The velocity, V, of a particle after t seconds, is \(V = 3t^{2} + 2t - 1\). Find the acceleration of the particle after 2 seconds.

The velocity, V, of a particle after t seconds, is \(V = 3t^{2} + 2t - 1\). Find the acceleration of the particle after 2 seconds.

Answer Details

The acceleration, A, of a particle is the rate of change of its velocity, V. So, we need to differentiate the given equation of velocity with respect to time, t to get the equation of acceleration. \[\frac{dV}{dt} = \frac{d}{dt}(3t^2 + 2t - 1)\] \[\frac{dV}{dt} = 6t + 2\] Therefore, the acceleration of the particle after 2 seconds is given by substituting t = 2 into the equation of acceleration as follows: \[A = 6t + 2\] \[A = 6(2) + 2\] \[A = 14 ms^{-2}\] Hence, the correct option is (c) 14\(ms^{-2}\).