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(D) Given acceleration $a = v \frac{dv}{dx} = -\alpha x^2$.Separating the variables,we get $v dv = -\alpha x^2 dx$.Integrating both sides from the ini

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

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(D) Given acceleration $a = v \frac{dv}{dx} = -\alpha x^2$.Separating the variables,we get $v dv = -\alpha x^2 dx$.Integrating both sides from the initial state $(x=0, v=v_0)$ to the final state $(x=d, v=0)$:$\int_{v_0}^{0} v dv = \int_{0}^{d} -\alpha x^2 dx$.Evaluating the integrals:$[\frac{v^2}{2}]_{v_0}^{0} = -\alpha [\frac{x^3}{3}]_{0}^{d}$.$0 - \frac{v_0^2}{2} = -\alpha (\frac{d^3}{3})$.$\frac{v_0^2}{2} = \frac{\alpha d^3}{3}$.Solving for $d$:$d^3 = \frac{3v_0^2}{2\alpha}$.$d = (\frac{3v_0^2}{2\alpha})^{1/3}$.

$A$ particle is projected with velocity $v_0$ along the $x-$axis,and its deceleration is given by $a = -\alpha x^2$. The distance at which the particle stops is

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