= Solution
If $y''=0$ over the subsequent segment, its slope is a constant $s$ and the <equation of motion> reduces to
$$
\boxed{\ddot x=-\frac{gs}{1+s^2}.}
$$
There is uniform horizontal <acceleration>, not generally zero <acceleration>. Starting that segment at $(t_0,x_0,\dot x_0)$,
$$
x(t)=x_0+\dot x_0(t-t_0)-\frac{gs}{2(1+s^2)}(t-t_0)^2.
$$
For a downhill straight track with $s=-\tan\theta$, this gives $\ddot x=g\sin\theta\cos\theta$, agreeing with <10/b>. Only $s=0$ gives constant horizontal <velocity>.
The condition $y''\to0$ must describe an actually straight limiting segment, with bounded <velocity>, for that subsequent constant-<acceleration> conclusion. If $y''$ vanishes only at an isolated inflection point, the boxed expression describes the instantaneous <acceleration> there; it does not make the rest of the track straight. Likewise, a vanishing $y''$ alone does not make the product $y''\dot x^2$ negligible in a limit where $\dot x$ diverges.
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