Question

A girl walks $4\, km$ towards West. Then, she walks $3\, km$ 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 $OA =-4 \hat{ i }$
$A B=\hat{i}|A B| \cos 60^{\circ}+\hat{j}|A B| \sin 60^{\circ}$

( $A B \cos 60^{\circ}$ is component of $A B$ along $X$-axis and $A B$ isin $60^{\circ}$ is component of $A B$ 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
$ O B=O A+A B =(-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 $\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$.