Question

A man moves 20 m north, then 10 m east and then $10\sqrt{2}m$ south-west. His displacement is

• A. 20 m north
• B. $10\sqrt{2}m$ north-west
• C. 10 m north
• D. $10\sqrt{2}m$ south-east

Answer 1

Answer: (C)

The situation is shown in figure.

$\overrightarrow{OA}$ represents $20m$ north, $\overrightarrow{AB}$ represents $10m$ east and $\overrightarrow{BC}$ represents $10\sqrt{2}m$ south-west. If $\hat{i}$ and $\hat{j}$ be the unit vectors along east and north respectively, then
$\overrightarrow{OA}=20\hat{j}m,\overrightarrow{AB}=10\hat{i}m$
$\overrightarrow{BC}=\left(-10\sqrt{2}\cos 45^{\circ }\hat{i}-10\sqrt{2}\sin 45^{\circ }\hat{j}\right)m$
$=\left(-10\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\hat{i}-10\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\hat{j}\right)m=(-10\hat{i}-10\hat{j})m$
$\therefore$ Displacement $\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}$
or $\overrightarrow{OC}=20\hat{j}m+10\hat{i}m+(-10\hat{i}-10\hat{j})m$
$=10\hat{j}m=10m$ due north