Graphs of Motion question from JEE Main 2026 (24 Jan, Shift 2)

Answer: (d)

1. From the graph $v = v_0 - kx$ (straight line, slope $-k$).
2. $a = v\frac{dv}{dx} = (v_0 - kx)(-k) = -kv_0 + k^2x$
3. So $a$-$x$ is a straight line with negative intercept $-kv_0$ and positive slope $k^2$, crossing zero where $v = 0$.
4. So the answer is (d).