The motion of a body falling from rest in a resting medium is described by the equation dvdt=a−bv where a and b are constant. The velocity at any time ‘t’ is given by

Given dvdt=a−bv⇒dv=a−bvdt⇒∫0vdva−bv=∫0tdt⇒Lna−bv−bvo=tt0⇒Lna−bvv0=−bt⇒Lna−bva=−bt⇒a−bva=e−bt⇒1−bav=e−bt∴bav=1−e−bt∴v=ab1−e−bt

The motion of a body falling from rest in a resting medium is described by the equation dvdt=a−bv where a and b are constant. The velocity at any time ‘t’ is given by

Given dvdt=a−bv⇒dv=a−bvdt⇒∫0vdva−bv=∫0tdt⇒Lna−bv−bvo=tt0⇒Lna−bvv0=−bt⇒Lna−bva=−bt⇒a−bva=e−bt⇒1−bav=e−bt∴bav=1−e−bt∴v=ab1−e−bt

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The motion of a body falling from rest in a resting medium is described by the equation $\frac{d v}{d t} = a - b v$ where a and b are constant. The velocity at any time ‘t’ is given by

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$v = \frac{a}{b} (1 - e^{- b t})$

$v = \frac{b}{a} (1 - e^{- b t})$

$v = \frac{a}{b} (1 + e^{- b t})$

$v = \frac{b}{a} e^{b t}$

answer is A.

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Detailed Solution

Given $\frac{d v}{d t} = a - b v$

$\Rightarrow d v = (a - b v) d t$

$\Rightarrow \int_{0}^{v} \frac{d v}{a - b v} = \int_{0}^{t} d t$ $\Rightarrow {[\frac{L_{n} (a - b v)}{- b}]}^{v}_{o} = {[t]}^{t}_{0}$

$\Rightarrow {[L_{n} (a - b v)]}^{v}_{0} = - b t$ $\Rightarrow L_{n} (\frac{a - b v}{a}) = - b t$

$\Rightarrow \frac{a - b v}{a} = e^{- b t}$ $\Rightarrow 1 - \frac{b}{a} v = e^{- b t}$

$∴ \frac{b}{a} v = 1 - e^{- b t}$

$∴ v = \frac{a}{b} [1 - e^{- b t}]$

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The motion of a body falling from rest in a resting medium is described by the equation dvdt=a−bv where a and b are constant. The velocity at any time ‘t’ is given by

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The motion of a body falling from rest in a resting medium is described by the equation $\frac{d v}{d t} = a - b v$ where a and b are constant. The velocity at any time ‘t’ is given by