Graphs of Motion question from JEE Main 2021 (16 Mar, Shift 1)

Answer: (a)

1. For $0 \le x \le 200\text{ m}$: $v = 10 + \frac{x}{5}$, so $\frac{dv}{dx} = 0.2\text{ s}^{-1}$.
2. $a = v\frac{dv}{dx} = 0.2\left(10 + \frac{x}{5}\right) = 2 + \frac{x}{25}$: linear, starting at $2\text{ m/s}^2$.
3. For $200 < x \le 400\text{ m}$: $v$ = constant, so $a = 0$.
4. Linear rise then zero matches graph (a). So the answer is (a).