A farmer moves along the boundary of a square field of side 10 m in 40 sec. what will be the magnitude of displacement of the farmer at the end of 2 min 20 sec?? plz fast
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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,
AC2= AB2 +BC2
AC2= (10)2 +(10)2
AC2= 100+100
AC2=200
AC = (200)1/2
AC= 10× (2) ½
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,
AC2= AB2 +BC2
AC2= (10)2 +(10)2
AC2= 100+100
AC2=200
AC = (200)1/2
AC= 10× (2) ½
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