WAEC - Further Mathematics (2011 - No. 35)
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.
2m
4m
6m
8m
Giải thích
\(v(t) = (3t^{2} - 2t) ms^{-1}\)
\(s(t) = \int v(t) \mathrm {d} t\)
= \(\int (3t^{2} - 2t) \mathrm {d} t = t^{3} - t^{2}\)
\(s(2) = 2^{3} - 2^{2} = 8 - 4 = 4m\)