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Let $\phi$ be a real, decreasing differentiable function defined on $[a, b]$. An object is dropped at the point $(a, \phi(a))$ and is accelerated only by gravity. How long does it take to reach $(b, \phi(b))$?

Using simple geometry, I believe I've correctly shown that the horizontal acceleration as a function of the horizontal position, x, is given by:

$$x''(t) = g^2\frac{\phi'(x)^2}{(1 + \phi'(x))^4}$$

Something like that anyway - I don't have my notes with me right now. The exact form of the expression doesn't matter, the point is it's a constant function (with respect to $t$) given by $\phi'(x)$.

Now I'd like to integrate this twice to get $x(t)$ (setting $x'(0) = 0$, $x(0) = a$), then set that equal to $b$ and solve for $t$. But I hit a snag. Conceptually, I can see that I need to do some sort of integration with respect to $x$ - thus "adding up" all of the values that the acceleration takes along the curve. But wait, maybe the integration should be with respect to $t$? After all, acceleration is a derivative with respect to $t$, not $x$. So how do I deal with the interactions between these two variables?

Could somebody explain the relevant methodology?

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You have a differential equation for $x(t)$. The solution of these is a massive subject. – Ross Millikan Apr 22 '13 at 15:59

I think the correct equation is $$x''(t)=-g\frac{\phi'(x)}{ 1+(\phi'(x))^2 } \tag1$$

The exact form of the expression doesn't matter

I beg to differ. The exact form matters very much when we deal with nonlinear equations. (The right hand side is most likely a nonlinear function of the unknown $x$.) It would take a minor miracle to get an explicit solution of (1), and whether or not a miracle occurs in your problem depends on what $\phi$ is.

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