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 correct answer is C. 14\(ms^{-2}\)
\(accl = \frac{\mathrm d V}{\mathrm d t}\)
\(V = 3t^{2} + 2t - 1 \therefore a = \frac{\mathrm d V}{\mathrm d t} = 6t + 2\)
\(a \text{(after 2 seconds)} = (6\times 2) + 2 = 12+2 = 14ms^{-2}\)
Previous question Next questionWhat is Exam without Practice? With our customizable CBT practice tests, you’ll be well-prepared and ready to excel in your examsStart Practicing Now