The velocity, V, of a particle after t seconds, is \(V = 3t^{2} + 2t - 1\). Find the acceleration of the particle after 2 seconds.
\(accl = \frac{\mathrm d V}{\mathrm d t}\)
\(V = 3t^{2} + 2t - 1 \therefore a = \frac{\mathrm d V}{\mathrm d t} = 6t + 2\)
\(a \text{(after 2 seconds)} = (6\times 2) + 2 = 12+2 = 14ms^{-2}\)
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