Question

A farmer moves along the boundary of a square field of side $10m$ in $40s$. What will be the magnitude of displacement of the farmer at the end of $2$ minutes $20$ seconds from his initial position?

Answer 1

Total distance traveled by farmer in $40s=4\times 10=40m$

Total time taken $=2min\times 20sec=140sec$

Total number of rounds $=\frac{140}{40}=3.5$

If farmer starts from A, then after 3.5 rounds, he reaches to point ${}^{\prime }{C}^{\prime }$.

Therefore, displacement $=AC$.

Now, $ABCD$ is a square, then $△ABC$ is a right angled $△$.

Using Pythagoras theorem,

${AC}^2={AB}^2+{BC}^2\Rightarrow AC=\sqrt{{10}^2+{10}^2}=\sqrt{200}=10\sqrt{2}$

$AC=10\times 1.414=14.14m$

Hence, displacement $=14.14m$