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