A particle starts from rest and moves in a straight line such that its acceleration after t secs is given by \(a = (3t - 2) ms^{-2}\). Find the distance covered after 3 secs.
Given, \(a(t) = (3t - 2) ms^{-2}\), the first integration of a(t), with respect to t, gives v(t) (the velocity). The second integration of a(t) or first integration of v(t) gives s(t).
\(v(t) = \int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)
\(s(t) = \int (\frac{3}{2}t^{2} - 2t) \mathrm {d} t\)
= \(\frac{t^{3}}{2} - t^{2}\)
When t = 3s,
\(s(3) = \frac{3^{3}}{2} - 3^{2} = \frac{27}{2} - 9 \)
= \(\frac{9}{2}\)
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