Question

A girl walks $4km$ towards West. Then, she walks $3km$ in a direction $30^{\circ }$ East to North and stops. The girls displacement from her initial point of departure is

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

Answer 1

Answer: (C)

Let $O$ and $B$ be the initial and final positions of the girl respectively.
Then, the girls'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 $AB$ isin $60^{\circ }$ is component of $AB$ along $Y$-axis)
$=\hat{i}\left(3\times \frac{1}{2}\right)+\hat{j}\left(3\times \frac{\sqrt{3}}{2}\right)=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
By the triangle law of vector addition, we have
$OB=OA+AB=(-4\hat{i})+\left(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\right)$
$=\left(-4+\frac{3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
$=\left(\frac{-8+3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
Hence, the girl's displacement from her initial point of departure is $\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$.