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 ...

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})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = \frac{α}{β} (1 - e^{- β t})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

v = \frac{β}{α} e^{- β t}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = \frac{α}{β} e^{- β t}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

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.

Q. The displacement

x

of a particle varies with time

t

x = a e^{- α t} + b e^{β t}

, where

a, b, α

and

β

are positive constants. The velocity of the particle will

I f \int_{2}^{e} [\frac{1}{l o g x} - \frac{1}{(l o g x)^{2}}] d x = α + \frac{β}{l o g 2}, t h e n

Q. If

A (α, β) = ⎡ ⎢ ⎣ \begin{matrix} cos α & sin α & 0 - sin α & cos α & 0 0 & 0 & e^{β} \end{matrix} ⎤ ⎥ ⎦

, then

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

The correct option is D v=αβ(1−e−βt)dvdt=α−βvdvdt=α{1−βvα}dv(1−βv/α)=αdtIntegratingbothsidesweget,ln(1−βαv)×−αβ=αt1−βαv=e−βtv=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. Assuming particle starts from rest, it's velocity as a function of time will be