Q 11-02-107JEE MainJEE Main 2021 (18 Mar, Shift 2)Medium
The velocity-displacement graph of a particle is shown in the figure.
The acceleration-displacement graph of the same particle is represented by:
Answer: (C) see figure
From the graph $v = v_0\left(1 - \dfrac{x}{x_0}\right)$, so $\dfrac{dv}{dx} = -\dfrac{v_0}{x_0}$.
$$a = v\frac{dv}{dx} = -\frac{v_0^2}{x_0}\left(1 - \frac{x}{x_0}\right) = -\frac{v_0^2}{x_0} + \frac{v_0^2}{x_0^2}x$$
This is a straight line with positive slope, starting at a negative value $-v_0^2/x_0$ at $x = 0$ and reaching zero at $x = x_0$, which is graph (3).
Solution by Sreeraj P, M.Sc Physics