Question

A girl walks $4km$ towards West, then she walks $3km$ in a direction $30^{\circ }$ East of North and stops. Then, the girl’s displacement from her initial point of departures is

• A. $-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
• B. $\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$
• C. $-\frac{1}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
• D. None of these

Answer 1

Answer: (A)

Let $O$ and $B$ be the initial and final positions of the girl, respectively.
Then, the girl's position 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 of $AB$ 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
$OB=AO+AB$
$=(-4\hat{i})+\left(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\right)$
$=\left(-4+\frac{3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
$=\left(\frac{-8+3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
Hence, the girl's displacement from her initial point of departure is
$\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$