Example of curve in $R^2$ which kills driver affirmation Given a (smooth) curve $C$ in $\mathbb{R}^2$, there exists at least one parameterization of a curve $f(t)=(x(t),y(t))$.
Given also a speed $v(t_0)$ (necessarily tangent at the curve a $t=t_0$) at some instant $t_0$ of an object moving on that curve.
I would like to find an example of a curve $C$ so that given $t_0,v(t_0)$ we can say this situation is impossible.
In short, suppose we have a car which crashed in a dangerous bend. The driver says to the police, my speed was $v(t_0)$. The police said: given the trace of the road, your affirmation is impossible.
It seems to me that the police can't never deny the driver. 
Indeed, let $f(t)=(x(t),y(t))$ a parameterization of $C$.
Then $g(t)=f(kt)$ where $k$ is a constant has also $C$ as a trace.
The speed of $g$ is $||g'(t)||=k||f'(t)||$ so we can choose any value of the speed.

You are correct (and you've actually already answered your own question). Given any curve $C$ that is parametrized by $f(t) = (x(t), y(t))$, there are infinitely many speeds in which to walk it, as you've shown: just reparameterize the curve. 
