A rolling stone on a rollercoaster 
A ball rolls from point $A$ ($v_0 = 0 \; m / s$) to the right along a friction-less path. We can then calculate the speed of the ball at the points $B$ and $C$ using the law of conservation of mechanical energy. Do the mathematics corresponding to the Newtonian model of the mechanics allow us to calculate the speed, even if we don't have information of the time required to travel between the points and the mass of the ball?
 A: Using conservation of energy, time does not enter into the relationships, and both gravitational potential energy and kinetic energy are proportional to the mass, so mass cancels out.  Therefore as you indicate you can calculate the speed at B and C, given the speed at A (zero here) using the conservation of energy.  Since gravity is a conservative force, the work done between two points is independent of the path taken.
Using a force/acceleration approach (Newtonian approach), you need to know the force along the path and the distance over which the force acts.  Your diagram provides the $x$ position as well as the elevations of A, B, and C, but not the shape of the path along these points.  To calculate the speed and the time of travel, you need the equation of the path along A to B to C to evaluate the component of the gravitational force along the path and the distance traveled along the path.
A: Yes.  The beauty of conservation of energy is that you can determine the speed in this case simply by considering the changes to gravitational potential energy.
A: Let's suppose there is some unknown trajectory $x(t)$ with velocity $v(t) = \frac{dx}{dt}$. I'll use the $\int dt$ symbol to represent an integral over the trajectory, integrating over time, from beginning to end. I'll use $\int dx$ to represent the same integral, but integrating over position. I'll call beginning "A" and end "B".
$\frac{1}{2}v(B)^{2} - \frac{1}{2}v(A)^{2} = \int \left(\frac{d}{dt}\frac{1}{2}v^{2}\right) dt = \int v \frac{dv}{dt} dt = \int \frac{dx}{dt}\frac{dv}{dt}dt = \frac{1}{m}\int F(t) \frac{dx(t)}{dt}dt = \frac{1}{m} \int F(x(t)) 
 dx(t)$
Now notice that we have a form of the integral which only depends on position, so we can introduce the potential $V(x)$: $\frac{1}{m}\int F(x)dx = -V(B)/m + V(A)/m$
So then
$\frac{1}{2}v(B)^{2} - \frac{1}{2}v(A)^{2}  = -V(B)/m + V(A)/m$.
In the middle, we used the Newtonian facts that $v = \frac{dx}{dt}$ and $\frac{dv}{dt} = a = F/m$. Essentially, I am showing you that whenever you use conservation of energy, it is equivalent to doing an integral over the trajectory: you do the proof once, then you use the result thereafter to shortcut the work, finding that you no longer need the time information in the trajectory!
