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.

  • A \(4 ms^{-2}\)
  • B \(6 ms^{-2}\)
  • C \(8 ms^{-2}\)
  • D \(10 ms^{-2}\)
waec 2011furthermathematics

The correct answer is D. \(10 ms^{-2}\)

\(v(t) = (3t^{2} - 2t) ms^{-1}\)

\(a(t) = \frac{\mathrm d v}{\mathrm d t} = (6t - 2) ms^{-2}\)

\(a(2) = 6(2) - 2 = 12 - 2 = 10 ms^{-2}\)

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