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

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

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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$$\frac{dx}{dt} = \frac{t}{x} = v$ (velocity).Now,differentiate velocity $v$ with respect to $t$ to find acceleration $a$:$a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{t}{x} \right)$Using the quotient rule $\frac{d}{dt} (u/v) = \frac{v(du/dt) - u(dv/dt)}{v^2}$:$a = \frac{x(1) - t(\frac{dx}{dt})}{x^2}$Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:$a = \frac{x - t(\frac{t}{x})}{x^2}$$a = \frac{x - \frac{t^2}{x}}{x^2}$$a = \frac{x}{x^2} - \frac{t^2}{x^3}$$a = \frac{1}{x} - \frac{t^2}{x^3}$.

Column $I$	Column $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d\|v\|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left\|\frac{dr}{dt}\right\|$	$(s)$ Magnitude of velocity

$A$ point moves in a straight line so that its displacement $x \ m$ at time $t \ s$ is given by $x^2 = 1 + t^2$. Its acceleration in $m/s^2$ at a time $t \ s$ is

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Column $I$ Column $II$
$(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration
$(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$ $(r)$ Velocity
$(D)$ $\left|\frac{dr}{dt}\right|$ $(s)$ Magnitude of velocity

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