Question

A girl walks 4 km towards west then, she walks 3 km in a direction ${30}^{\circ }$ 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 girls'sposition can be shown as in the figure. Now, we have OA = $4\hat{i}$
$AB=\hat{i}|AB|cos{60}^{\circ }+\hat{j}|AB|sin{60}^{\circ }$
(AB cos ${60}^{\circ }$ is component AB, along X-axis and AB sin ${60}^{\circ }$ is component of AB = along Y-axis)
=$\hat{i}3\times \frac{1}{2}+\hat{j}3\times \frac{\sqrt{3}}{2}=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
By the triangle law of vector addition, we have
$OB=AO+AB=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j})=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
$=\left(\frac{-8+3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
Hence, the girl's displacement from her initial point of departure is $\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}.$