Q 11-02-070JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
The relation between time $t$ and distance $x$ is $t = \alpha x^2 + \beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration $a$ and velocity $v$ is:
Answer: (A) $a = -2\alpha v^3$
$\dfrac{dt}{dx} = 2\alpha x + \beta = \dfrac1v$, so $v = (2\alpha x + \beta)^{-1}$.
$$a = v\frac{dv}{dx} = v\cdot\left[-2\alpha(2\alpha x + \beta)^{-2}\right] = -2\alpha v\cdot v^2 = -2\alpha v^3$$
Solution by Sreeraj P, M.Sc Physics