Quantum mechanically speaking...
Objects only have instantaneous velocity insofar as they don't have instantaneous position and mass. Velocity is merely the ratio of momentum to mass; the uncertainty can be calculated.
We have $p = mv$; from the rules of propagation of uncertainty we get
$$\sigma_p = mv \sqrt{\left(\frac{\sigma_m}{m}\right)^2 + \left(\frac{\sigma_v}{v}\right)^2 + 2\frac{\mbox{COV}(m,v)}{mv}}$$
where $\mbox{COV}(m,v)$ is the covariance between the mass and velocity. I am confident that this term can be worked out explicitly, but it has been a couple years since I took quantum mechanics and I don't have the time to do it right now. It should also be possible to write the mass in terms of the velocity depending on what you know about the system and observer at least. You can then substitute this into the uncertainty equation
$$\sigma_x \sigma_p \geq \frac{\hbar}{2} $$
This will set a limit on the uncertainty of the velocity in terms of the uncertainties of mass and position. Clearly, however, the uncertainty cannot go to zero and the ultimate resolution of your question then comes down to your definition of "instantaneous."