 Solution

If $y^{\prime \prime }=0$ over the subsequent segment, its slope is a constant $s$ and the equation of motion reduces toThere is uniform horizontal acceleration, not generally zero acceleration. Starting that segment at $(t_0,x_0,{\dot{x}}_0)$,For a downhill straight track with $s=-\tan \theta$, this gives $\ddot{x}=g\sin \theta \cos \theta$, agreeing with b. Only $s=0$ gives constant horizontal velocity.

The condition $y^{\prime \prime }\to 0$ must describe an actually straight limiting segment, with bounded velocity, for that subsequent constant-acceleration conclusion. If $y^{\prime \prime }$ 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 $y^{\prime \prime }$ alone does not make the product $y^{\prime \prime }{\dot{x}}^2$ negligible in a limit where $\dot{x}$ diverges.