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 $\mathrm{t}$ is $x^{-\mathrm{n}}$ where $\mathrm{n}=$ _________.

<p>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 $n$.</p> <p>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:</p> <p>$2x\frac{dx}{dt} = 2t$</p> <p>This simplifies to:</p> <p>$\frac{dx}{dt} = \frac{t}{x}$</p> <p>Now, let's differentiate this velocity to find the acceleration:</p> <p>$a = \frac{d^2x}{dt^2} = \frac{d}{dt}\left(\frac{t}{x}\right)$</p> <p>To differentiate $\frac{t}{x}$ with respect to $t$, we'll use the quotient rule:</p> <p>$$\frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$$</p> <p>Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:</p> <p>$$a = \frac{x(1) - t(\frac{t}{x})}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$$</p> <p>Recalling that the displacement equation given was $x^2 = 1 + t^2$, substitute this into our expression for acceleration:</p> <p>$a = \frac{1 + t^2 - t^2}{x^3} = \frac{1}{x^3}$</p> <p>Thus, the acceleration of the particle at any time $t$ is $x^{-3}$, which means our value for $n$ is $3$.</p>