Question

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

Answer 1

$\large{\boxed{\sf{\underline{\pink{SoLuTion.. }}}}}$

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 = 1×140=140m

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

Thus the displacement s =

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

(from Pythagoras theorem)

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

Hope it helps

Answer 2

Answer:

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 = 1×140=140m

Now the total number of rotation the farmer makes to cover a distance of 140 meters =

\frac{total \: distance}{perimeter}perimetertotaldistance

= 3.5

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

Thus the displacement s =

\sqrt{10 ^{2} + {10}^{2} }102+102

(from Pythagoras theorem)

s = 10 \sqrt{2} = 14.14ms=102=14.14m

Explanation:

hope it will help you dear..