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(A) Given that retardation $a$ is proportional to displacement $x$,so $a = -kx$ (where $k$ is a positive constant).Using the relation $a = v \frac{dv}

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$x^2$

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(A) Given that retardation $a$ is proportional to displacement $x$,so $a = -kx$ (where $k$ is a positive constant).Using the relation $a = v \frac{dv}{dx}$,we have $v \frac{dv}{dx} = -kx$.Integrating both sides: $\int v dv = \int -kx dx$.This gives $\frac{v^2}{2} = -\frac{kx^2}{2} + C$.At $x = 0$,let the initial velocity be $u$,so $C = \frac{u^2}{2}$.Thus,$\frac{v^2}{2} = \frac{u^2}{2} - \frac{kx^2}{2}$,which implies $u^2 - v^2 = kx^2$.The loss in kinetic energy is $\Delta K = \frac{1}{2}m(u^2 - v^2) = \frac{1}{2}m(kx^2)$.Since $m$ and $k$ are constants,the loss in kinetic energy is proportional to $x^2$.

$A$ particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to:

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