The relation between time '$$t$$' and distance '$$x$$' is $$t = \alpha x^2 + \beta x$$, where $$\alpha$$ and $$\beta$$ are constants. The relation between acceleration $$a$$ and velocity $$v$$ is:

The time-distance relation is given as $$t = \alpha x^{2} + \beta x$$, where $$\alpha$$ and $$\beta$$ are constants.

First differentiate with respect to $$x$$:
$$\frac{dt}{dx} = 2\alpha x + \beta$$ $$-(1)$$

Velocity is $$v = \frac{dx}{dt} = \frac{1}{\dfrac{dt}{dx}} = \frac{1}{2\alpha x + \beta}$$ $$-(2)$$

Acceleration is $$a = \frac{dv}{dt}$$.
Using the chain rule, $$a = \frac{dv}{dx}\,\frac{dx}{dt} = v \,\frac{dv}{dx}$$ $$-(3)$$

Differentiate $$v$$ from $$(2)$$ with respect to $$x$$:
$$\frac{dv}{dx} = \frac{d}{dx}\left[(2\alpha x + \beta)^{-1}\right] = -1 \cdot (2\alpha)\,(2\alpha x + \beta)^{-2} = -\frac{2\alpha}{(2\alpha x + \beta)^{2}}$$ $$-(4)$$

Substitute $$(2)$$ and $$(4)$$ into $$(3)$$:
$$a = v\left(-\frac{2\alpha}{(2\alpha x + \beta)^{2}}\right) = -\frac{2\alpha\,v}{(2\alpha x + \beta)^{2}}$$ $$-(5)$$

From $$(2)$$, $$v = \dfrac{1}{2\alpha x + \beta} \;\Rightarrow\; 2\alpha x + \beta = \frac{1}{v}$$ $$-(6)$$

Insert $$(6)$$ into $$(5)$$ to eliminate $$x$$:
$$a = -\frac{2\alpha\,v}{\left(\dfrac{1}{v}\right)^{2}} = -2\alpha\,v \, v^{2} = -2\alpha\,v^{3}$$

Thus the required relation is $$a = -2\alpha v^{3}$$.

Option A is correct.