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