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.

$10m$

$9m$

$\frac{13}{3} m$

$\frac{9}{2} m$

Correct answer is D

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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