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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 = 2t^2 + 3t +1 |
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Answer» Solution :`s = 2t^2 + 3T + 1` `DS/dt = 4t + 3, d^2s/dt^2 = 4` `(ds)/dt]_(t=2) = 4 XX 2 +3 = 11`, `(d^2s)/dt^2] (t = 2) = 4` `therefore` Velocity is 11 units/sec and acceleration is 4 `units/sec^2`. |
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