Question

a farmer moves along the boundary of a square field of side 10meter in 4sec . what were be the magnitude of displacement of the farmer at the end of 2 min. 20 sec. from his initial position ?​

Answer 1

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 =

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

$\huge\bold\red{Answer}$

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 =

perimeter

totaldistance/perimeter

=3.5

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

Thus the displacement=.$\sqrt{10²+10²}$ from Pythagoras theorem.

s=10 $\sqrt{2}$

=14.14m

Hope it's helpful ❤️

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