Velocity is given by $v=4 t(1-2 t)$, then find time at which velocity is maximum

To find value of time at which velocity is maximum, taking differentiation of v with respect to time $\Rightarrow \quad \frac{\mathrm{dv}}{\mathrm{dt}}=0$ Here, $v=4 t(l-2 t)$ $\begin{array}{lc} & \mathrm{v}=4 \mathrm{t}-8 \mathrm{t}^2 \\ \Rightarrow & \frac{\mathrm{d}}{\mathrm{dt}}\left(4 \mathrm{t}-8 \mathrm{t}^2\right)=0 \\ \Rightarrow & 4-16 \mathrm{t}=0 \\ \Rightarrow & \mathrm{t}=\frac{1}{4} \mathrm{~s}=0.25 \mathrm{~s}\end{array}$ Again taking differentiation, $\Rightarrow \quad \frac{\mathrm{d}^2 v}{\mathrm{dt}^2}=-16 \lt 0$ So, at $\mathrm{t}=0.25 \mathrm{~s}$ velocity is maximum.