How is the Dirac delta function used in classical mechanics? If the contact force applied to a physical object (ex. empty bucket) is given by the Heaviside function:
$$
  F(t) = F_0~H(t)=\begin{cases}
    0, t<0 \\
    F_0, t \geq 0\\
  \end{cases}
$$
Then, assuming constant mass, the rate of change of acceleration is given by $$\lim_{t \to 0} \frac{da}{dt} \rightarrow \infty$$ since $$\frac{dF}{dt}=\delta(t),$$ the Dirac delta function. 
How am I supposed to interpret this result physically? My intuition suggests that a mechanical system would have a minimum reaction time and that we can't have an 'instantaneous' force. 
 A: As a general rule, whenever you are setting sudden discontinuities (as in your example with the Heaviside function) it is not a surprise that these may reflect in discontinuous distributions, or derivatives, or infinities here or there. Do keep in mind that it is just a result of the mathematical simplifications that we are introducing (although distributions may look scary, they are actually the easiest we can do to model).
In physics and in nature there are no sudden discontinuities, everything changes (more or less) smoothly; as a consequence $0$ and $\infty$ are to be interpreted as very small and very big, instead. A correct and precise smooth approximation would yield finite results, instead.
Example: a standard example of the above is the Coulomb law between two point particles that sit very far away from each other:
$$
\textbf{F}_{1,2} = k \frac{q_1q_1}{r^2}\textbf{u}_{1\to 2}
$$
what happens when $r\to 0$, is the force diverging? In such a case the above approximation does not hold anymore and the two charges cannot be considered point like and very far away from each other: charge distribution comes into play and the effect of the density modifies the Coulomb law to a form that is totally regular in $r=0$ (there are plenty of exercises on that in the literature).
