A particle moves according to the equation dv dt - - v where - and - are constants- Assuming particle starts from rest- it-s velocity as a function of time will be

The correct option is D v=αβ (1 e ^ β t ) dv dt =α β v dv dt =α{ 1 β v α #160;} dv (1 β v/α ) =α dt Integrating both si ...

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Standard XII

Physics

Uniform and Non Uniform

A particle mo...

Question

A particle moves according to the equation $\frac{d v}{d t} = α - β v$ where $α$ and $β$ are constants. Assuming particle starts from rest, it's velocity as a function of time will be

v = \frac{β}{α} (1 - e^{- β t})

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v = \frac{α}{β} (1 - e^{- β t})

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v = \frac{β}{α} e^{- β t}

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v = \frac{α}{β} e^{- β t}

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Solution

The correct option is D $v = \frac{α}{β} (1 - e^{- β t})$
$\frac{d v}{d t} = α - β v \frac{d v}{d t} = α {1 - \frac{β v}{α}} \frac{d v}{(1 - β v / α)} = α d t I n t e g r a t i n g b o t h s i d e s w e g e t, l n (1 - \frac{β}{α} v) \times \frac{- α}{β} = α t 1 - \frac{β}{α} v = e^{- β t} v = \frac{α}{β} (1 - e^{- β t})$

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A particle moves according to the equation dvdt=α−βv where α and β are constants. Assuming particle starts from rest, it's velocity as a function of time will be

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A particle moves according to the equation $\frac{d v}{d t} = α - β v$ where $α$ and $β$ are constants. Assuming particle starts from rest, it's velocity as a function of time will be