A farmer movers along the boundary of a square field of side 10m in 40seconds. What will be the distance and displacement of farmer at the end of 2 min 20 sec from him initial position?

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Solution

Total distance 40m in 40 sec.
Total time taken by the farmer = 2 min 20 sec. =140 seconds.

Total rounds completed= 140/40=3.5 rounds

That means if the farmer starts from point A of the square field, he reaches point C.

Therefore, displacement is AC.

Assume ABC is a right angled triangle.

Therefore,

AC^2= AB^2 +BC^2

AC^2= (10)^2 +(10)^2

AC^2= 100+100

AC^2=200

AC = (200)^1/2

AC= 10× (2)^ ½
Distance = 3.5*40=140m

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