The relation between time ‘t’ and distance x for a moving particle is $t=\alpha x^2+\beta x$ where $\alpha$ and $\beta$ are constants, If v is the velocity at distance x, then the retardation of the particle is

$$\begin{gathered} given relation time is t=\alpha x^2+\beta x \\ differentiate w.r.t. displacement x \\ \begin{array}{l}\frac{dt}{dx}=2\alpha x+\beta \\ velocity=v=\frac{dx}{dt}=\frac{1}{2\alpha x+\beta }---\left(1\right)\\ \therefore v={\left(2\alpha x+\beta \right)}^{-1}\\ differentiate w.r.t. x\end{array} \\ \begin{array}{l}\frac{dv}{dx}=-1{\left(2\alpha x+\beta \right)}^{-2}\left(2\alpha \right)\\ substitute equation \left(1\right)\\ \frac{dv}{dx}=-\frac{2\alpha }{{\left(2\alpha x+\beta \right)}^2}=-2\alpha v^2\\ \end{array} \\ we know,a=\frac{v.dv}{dx} \left[\sin ce a=\frac{dv}{dt}=\frac{dv}{dt}\frac{dx}{dx}=\frac{dx}{dt}\frac{dv}{dx}=v\frac{dv}{dx}\right] \\ a=-2\alpha v^3 \\ retardation of the paticle=2\alpha v^3 \end{gathered}$$