How does the uncertainty principle make sense of the fact that momentum for massive particles depends in part on position? The momentum of an object is in part dependent on the change in position meaning the final position minus the initial position.  The  equation for momentum is
$$p=\frac{m \Delta x}{t\sqrt{1-(\Delta x/tc)^2}}$$
where $\Delta x$ means change in position, $c$ means speed of light, and $t$ means time.
The uncertainty principle states that the uncertainty of position and the uncertainty of momentum are inversely related.  How are the uncertainty principle and the equation for momentum consistent with each other?
 A: They are consistent with each other because quantum mechanical momentum is not the change in position.
There is no quantum notion of velocity. Classically, velocity is the time derivative of $x(t)$ along a particular trajectory. The quantum theory has no notion of a real-valued trajectory $x(t)$. Quantum states, in general, are not position eigenstates, they don't have a position from which you could compute the velocity.
Quantum mechanical momentum is the quantization of the classical Hamiltonian momentum, not of classical velocity. It is the Hamilton equation of motion
$$ \dot{x} = \frac{\partial H}{\partial p} = \{q,H\}$$
that relates $\dot{x}$ to $p$ for the usual Hamiltonian quadratic in the momenta, but this equation is purely classical. It doesn't exist in the quantum theory except as an equation of expectation values (by Ehrenfest's theorem)
$$ \frac{\mathrm{d}}{\mathrm{d}t}\langle x\rangle = \frac{1}{\mathrm{i}\hbar}\langle[x,H]\rangle$$
which makes no statement about the standard deviations.
There is another sense in which $\dot{x} = p$ holds in the quantum theory, which is that the quantum mechanical equation of motion in the Heisenberg picture is really the quantum version of the classical Hamiltonian equations of motion, i.e. all operators are time-dependent, and so writing
$$ \dot{x} = \frac{i}{\hbar}[H,x]$$
does makes sense. However, the object $\dot{x}$ is not the classical notion of velocity - the time derivative is here not along a classical trajectory, but along the one-parameter group of operators $\{\mathrm{e}^{-\mathrm{i}tH}x\mathrm{e}^{\mathrm{i}tH}\}$ that are the time-evolution of the operator $x$. In particular, it does not make sense to speak of the velocity being the "change in position" divided by the time, because, again, most quantum states don't have a well-defined position and because something that is a position eigenstate at initial time will usually not stay one.
A: The velocity $\hat v$ is a well-defined operator for particles, equal to $\hat p/m$ in non-relativistic physics. The non-relativistic, non-linear correction factors could be added but that would create a can of worms because the correct way to describe relativistic particles requires quantum field theory where $\hat x,\hat v,\hat p$ aren't quite well-defined because the number of particles may change.
In non-relativistic physics, it is also true that $\hat v = d \hat x / dt$, at least in the Heisenberg picture where this equation holds as an operator equation.
Now, the equation with $m\cdot \Delta x$ in the original question only holds if $\hat p$ means the "average" momentum (and similarly velocity etc.) during the relevant interval of time.
But we may ignore these things by taking $\Delta x$ infinitesimal. Then, ignoring the non-relativistic corrections,
$$ p = \frac{\Delta x\cdot m}{\Delta t} $$
Here, the $\Delta$ means the "change of the quantity" in the infinitesimal time $\Delta t$. 
The key point. This equation is in no way "incompatible" with the uncertainty principle. The uncertainty principle says that if we know $x$ accurately e.g. at the initial moment of the infinitesimal interval, we can't know $p$ accurately. Because $p$ is rewritten in terms of $\Delta x$, the uncertainty principle simply means that if we know $x$ at the initial moment accurately, we can't know $\Delta x$ accumulated in the short period of time.
If $\delta$ means the statistical uncertainty, we will have
$$ \delta x \cdot \delta(\Delta x) \geq \frac{\hbar\cdot \Delta t}{2m} $$
For example, the uncertainty principle says that if the initial $x(t)$ were completely sharply determined, the position $x(t+\Delta t)$ would already be completely undetermined. 
A: The uncertainty principle states that product of the standard deviations of two observables is bounded below by a multiple of the absolute value of the expectation value of the commutator of the two operators. In detail, if $\Delta A=\sqrt{\langle \Psi|\left(A^2-\langle \Psi|A|\Psi\rangle^2\right)|\Psi\rangle}$ and $\Delta B=\sqrt{\langle \Psi|\left(B^2-\langle \Psi|B|\Psi\rangle^2\right)|\Psi\rangle}$ then $$\Delta A \Delta B \geq \left|\frac{\langle AB-BA\rangle}{2i}\right|=\left|\frac{\langle\Psi| AB-BA |\Psi\rangle}{2i}\right|
$$
For a canonical momentum conjugate to a specific coordinate, the commutator is a simple scalar and so has an expectation value that doesn't depend on the state and you can get the product of the two standard deviations to be bounded below by $\hbar/2.$
The canonical momentum only sometimes is equal to the mechanical momentum. And your formulation with a $t$ in the denominator is just not good at all. And they don't have to be inversely related to have a lower bound. To better understand the uncertainty principle you can read my answer to https://physics.stackexchange.com/a/169757
