Question

A particle undergoes three successive displacements given by $S_1=\sqrt{2}m$ north-east, $S_2=2m$ due south and $S_3=4m$, ${30}^{\circ }$ north of west, then magnitude of net displacement is

• A. $\sqrt{14+4\sqrt{3}}m$
• B. $\sqrt{14-4\sqrt{3}}m$
• C. $\sqrt{4}m$
• D. $1-2\sqrt{3}m$

Answer 1

The correct option is B $\sqrt{14-4\sqrt{3}}m$

Let us first write down all the three displacements in vector form
$\overrightarrow{S_1}=\left(\sqrt{2}cos{45}^{\circ }\right)\hat{i}+\left(\sqrt{2}sin{45}^{\circ }\right)\hat{j}$
$=1\hat{i}+1\hat{j}$
$\overrightarrow{S_2}=-2\hat{j}$
and $\overrightarrow{S_3}=(-4cos{30}^{\circ })\hat{i}+\left(4sin{30}^{\circ }\right)\hat{j}$
$\overrightarrow{S_3}=-2\sqrt{3}\hat{i}+2\hat{j}$
Total displacement $=\overrightarrow{S_1}+\overrightarrow{S_2}+\overrightarrow{S_3}=(1-2\sqrt{3})\hat{i}+1\hat{j}$
Hence, magnitude of total displacement = $\sqrt{(1-2\sqrt{3}{)}^2+1^2}=\sqrt{14-4\sqrt{3}}m$