Question

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

• A. 14.14 m
• B. 5 m
• C. 15.06 m
• D. 14.19 m

Answer 1

Answer: (A)

14.14 m

The perimeter of square field
$ABCD=4\times 10m=40m$.
Time for moving around the $10m$ square field once $=40s$.
Time for journey of farmer $=2min$ and $20s=140s$.
Number of times the farmer moved around the square field $=\frac{140}{40}=3\frac{1}{2}$times.
For going once around the 'square field, the displacement $=0$
For going thrice around the square field, the displacement $=0$
For going $\frac{1}{2}$ around the square field, the distance covered $=40\times \frac{1}{2}=20m$.
It is obvious from the figure, that if the farmer starts from point $A$, then he will cover $10m$ along $AB$ and then $10m$ along $BC$.
Therefore, displacement of farmer from the point $A$ to point $C$ is
$AC=\sqrt{A{B}^2+B{C}^2}$

$=\sqrt{{10}^2+{10}^2}$

$\approx 14.14m$