The relation between time and distance is $\mathrm{t}={\mathrm{\alpha x}}^2+\mathrm{\beta x}$, where $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ are constants. The retardation is

$$\begin{gathered} \mathrm{Given}, \mathrm{t}={\mathrm{\alpha x}}^2+\mathrm{\beta x} \\ \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{\alpha x}+\mathrm{\beta } \\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\frac{1}{2\mathrm{\alpha x}+\mathrm{\beta }} \end{gathered}$$
As, acceleration, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}}\frac{\mathrm{dx}}{\mathrm{dt}}$
$$\begin{gathered} \Rightarrow \mathrm{a}=\mathrm{v}.\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{2\mathrm{\alpha x}+\mathrm{\beta }}\left(\frac{-2\mathrm{\alpha }}{{(2\mathrm{\alpha x}+\mathrm{\beta })}^2}\right) \\ =-2\mathrm{\alpha v}.{\mathrm{v}}^2=-2{\mathrm{\alpha v}}^3 \\ \therefore \mathrm{Retardation}=2{\mathrm{\alpha v}}^3 \end{gathered}$$