Q 11-02-062JEE MainJEE Main 2024 (6 Apr, Shift 2)Medium
A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2 = 1 + t^2$. Its acceleration at any time $t$ is $x^{-n}$, where $n = $ ______.
Numerical value type. Enter your answer.
Answer: 3
Differentiate: $2xv = 2t \Rightarrow v = \dfrac tx$.
$$a = \frac{dv}{dt} = \frac{x - tv}{x^2} = \frac{x - t^2/x}{x^2} = \frac{x^2 - t^2}{x^3} = \frac{1}{x^3}$$
So $n = 3$.
Solution by Sreeraj P, M.Sc Physics