Question

A farmer moves along the boundary of a square field side 10 m in 40 s. What will be the magnitude of displacement and distance of the farmer at the end of 2 minutes 20 seconds .

Answer 1

Given -

Farmer moves along the boundary once in 40s.

Hence, time,t = 2 min 20s

⠀⠀⠀⠀⠀⠀ ⠀⠀= 140s

To find -

Magnitude of displacement

Solution -

He will complete the n number of revolutions of field, where,

$n = \frac{140}{40} = 3 \frac{1}{2}$

It means that if the farmer haa started to move from point A then after completing three and the half rounds he will reach diagonally opposite point C therefore-

By using Pythagoras theorem we will find the magnitude of displacement.

$ac = \sqrt{ad \: ^{2} + dc \: ^{2} }$

$ac = \sqrt{ {(10)}^{2} + {(10)}^{2} }$

$ac = 10 \sqrt{2}$

_____________________________________

$\therefore$ the magnitude of displacement will be $10 \sqrt{2} \: m$

Answer 2

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 =

$\frac{total \:distance}{perimetre}$

$= 3.5$

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

Thus the displacement s=

$10 ^{2} + 10 {}^{2}$

from Pythagoras theorem.

s =10

$= \sqrt{2}$

$= 14.14m$