A girls walks \(4 \mathrm{~km}\) towards West. Then, she walks \(3 \mathrm{~km}\) 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

  \(-\frac{5}{2} \hat{i}+\frac{3}{2} \hat{j}\)
• Correct answer
  C

  \(-\frac{5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}\)
• D

  None of these

Solution

Let \(\mathrm{O}\) and \(\mathrm{B}\) 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 \(\mathrm{OA}=-4 \hat{\mathrm{i}}\) \(\mathrm{AB}=\hat{\mathrm{i}}|\mathrm{AB}| \cos 60^{\circ}+\hat{\mathrm{j}}|\mathrm{AB}| \sin 60^{\circ}\)
(AB \(\cos 60^{\circ}\) is component of \(\mathrm{AB}\) along \(\mathrm{X}\)-axis and \(\mathrm{AB} \sin 60^{\circ}\) is component of \(\mathrm{AB}\) along \(\mathrm{Y}\) axis) \(=\hat{\mathrm{i}}\left(3 \times \frac{1}{2}\right)+\hat{\mathrm{j}}\left(3 \times \frac{\sqrt{3}}{2}\right)=\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\) By the triangle law of vector addition, we have \(\begin{aligned} & \mathrm{OB}=\mathrm{OA}+\mathrm{AB}=(-4 \hat{\mathrm{i}})+\left(\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\right) \\ & =\left(-4+\frac{3}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}} \\ & =\left(\frac{-8+3}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}=\frac{-5}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}} \end{aligned}\) Hence the girl's displacement from her initial point of departure is \(\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}\)

Hence, the correct answer is option: 'C'.