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(D) Given that the deceleration $a = -\alpha x^2$. We know that $a = v \frac{dv}{dx}$.Substituting this into the equation,we get $v \frac{dv}{dx} = -\

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$(\frac{3v_0^2}{2\alpha})^{1/3}$

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(D) Given that the deceleration $a = -\alpha x^2$. We know that $a = v \frac{dv}{dx}$.Substituting this into the equation,we get $v \frac{dv}{dx} = -\alpha x^2$.Separating the variables,we have $v \, dv = -\alpha x^2 \, dx$.Integrating both sides with limits from initial velocity $v_0$ to final velocity $0$ and position from $0$ to $S$:$\int_{v_0}^{0} v \, dv = -\alpha \int_{0}^{S} x^2 \, dx$.Evaluating the integrals:$[\frac{v^2}{2}]_{v_0}^{0} = -\alpha [\frac{x^3}{3}]_{0}^{S}$.$0 - \frac{v_0^2}{2} = -\alpha (\frac{S^3}{3})$.$\frac{v_0^2}{2} = \frac{\alpha S^3}{3}$.Solving for $S$,we get $S^3 = \frac{3v_0^2}{2\alpha}$,which implies $S = (\frac{3v_0^2}{2\alpha})^{1/3}$.

$A$ particle is projected with velocity $v_0$ along the $x$-axis. The deceleration of the particle is proportional to the square of the distance from the origin,i.e.,$a = -\alpha x^2$. The distance at which the particle stops is:

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