How far do you get in time t if your velocity is given by the incline? Suppose we have a person walking in a straight line across a landscape, starting at 0. The function $h$ gives the elevation of the terrain for each point along the line.
Suppose that the speed of the person is given by the incline of the terrain she is walking across, i.e. for each point $x$ along the line, the speed is given by $u(x) = \sigma(h'(x))$, where $\sigma$ is the logistic function (of course, any monotone function $\sigma':\mathbb{R} \rightarrow \mathbb{R}^+$ could be used instead). 
The question is, what is the function $d$, that gives us for each time $t$ the distance covered in that time?
Here is my partial solution. If we knew the velocity as a function $v$ of time. We could get $d(t) = \int_0^t v(y) \: dy$. We can see that $v(t) = u(d(t))$, which gives the following recursive definition $$d(t) = \int_0^t \sigma(h'(d(y)))\: dy$$
Is my approach flawed? Is there an analytic solution for $d$? If no, are there special cases except linear $h$ where such a solution exists (piecewise linear, perhaps)?
 A: I'm using the assumption that the function $\sigma$ gives horizontal speed - that is, it is the speed at which the walker covers $x$ distance as opposed to the magnitude of velocity including the vertical component.  If it is actually velocity then it gets harder.
For a straight hill of slope $h'$ and map distance $x$, then, the time taken is $x / \sigma(h')$.  Similarly, for a rolling hill of slope $h'(x)$, the time taken for a small distance $dx$ at a point $x$ is $$\frac{dx}{\sigma(h'(x))}$$  Integrating this we get a function for time taken for a given distance:
$$t(d) = \int_0^d\frac{1}{\sigma(h'(x))}dx$$
This is unfortunately going to be untenably nasty most of the time: there is no nice way to take the antiderivative of a division in general, and you will be stuck with numerical work.
From here, we must invert this function: given $t(d)$ we must find $d(t)$.  This is again not likely to be tractable, but fortunately since $\sigma$ is nonnegative we can at least try somewhat.
Piecewise linear landscapes will certainly be plausible but will require that you keep track of your progress at each break point.
