1.

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

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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