Question

A farmer moves along the boundary of a square field of side 10m in 40sec .What will be the magnitude of displacement of the farmer at the end of 2min 20 sec from his initial position ?

Answer 1

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 =

perimeter

totaldistance

=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 2

=14.14m

Answer 2

Answer:

14.14 m

Explanation:

Given side of square =10m,

Perimeter P= 4×10 = 40m

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

So in 1 second, the farmer covers a distance of 1 m

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 = perimeter ÷ total distance

=3.5

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

Thus the displacement s= √(10²+10²)

By Pythagoras theorem,

s=10√2

=14.14m