What kind of a curve can represent a physical trajectory It is very well known that conics, spirals, etc. can represent a realistic trajectories of point particles. However, a physical trajectory can also intersect itself, have a cusp, and other kinds of singularities. Therefore, it is natural to assume that physical trajectories are a subset of algebraic curves. But an algebraic curve can consist of several disconnected curves (e.g. hyperelliptic curve) which cannot possibly describe a real trajectory of a particle (we, of course, forbid teleportation).
So, what would be the most general mathematical curve describing a physical trajectory?
 A: If you give me machines that can generate forces, then the answer is "lots". Take your curve, and consider its projection $t \mapsto u(t)$ to the $x$ axis. Let's assume the curve describes the motion of a point-particle of mass $m$.  By computing $u''(t)/m$, you can find the force, $F_x(t)$ that must be applied to the particle in the $x$-direction. If you assume, for example, that the particle is charged and floats between two nearly infinite planes whose voltage you can adjust arbitrarily, you just aadjust the voltage so that it varies as $F_x$. You do the same in the $y$-direction, perhaps by dragging large masses to be near/far from your particle, so that only gravitational forces act in the $y$ direction. 
The only constraint here is that $u$ and $v$ are twice differentiable, which is essential if the first law is to hold: $F = ma$ makes no sense if $a$ is undefined. 
Note that even simple force profiles like $F_x(t) = \sin t$, $F_y(t) = 1$ lead to non-algebraic curves, so the answer is certainly richer than just "algebraic curves". 
By the way, the image of $(u, v)$ may contain cusps, but the plot of $(t, u(t), v(t))$ will generally not, except at places where $u'$ and $v'$ are both zero. 
All this describes idealized physics of a kind of Physics 1 world (in which point particles exists), and ignores relativistic effects completely. Otherwise I'd need to say that $u'(t)^2 + v'(t)^2 < c^2$ for all $t$, where $c$ is the speed of light, and (possible) some other things as well. 
