Question

A farmer moves along the boundary of square field of side 5m in 20 sec. What will be the magnitude of displacement of the farmer at the end of 1 minute 10 seconds.​

Answer 1

Solution :

ABCD is a square field of side 5 m.

Time for one round = 20 s,

total time 1 min 10 s= (1 x 60 + 10) s = 70 s

Number of rounds completed

==> 70/20

==> 3 : 5

if farmer starts from A, he will complete 3 rounds (A > B > C > D > A) at A. In the last 0 - 5 round starting from A, he will finish at C

Displacement = AC, where

AC = √AB² + BC² m

==> √10² +10²

==> 10√2 m

Answer 2

Question:-

A farmer moves along the boundary of square field of side 5m in 20 sec. What will be the magnitude of displacement of the farmer at the end of 1 minute 10 seconds.

Answer:-

➭ 10√2

Solution:-

• Given side of square =10m, thus perimeter P=40mTime taken to cover the boundary of 40 m =4s

• Thus in 1 second, the farmer covers a distance of 1 mNow 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 = totaldistance pperimete =3.5 At this point, the farmer is at a point say B from the origin O.

Thus the displacement:-

$s \: = \sqrt{ {10}^{2}+ {10}^{2} } \\ from \: pythagoras \: therom \\ s \: = 10 \sqrt{2 } = 14.14m$