Question

Maganbhai moves along the boundary of a square field of side 20meter in 80seconds what will be the magnitude of displacement of maganbhai at end of 4min 40seconds from his initial position

Answer 1

Answer:-10m

Explanation:we obtain speed=0.25m/s by using formula of s=d/t now time=4min40s

(4×60)+40=280s

S=d/t. Where S=0.25,t=280 we get distance= 70m but in the square each side is 20m so the perimeter is 80m so we can consider it as origin,

Displacement= final position- initial position, therefore 70m-80m= -10m,

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