The relation between time t and distance x is t = ax² + bx, where a and b are constants. The acceleration is

t = ax² + bx on differentiating w.r.t. time
$\begin{array}{l}1=2\mathrm{ax}\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{b}\frac{\mathrm{dx}}{\mathrm{dt}}\\ 1=(2\mathrm{ax}+\mathrm{b})\mathrm{V}\Rightarrow \mathrm{V}=\frac{1}{2\mathrm{ax}+\mathrm{b}}\\ \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{-1}{(2\mathrm{ax}+\mathrm{b}{)}^2}\cdot 2\mathrm{av}=-2{\mathrm{av}}^{3 }\left(\mathrm{v}=\frac{1}{(2\mathrm{ax}+\mathrm{b})}\right)\end{array}$