Question

A farmer moves along the boundary of a square field of side 10m in 40s what will be the magnitude of displacement of the farmer
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Answer 1

Explanation:

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 =totaldistance/perimeter

=3.5

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

Thus the displacement s= √10²+√10²

from Pythagoras theorem.

s=10 √2 =14.14m

Answer 2

Question :-

1. A farmer moves along the boundary of a square field of side 10m in 40s what will be the magnitude of displacement of the farmer

_____________

Solution

• side of square field = 10m

⠀⠀⠀Therefore , perimeter = 4 × side

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 4 × 10

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 40m

Farmer moves along the boundary in 40 seconds.

• Time = 2 minutes 20 seconds

⠀⠀⠀⠀⠀⠀⠀= 2 ×60s +20

⠀⠀⠀⠀⠀⠀⠀= 140 seconds

since in 40s a farmer moves 40m

therefore ,

⠀⠀⠀In 1s distance covered by farmer

⠀⠀⠀= $\dfrac{40}{40}$

⠀⠀⠀= 1m

⠀⠀⠀In 140s distance covered by farmer

⠀⠀⠀= 1 × 140

⠀⠀⠀= 140m

Now ,

⠀⠀⠀Number of rotation to cover 140 along the boundary

⠀⠀⠀= $\dfrac{Total\: distance}{perimeter}$

⠀⠀⠀= $\dfrac{140}{40}$

⠀⠀⠀= 3.5 round