Roller coaster down a Bezier Curve Say we have the following Bezier curve:
$$x(t)=110t^3-120t^2+60t$$
$$y(t)=100t^3-150t^2+50$$
$$Where \quad 0\leq t  \leq 1$$

This curve maps out the shape of the roller coaster track. Now assuming there are no frictional forces acting on the rollercoaster, only gravity, how could one find an equation which models the motion of the roller coaster down said function with respect to time?
Here is my method (which didn't work):
So I begin with the equation for acceleration down a slope. Then differentiate the bezier curve to find the slope at any point along the function.
$$a=g\sin{\theta}$$
$$\theta = \tan^{-1}{\left |\frac{dy}{dx}\right |} = \tan^{-1}{\left |\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right |}$$
$$=\tan^{-1}{\left |\frac{300t^2-300t}{330t^2-240t+60}\right |}$$ 
$$\therefore \theta = \tan^{-1}{\left | \frac{10t^2-10t}{11t^2-8t+2} \right |}$$
So this expression for $\theta$ should give the instantaneous angle of the slope at any point along the curve. Now substituting this expression for theta within $a=g\sin{\theta}$.
$$a(t)=g\sin{\left | \frac{10t^2-10t}{11t^2-8t+2} \right |}$$
Here is where I am stuck. I understand that the above equation is not with respect to "real" time. The bezier curve simply maps out the general shape of the track and is not a displacement function with respect to time. 
So how could one generate a function which modelled the acceleration, and therefore the velocity and displacement, with respect to time? I swear there should be a simplistic way to do this. 
Thank you!
 A: Since our curve $\gamma$ is given in terms of some arbitrary parameter $t$ we shall express "everything" in terms of this $t$, in particular the "real time" $\tau=\tau(t)$. Let $v$ be the "physical speed", i.e., the speed with respect to $\tau$, of the roller coaster. Then
$$v(t)={ds\over d\tau}={s'(t)\over \tau'(t)}\ ,\tag{1}$$
whereby 
$$s'(t)=\sqrt{x'^2(t)+y'^2(t)}$$
is the derivative of the arc length function $t\mapsto s(t)$ along $\gamma$. On the other hand  conservation of energy implies 
$${m\over2} v^2(t)=mg\bigl(50-y(t)\bigr)\ ,$$
so that we have
$$v(t)=\sqrt{2g\bigl(50-y(t)\bigr)}\ .\tag{2}$$
Introducing this into $(1)$ we obtain the following explicit formula for the unknown function $t\mapsto\tau(t)$:
$$\tau'(t)=\sqrt{\strut x'^2(t)+y'^2(t)\over2g\bigl(50-y(t)\bigr)}\qquad(0\leq t\leq 1)\ .$$
I don't think that you can express $\tau(t)$ in elementary terms. If you are just interested in the "physical speed" of the car at various points of the track you can use $(2)$ directly.
