Question

A farmer moves along the boundary of a square field of side $10\text{m}$ in $30\text{s}$. What will be the magnitude of displacement of the farmer at the end of $1\text{min}45\text{sec}$?

• A. $10\text{m}$
• B. $10\sqrt{2}\text{m}$
• C. $20\text{m}$
• D. Zero

Answer 1

The correct option is B $10\sqrt{2}\text{m}$

In the given figure $ABCD$ is a square field of side $10\text{m}$
Time for one round is $30\text{s}$ and total time of motion is $1\text{min}45\text{sec}=105\text{s}$
So, number of round completed is $\frac{105}{30}=3.5$
Thus, if farmer starts from $A$, he will complete $3$ rounds at $A$. In the last $0.5$ round starting from $A$, he will finish at $C$.
$\therefore$ Displacement $AC=\sqrt{(AB{)}^2+(BC{)}^2}=\sqrt{{10}^2+{10}^2}=10\sqrt{2}\text{m}$