A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Determine the acceleration when t = 2 secs.
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Correct Answer: Option D
Explanation:
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(a(t) = \frac{\mathrm d v}{\mathrm d t} = (6t - 2) ms^{-2}\)
\(a(2) = 6(2) - 2 = 12 - 2 = 10 ms^{-2}\)
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(a(t) = \frac{\mathrm d v}{\mathrm d t} = (6t - 2) ms^{-2}\)
\(a(2) = 6(2) - 2 = 12 - 2 = 10 ms^{-2}\)