A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Calculate the distance covered in the first 2 seconds.
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Correct Answer: Option B
Explanation:
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(s(t) = \int v(t) \mathrm {d} t\)
= \(\int (3t^{2} - 2t) \mathrm {d} t = t^{3} - t^{2}\)
\(s(2) = 2^{3} - 2^{2} = 8 - 4 = 4m\)
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(s(t) = \int v(t) \mathrm {d} t\)
= \(\int (3t^{2} - 2t) \mathrm {d} t = t^{3} - t^{2}\)
\(s(2) = 2^{3} - 2^{2} = 8 - 4 = 4m\)