All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

For a uniformly decelerating motion that reverses direction, velocity decreases linearly with time, position varies parabolically, and distance must monotonically increase with a continuously non-negative slope.

From graph (A):

$$v = u - at \implies \text{motion with uniform negative acceleration}$$

Analyzing Option (C):

$$\text{Slope of distance-time graph} = \text{speed} = \vert{}v\vert{}$$

$$\text{At } v = 0 \implies \text{speed} = 0 \implies \text{slope must be zero}$$

$$\text{For } t > t_{\text{reversal}} \implies \vert{}v\vert{} \text{ increases } \implies \text{slope must increase}$$

$$\text{Graph (C) shows slope decreasing to zero at the end, which is incorrect for distance.}$$