The relation between time $$t$$ and distance $$x$$ is $$t = a{x^2} + bx$$    where $$a$$ and $$b$$ are constants. The acceleration is-

Solution :
$$\eqalign{ & t = a{x^2} + bx;\,{\text{Different with respect to time }}\left( t \right) \cr & \frac{d}{{dt}}\left( t \right) = a\frac{d}{{dt}}\left( {{x^2}} \right) + b\frac{{dx}}{{dt}} = a.2x\frac{{dx}}{{dt}} + b.\frac{{dx}}{{dt}} \cr & 1 = 2axv + bv = v\left( {2ax + b} \right) \Rightarrow 2ax + b = \frac{1}{v} \cr & {\text{Again differentiating, }}2a\frac{{dx}}{{dt}} + 0 = - \frac{1}{{{v^2}}}\frac{{dv}}{{dt}} \cr & \Rightarrow \frac{{dv}}{{dt}} = f = - 2a{v^3}\,\,\,\,\left( {\because \,\frac{{dv}}{{dt}} = f = acc} \right) \cr} $$