More on the Feynman Path Integral Formula  in Brian Cox' Lecture and its Consequences This is a continuation of this question about Brian Cox' lecture Night with the Stars.
I know the main steps to get from $K(q",q',T)=\sum_{paths}Ae^{iS(q",q',T)/h}$ to $\Delta t > \dfrac{m(\Delta x)^2}{h}$ as stated below, but can you expand? (just read below)
PART 1
The action function $S(q",q',T)$ is given by $ S = \displaystyle\int dt\left( \dfrac{1}{2} m v^2 -U\right)$. For a classical path that goes uniformly from one point to the other you have $v = \dfrac{\Delta x}{\Delta t}$ and so you get $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$. What are the processes and steps taken to you get to $S \propto m \left(\dfrac{\Delta x}{\Delta t}\right)^2\Delta t=m\dfrac{(\Delta x)^2}{\Delta t}$? (explained clearly please).
PART 2
$S/h$ appears as a complex phase term. To make it small we set $S/h < 1$, and we can then deduce that $\Delta t > \dfrac{m(\Delta x)^2}{h}$.
What are the processes and steps taken to then get to $\Delta t > \dfrac{m(\Delta x)^2}{h}$?
 A: Part 1:
Let's say the velocity in the integral is constant in time, and the integral is from 0 to $\Delta t$. We now have a trivial integral of a constant. So (ignoring U)
$$
S = \int_0^{\Delta t} \frac{1}{2}mv^2\ dt\\
=\left[\frac{1}{2}mv^2t\right]_{t=0}^{t=\Delta t}\\
=\frac{1}{2}mv^2\Delta t
$$
So, substituting $v = \dfrac{\Delta x}{\Delta t}$, and ignoring any constants we have before them, except for mass, we get
$$
S \propto mv^2 \Delta t\\
  \propto m\left(\frac{\Delta x}{\Delta t}\right)^2 \Delta t\\
  \propto m\frac{\Delta x^2}{\Delta t}
$$
So $S \propto m \dfrac{\Delta x^2}{\Delta t}$ as required.
Part 2:
We have
$$
S \propto m \dfrac{\Delta x^2}{\Delta t}
$$, so then, saying the constant of proportionality is $k$, and that it is approximately 1 (we'll want this not to be huge in a minute), we get
$$
S = k m \frac{\Delta x^2}{\Delta t}
$$
Now, setting $S/h < 1$, we get
$$
\frac{S}{h} = k m \frac{\Delta x^2}{h \Delta t} < 1
$$
Setting $k = 1$, we can now say
$$
m \frac{\Delta x^2}{h \Delta t} < 1
\Rightarrow m \frac{\Delta x^2}{h} < \Delta t
$$
as required.
