Hamiltonian 2nd positional derivative analogous to acceleration?
It's related to kinetic energy
$$T = \frac{1}{2} mv^2 $$

which can be rewritten in terms of momentum $p$
$$ T = \frac{p^2}{2m} $$

Replacing $p$ with its quantum mechanical operator
$$ p = -i\hbar\frac{\partial}{\partial x} $$

gives
$$\hat{T} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} $$

I'm not sure why you bring up acceleration. Acceleration $a$ is the second time derivative of position, $$a = \frac{d^2 x}{d t^2}$$