Question
A point p moves in counter -clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out the length s = t3 + 5, where s is metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2 s is nearly
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13 ms-2
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12 ms-2
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7.2 ms-2
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14 ms-2
Solution
D.
14 ms-2
Given that, s =t3 +5
therefore speed v, = ds/st = 3t2
and rate of change of speed, at = dv/dt = 6t
∴ Tangential acceleration at t =2s
at = 6 x 2 = 12 ms-1
and at t = 2s, v = 3 (2)2 = 12 ms-1
∴ Centripetal acceleration, ac = v2/R = 144/20 ms-2
∴ Net acceleration = 
