Physics - Kinematics Question with Solution | TestHub

Given the instantaneous position is

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

The velocity of the particle is given by

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

And, the acceleration of the particle is given by

$\begin{array}{rcl}a& =& \frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{t}}\\ & =& 6t-12 ...\left(3\right)\end{array}$

When $a=0$, it implies

$$\begin{gathered} 6t-12=0 \\ \Rightarrow t=2 \mathrm{s} \end{gathered}$$

At $t=2 \mathrm{s}$, the velocity can be calculated as follows:

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