The distance at the which the particle stops is
MY SOLUTION $$a=v\frac{dv}{dx}$$ $$vdv=\alpha x^2 dx$$ Integrating $$\int_{v_0}^0 v\,dv=\int_0^x \alpha x^2\,dx$$ $$-\frac{(v_0)^2}{2}=\alpha \frac{x^3}{3}$$ $$x=(\frac{-3v_0^2}{2\alpha})^\frac13$$
The right answer is $[\frac{3v_0}{2\alpha}]^\frac13$. What’s going wrong?
As the other answer pointed out, the acceleration is $a=-\alpha x^2$. And using dimension analysis, we can see that the dimension of the "right answer" is not right, but yours is, because: $$[x]=L$$ and $$[v]=LT^{-1}$$ and $$[a]=LT^{-2}$$ Which implies that $$[\alpha]=T^{-2}L^{-1}$$ So $$\left[\frac{v_0}{\alpha}\right]=L^{2}T$$ and $$\left[\frac{v_0^2}{\alpha}\right]=L^{3}$$ In fact, we can calculate the only possible combination of $v_0$ and $\alpha$ to get length dimension: We want that $$[v_0^n \alpha^m]=L$$ $$L^{n-m}T^{-n-2m}=L$$ Which means that $n-m=1$ and $-n-2m=0$, i.e. $n=2/3$ and $m=-1/3$. And the possible dimensionless combination is $$[v_0^p \alpha^q]=1$$ $$L^{p-q}T^{-p-2q}=1$$ Which means that $p-q=0$ and $-p-2q=0$, i.e. $p=q=0$. So the solution must have the following form: $$x=C v_0^{2/3} \alpha^{-1/3}$$ where $C$ is a dimensionless constant.