QJEEM24Ph9086: JEE Main 2024 Physics - Class 11 JEEM24Ph9086 Solution: A particle moves in a straight line so that its displacement at …

Given the displacement of the particle $x^2 = 1 + t^2$ , we want to find the acceleration, which is the second derivative of displacement with respect to time, $a = \frac{d^2x}{dt^2}$ , and we are given that the acceleration at any time $t$ is $x^{-n}$ , for us to find the value of $n$ .

First, let's find the first derivative of displacement with respect to time, which gives us the velocity. Differentiating $x^2 = 1 + t^2$ with respect to t, we get:

$2x\frac{dx}{dt} = 2t$

This simplifies to:

$\frac{dx}{dt} = \frac{t}{x}$

Now, let's differentiate this velocity to find the acceleration:

$a = \frac{d^2x}{dt^2} = \frac{d}{dt}\left(\frac{t}{x}\right)$

To differentiate $\frac{t}{x}$ with respect to $t$ , we'll use the quotient rule:

$\frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$

Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:

$a = \frac{x(1) - t(\frac{t}{x})}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$

Recalling that the displacement equation given was $x^2 = 1 + t^2$ , substitute this into our expression for acceleration:

$a = \frac{1 + t^2 - t^2}{x^3} = \frac{1}{x^3}$

Thus, the acceleration of the particle at any time $t$ is $x^{-3}$ , which means our value for $n$ is $3$ .