A particle is moving in a straight line. The variation of position...

Given the instantaneous position is

$x = {t}^{3} - 6{t}^{2} + 20t + 15 . . . \left(1\right)$

The velocity of the particle is given by

$\begin{cases}v & = & \dfrac{dx}{dt} \\ & = & 3{t}^{2} - 12t + 20 . . . \left(2\right)\end{cases}$

And, the acceleration of the particle is given by

$\begin{cases}a & = & \dfrac{dv}{dt} \\ & = & 6t - 12 . . . \left(3\right)\end{cases}$

When $a = 0$, it implies

$6t - 12 = 0 ⇒ t = 2s$

At $t = 2s$, the velocity can be calculated as follows:

$\begin{cases}v & = & 3 ( 2{)}^{2} - 12 ( 2 ) + 20 \\ & = & 8m{s}^{- 1}\end{cases}$