Given relation

$t = \alpha x^2 + \beta x$

We need the relation between acceleration $a$ and velocity $v$.

Differentiate with respect to $x$

Since $t$ is given as a function of $x$,

$\frac{dt}{dx} = 2\alpha x + \beta$

But

$v = \frac{dx}{dt} = \left(\frac{dt}{dx}\right)^{-1}$

so,

$v = \frac{1}{2\alpha x + \beta}$

Express $x$ in terms of $v$

$2\alpha x + \beta = \frac{1}{v}$

Find acceleration

We use

$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}$

From

$v = (2\alpha x + \beta)^{-1}$

differentiate with respect to $x$:

$\frac{dv}{dx} = -(2\alpha)(2\alpha x + \beta)^{-2}$

Since

$(2\alpha x + \beta)^{-1} = v \quad \Rightarrow \quad (2\alpha x + \beta)^{-2} = v^2$

hence,

$\frac{dv}{dx} = -2\alpha v^2$

Therefore,

$a = v\frac{dv}{dx} = v(-2\alpha v^2) = -2\alpha v^3$

Compare with options

$a = -2\alpha v^3$

This matches Option C.