1. Given equation of motion

The retarding force is along negative $x$-direction and has magnitude proportional to $x^2$:

$m a = -\alpha x^2$

So,

$a = -\frac{\alpha}{m}x^2$

We need the distance $x$ where the particle stops, i.e. where $v=0$.

2. Use the relation between acceleration and velocity

Since acceleration depends on position, use

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

Thus,

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

3. Integrate using initial conditions

Initially, at $x=0$, the particle is projected with speed $v_0$. At the stopping point, let the distance be $x=s$, and there $v=0$.

So integrate:

$\int_{v_0}^{0} v\,dv = -\frac{\alpha}{m}\int_0^{s} x^2\,dx$

Left side:

$\int_{v_0}^{0} v\,dv = \left[\frac{v^2}{2}\right]_{v_0}^{0} = -\frac{v_0^2}{2}$

Right side:

$-\frac{\alpha}{m}\int_0^{s} x^2\,dx = -\frac{\alpha}{m}\left[\frac{x^3}{3}\right]_0^{s} = -\frac{\alpha s^3}{3m}$

Hence,

$-\frac{v_0^2}{2} = -\frac{\alpha s^3}{3m}$

Cancelling minus signs,

$\frac{v_0^2}{2} = \frac{\alpha s^3}{3m}$

4. Solve for stopping distance

$s^3 = \frac{3mv_0^2}{2\alpha}$

Therefore,

$s = \left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}$

5. Match with the options

This matches Option A:

$\boxed{\left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}}$

6. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.