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Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules. s = 3/2t + 1 |
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Answer» Solution :`s = 3/2t + 1 ` ds/dt = 6/(2t + 1)^2 (d^2s)/dt^2 = 24/ (2t + 1)^3` Now `ds/dt](t = 2) = 6/25` (d^2s)/dt^2](t = 2) = 24/125` `therefore` VELOCITY is - 6/25 units/sec and ACCELERATION is 24/125 `units/sec^2` . |
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