Question

A farmer moves along the boundary 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?

Answer 1

Given, side of the square field,s = 10 m
Therefore, perimeter = $4\times s=4\times 10=40m$
The farmer moves along the boundary in 40 s.
Seconds in $2m20s=2\times 60s+20s=140s$
Since in 40 s, the farmer moves 40 m, in 1 s distance covered by farmer = $\frac{40}{40}=1m$
Therefore, in 140 s, distance covered = $140m$.
Number of rounds covered $=\frac{140m}{40m}=3.5$ rounds
If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.
Therefore, displacement $AC=\sqrt{(AB{)}^2+(BC{)}^2}$
$=\sqrt{(10{)}^2+(10{)}^2}$
$=\sqrt{100+100}$
$=\sqrt{200}$
$=10\sqrt{2}$
$=10\times 1.414=14.14m$