarrow{AB}|cos60°+\hat{j}|\overrightarrow{AB}|sin 60°$
$=\hat{i}3\times\frac{1}{2}+j^{3}\times\frac{\sqrt{3}}{2}$
$=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
By the triangle law of vector addition,we have:
$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$
$=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{3}\hat{j})$
$=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
$=(-8+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
$=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
Hence,the girl's displacement from her initial point of departure is
$\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$