Question

A farmer moves along the
boundary of a square field of side10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?​

Answer 1

$\red{\boxed{Answer:}}$

14.14 m

$\red{\boxed{Step-by-step \: Explanation:}}$

Given side of square = $10 \: m$, thus Perimeter $P = 40 \: m$

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

40 sThus in 1 second, the farmer covers a distance of 1 m

40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = $1 \times 140 = 140 \: m$

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

= 3.5

At this point, the farmer is at a point say B from the origin O

At this point, the farmer is at a point say B from the origin OThus the displacement S = $\sqrt{ {10}^{2} + {10}^{2} }$ from Pythagoras Theorem.

= $10 \sqrt{2}$

$\implies{\red{\boxed{{14.14 \: m}}}}$