Graphs of Motion question from JEE Main 2021 (18 Mar, Shift 2)

Answer: (a)

1. From the graph $v = v_0\left(1 - \frac{x}{x_0}\right)$, so $\frac{dv}{dx} = -\frac{v_0}{x_0}$.
2. $a = v\frac{dv}{dx} = -\frac{v_0^2}{x_0}\left(1 - \frac{x}{x_0}\right)$
3. At $x = 0$: $a = -\frac{v_0^2}{x_0}$; at $x = x_0$: $a = 0$; linear in between.
4. Straight line from a negative value up to zero at $x_0$. So the answer is (a).