Graphs of Motion question from JEE Main 2017 (8 Apr, Shift 1)

Answer: (c)

1. With constant retardation $a$: $v^2 = u^2 - 2ax$, so $v = \sqrt{u^2 - 2ax}$.
2. $v$ stays positive and falls slowly at first, then steeply, reaching zero at $x = \frac{u^2}{2a}$ with a vertical tangent.
3. This matches the velocity-distance curve that bends down to the $x$-axis. So the answer is (c).