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.

A.

2m

B.

4m

C.

6m

D.

8m

Correct answer is B

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

\(s(t) = \int v(t) \mathrm {d} t\)

= \(\int (3t^{2} - 2t) \mathrm {d} t = t^{3} - t^{2}\)

\(s(2) = 2^{3} - 2^{2} = 8 - 4 = 4m\)