A particle moves according to the law a = – ky.  Find the velocity as a function of distance y where v0 is initial velocity and a is the acceleration.

# A particle moves according to the law a = - ky.  Find the velocity as a function of distance y where ${\text{v}}_{\text{0}}$ is initial velocity and a is the acceleration.

1. A

${\text{v}}^{2}={\text{v}}_{\text{0}}^{\text{2}}-\text{k}{y}^{2}$

2. B

${\text{v}}^{2}={\text{v}}_{\text{0}}^{\text{2}}-2\text{k}y$

3. C

${\text{v}}^{2}={\text{v}}_{\text{0}}^{\text{2}}-2\text{k}{y}^{2}$

4. D

${\text{v}}^{2}={\text{v}}_{\text{0}}^{\text{2}}+\text{k}{y}^{2}$

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### Solution:

$\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{dv}}{\text{dy}}\text{\hspace{0.17em}}\frac{\text{d}y}{\text{dt}}=-\text{k}y$

$⇒\underset{{\text{v}}_{\text{o}}}{\overset{\text{v}}{\int }}\text{v}\text{\hspace{0.17em}}\text{dv}=-\underset{\text{o}}{\overset{y}{\int }}\text{k}y\text{d}y$

$⇒\frac{{\text{v}}^{2}-{\text{v}}_{\text{o}}^{\text{2}}}{2}=-\text{\hspace{0.17em}}\text{k}\frac{{y}^{2}}{2}$

$\therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{v}}^{\text{2}}={\text{v}}_{\text{0}}^{\text{2}}-\text{k}{y}^{2}$

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