How to approximate acceleration from a trajectory's coordinates? If I only know $x$- and $y$- coordinates of every point on a trajectory without knowledge of time information, is there any way to approximate Cartesian acceleration angle at each point? Time interval between every two points is very small, ~0.03 second. 
 A: Yes.  Provided you are only interested in the direction of the acceleration, and not it's magnitude.  And further assuming your time samples are equally spaced, you can take the second derivative of the path and this will be proportional to the acceleration.
A decent method in practice would be to use a second order central finite difference scheme wherein you say that:
$$ a_x(t) = x(t-1) - 2x(t) + x(t+1) $$
and
$$ a_y(t) = y(t-1) - 2y(t) + y(t+1) $$
this will give you decent estimates for the cartesian components of acceleration at every time, caveat to an overall scaling in magnitude that you won't know without knowing the actual timing, but the direction should be alright.
A: In your case, lets $\Delta t = 0.03s $ 
By the method alemi explained, 
$$a_{x}(t)=\frac{x(t-\Delta t)-2x(t)+x(t+\Delta t)}{(\Delta t)^{2}}$$
and
$$a_{y}(t)=\frac{y(t-\Delta t)-2y(t)+y(t+\Delta t)}{(\Delta t)^{2}}$$
A: From the path you need to find the radius of curvature $\rho$ at each point. This would be kind of noisy unless you have really precise data. Your best bet into input all the x and y points into cubic spline in order to get what the derivatives $x'$ and $y'$ are (in units of length per frame). In addition, you need to get the kinematic accelerations $x''$ and $y''$. 
Then
$$\rho = \frac{ \left( x'^2 + y'^2 \right) ^ {\frac{3}{2} }}{x''\,y' - x'\,y''} $$
You can also estimate the speed by $$ v = \dot{s} \sqrt{ x'^2+y'^2 } $$ where $\dot{s}$ is the sample rate (frames/second).
The tangent acceleration to the path is $$ a_{T} = \dot{v} = \dot{s} \frac{x'\,x'' + y'\,y''}{\sqrt{x'^2+y'^2}} $$
and the transverse acceleration is $$a_{N} = \frac{v^2}{\rho} =  \frac{ \dot{s}^2  \left({x''\,y' - x'\,y''}\right) }{ \sqrt{ x'^2+y'^2 } }$$
