The body travelling with the uniform acceleration crosses two points A and B with velocities 20m/s and 80m/s respectively. If P and Q are the points on the line AB such that AP=PQ=BQ then the ratio of velocity of the boby at point P and Q is
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Answer:
AB = s
2 a s= (30² - 20² ) = 500 s = 250/a
AB/2 = s/2
2 a s/2 = (v² - 20²)
250 = v² - 400 => v = √650 m/s
Explanation:
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