The relationship between time ($$t$$) and distance ($$x$$) is given by $$t = \alpha x^2 + \beta x$$, where $$\alpha$$ and $$\beta$$ are constants. To find the relation between acceleration ($$a$$) and velocity ($$v$$), we can follow these steps:

First, we differentiate the given equation with respect to time:

$$\begin{aligned} & t = \alpha x^2 + \beta x \quad \text{(differentiating with respect to time)} \\ & \frac{dt}{dx} = 2\alpha x + \beta \\ & \frac{1}{v} = 2\alpha x + \beta \\ \end{aligned}$$

Next, we differentiate again with respect to time to find the acceleration:

$$\begin{aligned} & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha \frac{dx}{dt} \\ & \text{Since} \quad \frac{dx}{dt} = v, \quad \text{we have:} \\ & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha v \\ & \frac{dv}{dt} = -2\alpha v^3 \\ & \text{Therefore,} \quad a = -2\alpha v^3 \end{aligned}$$