Question

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

Answer 1

Given

Side of square $=10m$, thus perimeter $P=40m$

Time taken to cover the boundary of 40 m = 40 s

Thus in 1 second, the farmer covers a distance of 1 m.

Now distance covered by the farmer in 2 min 20 seconds = $1\times 140=140m$

Now the total number of rotations the farmer makes to cover a distance of 140 meters =$\frac{totaldistance}{perimeter}$

$=3.5$

At this point, the farmer is at a point which is diagonally opposite from the starting point.

Thus the displacement $s=\sqrt{{10}^2+{10}^2}$ from Pythagoras theorem.

s=$10\sqrt{2}=14.14m$