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(C) Given the displacement equation: $x^2 = 1 + t^2$.Differentiating both sides with respect to $t$: $2x \frac{dx}{dt} = 2t$,which simplifies to $x v

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$3$

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(C) Given the displacement equation: $x^2 = 1 + t^2$.Differentiating both sides with respect to $t$: $2x \frac{dx}{dt} = 2t$,which simplifies to $x v = t$,where $v$ is the velocity.Differentiating $x v = t$ with respect to $t$ using the product rule: $x \frac{dv}{dt} + v \frac{dx}{dt} = 1$.Since $\frac{dv}{dt} = a$ (acceleration) and $\frac{dx}{dt} = v$,we get $x a + v^2 = 1$.Substituting $v = \frac{t}{x}$ into the equation: $x a + (\frac{t}{x})^2 = 1$.$x a = 1 - \frac{t^2}{x^2} = \frac{x^2 - t^2}{x^2}$.Since $x^2 - t^2 = 1$ from the original equation,we have $x a = \frac{1}{x^2}$.Therefore,$a = \frac{1}{x^3} = x^{-3}$.Comparing this with $x^{-n}$,we find $n = 3$.

$A$ 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 $n = . . . . .$

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