Physics - Kinematics Question with Solution | TestHub

$x\left(t\right)=at+b{t}^2-c{t}^3$
velocity$v\left(t\right)=\frac{d}{dt}\left(x\right)=\frac{d}{dt}\left(at+b{t}^2-c{t}^3\right)$
$=a+2bt-3c{t}^2$
Acceleration$=\frac{d}{dt}v\left(t\right)=2b-6tc$
$$\begin{gathered} \text{acceleration}=0 \\ \Rightarrow 2b-6tc=0 \end{gathered}$$
$t=\frac{b}{3c}$
$\therefore$velocity when$t=\frac{b}{3c},$
$v_{\left(t=\frac{b}{3c}\right)}=a+2b\left(\frac{b}{3c}\right)-3{\left(\frac{b}{3c}\right)}^2$
$=a+\frac{2b^2}{3c}-\frac{b^2}{3c}$
$=a+\frac{b^2}{3c}$