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CHAMABHAI MOVES ALONG THE BOUNDARY OF A SQUARE FIELD OF SIDE 20 METER IN 80 SECONDS . WHAT WILL BE EHE MAGNITUDE OF DISPLACEMENT OF CHHAGAN BHAI AT THE END OF 4 MINUTES 40 SECONDS FROM HIS INITIAL POSITION
Asked by manish7517 | 17 Nov, 2019, 07:45: PM
Side of square = 20m
Perimeter of square = 20 × 4 =80 m
Time taken to complete 1 round = 80 s
Thus, the man is moving with velocity of 80m/80s =1m/s
Thus, distance covered in 4 min 40 s i.e. 4 ×60 + 40s =240 + 40 =280 s with same speed = 280 m
Number of rounds completed for covering 280 m = 280/80 = 3.5
With 3 rounds the displacement will be zero.
And for the rest of the 0.5 m the man will be at diagonally opposite end of his initial point.
So the magnitude of the displacement is,
S2= 20 + 202 = 400 + 400 = 800
S =√(800) m

Answered by Shiwani Sawant | 17 Nov, 2019, 11:23: PM

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