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A particle moves according to the law a = - ky. Find the velocity as a function of distance y where $v_{o}$ is initial velocity and $a$ is the acceleration.

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$v^{2} = v_{o}^{2} - k y^{2}$

$v^{2} = v_{o}^{2} - 2 k y^{2}$

$v^{2} = v_{o}^{2} + k y^{2}$

$v^{2} = v_{o}^{2} - 2 k y$

answer is A.

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Detailed Solution

$a = \frac{dv}{dt} = \frac{dv}{d y} \frac{d y}{dt} = - k y$

$\Rightarrow \int_{v_{o}}^{v} v dv = - \int_{o}^{v} k y d y$

$\Rightarrow \frac{v^{2} - v_{o}^{2}}{2} = - k \frac{y^{2}}{2}$

$∴ v^{2} = v_{o}^{2} -k y^{2}$

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