A player moves around a boundary of a square ground of side 50 m in 200 sec. The magnitude of displacement of the player at the end of 11 mins 40 sec from his initial position is ?
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Answer is 50√2
He moves 200m in 200sec thus 1m in 1sec
In 11min 40sec that is 700sec he will move 700m that is he will reach the opposite corner of the square.
Thus by pythagorus theorem
50²+ 50²= H²
5000 = H²
H = 50√2
He moves 200m in 200sec thus 1m in 1sec
In 11min 40sec that is 700sec he will move 700m that is he will reach the opposite corner of the square.
Thus by pythagorus theorem
50²+ 50²= H²
5000 = H²
H = 50√2
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side of ground = 50 m,perimeter= 50×4 = 200m
speed = 50/200 = 1/4 m /s
distance covered in 11min 40 secs
= s×t = 1/4 * (11×60 + 40) = 700/4 = 175 m
no. of rounds completed = 175/200 =.875
after 0.5 round the farmer will be diagonally opposite to his initial position from where he started .
so displacement AC =√2 ×50 = 50√2
= 70.71 m
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