A farmer moves along a boudary of square field of size 10 m in 40 secs what will be magnitude of didplacement of farmer at the end of 2 mins 20 secs
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Solution :-
Total distance 40m in 40 sec.
Total time taken by the farmer
= 2 min 20 sec. =140 sec.
Total rounds completed= 140/40=3.5 rounds
So, if the farmer starts from point A of the square field, he reaches point C and displacement is AC.
Assume ABC is a right angled triangle.
There
AC2= AB2 +BC2
AC2= (10)2 +(10)2
AC2= 100+100
AC2=200
AC = (200)1/2
AC= 10× (2) ½ Ans .
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I hope it's help you....!!!! :)✌️✌️
Total distance 40m in 40 sec.
Total time taken by the farmer
= 2 min 20 sec. =140 sec.
Total rounds completed= 140/40=3.5 rounds
So, if the farmer starts from point A of the square field, he reaches point C and displacement is AC.
Assume ABC is a right angled triangle.
There
AC2= AB2 +BC2
AC2= (10)2 +(10)2
AC2= 100+100
AC2=200
AC = (200)1/2
AC= 10× (2) ½ Ans .
==============
I hope it's help you....!!!! :)✌️✌️
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