A particle moves according to the equation dv - dt - v- where - and - are constants- Find the velocity as a funtion of time- Assume body starts from rest-A- v- -e- tB- v- -1-e- tC- v- -e- tD- v- -1-e- t

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Standard XII

Physics

Motion Under Variable Acceleration

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. Find the velocity as a funtion of time. Assume body starts from rest.

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Solution

The correct option is C

Take the relation given for acceleration and integrate with the given limits to obtain the desired function for velocity.

$\frac{d v}{d t}$ = $α$ - $β$ v $\Rightarrow$ $v \int 1$ $\frac{d v}{α - β v}$ = $t \int 0$ dt

v = ( $\frac{α}{β}$ ) (1 - $e^{- β t}$ )

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A particle moves according to the equation dvdt = α - β v , where α and β are constants. Find the velocity as a funtion of time. Assume body starts from rest.

The correct option is C Take the relation given for acceleration and integrate with the given limits to obtain the desired function for velocity. dvdt = α - β v ⇒ v∫1 dvα−βv = t∫0 dt v = (αβ) (1 - e−βt)

A particle moves according to the equation $\frac{d v}{d t}$ = $α$ - $β$ v , where $α$ and $β$ are constants. Find the velocity as a funtion of time. Assume body starts from rest.

The correct option is C

Take the relation given for acceleration and integrate with the given limits to obtain the desired function for velocity.

$\frac{d v}{d t}$ = $α$ - $β$ v $\Rightarrow$ $v \int 1$ $\frac{d v}{α - β v}$ = $t \int 0$ dt

v = ( $\frac{α}{β}$ ) (1 - $e^{- β t}$ )