Although this is a biomechanics problem it is of interest to see what kind of physics model would apply here. So I shall take the simplest model of movement against friction and gravity. I shall discuss some more aspects of it after some basics.
I have translated some of the numbers into SI and also will introduce some notation.
D = 1609 meters - the distance
T = 415 seconds - the time
H = 71.93 ~ 72 meters - the height increase
$V_{av}$ = 3.877 meters per second - the average velocity
$\theta$ = 2.56 degrees - the elevation angle
$cos \theta = 0.999$ and $tan \theta = 0.045$ (the number used in the V02 formula)
So the most basic equivalent model is that of pushing a mass M against a friction force $\mu$ for a distance D: level ($\theta=0$) and inclined ($\theta$> 0).
The equation for moving level is:
Work Done = $WD^{level}$ = Friction $\times$ D = $\mu mgD$
For moving at incline $\theta$ is:
Work Done = $WD^{incline}$ = Friction $\times$ D + WD on gravity over H
$= \mu mgD cos \theta + mgD sin \theta$
$= mgD (\mu cos \theta + sin \theta)$
If we note that $cos \theta$ is appoximately 1 we can write:
$WD^{incline} = \mu mgD (1 + (\mu)^{-1} tan \theta) = WD^{level} \times \alpha^2$ say.
Thus moving a fixed distance against a common value of friction the extra work done in going upward is $\alpha^2 = (1 + (\mu)^{-1} tan \theta)$. This is greater than one (as expected) and in our approximation is work against gravity modified by the inverse coefficient of friction.
Now we come to the biomechanics. Can we obtain information about the speed increase in a runner (or any powered object)? The parameter we have available here is the coefficient of friction. We also need to assume some model as to how the extra available energy will be used. Let us assume that all of the extra energy is transfered into kinetic energy: so we assume a linear model (in absence of any more detailed model).
Work Done $\propto$ Kinetic Energy = $1/2 m V^2$
In our case all the extra energy $\alpha^2$ is being added to the energy for the plane giving an increase of velocity:
$V_{new}^2 = V_{av}^2(1 + (\mu)^{-1} tan \theta)$
and thus a proportional decrease of time:
$T_{new} = T (1 + (\mu)^{-1} tan \theta)^{-1/2}$
We know the value of $tan \theta$ but do not know $\mu$. Just to put in a first value for $\mu$ which corresponds to rubber on concrete ie 1 we get:
$T_{new} = T \sqrt(1+0.045)^{-1} = 415 \times 0.978 = 406$ seconds.
This is somewhat slower than the V02 formula and we have not really a valid model of movement because this assumes a "continuous dragging" but it is a first approximation. A different approximation is to consider rubber tyres rolling. The coefficient of friction is defined differently, but could be as low as 0.01. This ignores all the other friction effects in the vehicle however.
Inverting the formula and putting in the result from the V02 answer (344 seconds), to derive a value for $\mu$ we get:
$\mu = 0.045 / ((T/T_{new})^2 -1) = 0.045 / (1.455 - 1) = 0.1$
So this "generalised friction" model predicts a $\mu$ value smaller than most static friction cases and in the direction of a "rolling model".
