Question

A girl walks $4 \,km$ towards West, then she walks $3\, km$ 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 $OA =4 \hat{ i }$
$AB =\hat{ i }| AB | \cos 60^{\circ}+\hat{ j }| AB | \sin 60^{\circ}$
$\left(A B \cos 60^{\circ}\right.$ is component of $A B$ along $X$ -axis and $A B \sin 60^{\circ}$ is component of $A B$ 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
$\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$