The motion of a body falling from rest in a medium is given by equation- dvdt-1-2v- The velocity at any time t is-

The correct option is D 12(1 e^ 2t) dvdt=1 2v dv1 2v=dt Integrating #160; log(1 2v)2=t 1 2v=e^ 2t v=12(1 e^ 2t) ...

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Standard XII

Physics

Instantaneous Velocity

The motion of...

Question

The motion of a body falling from rest in a medium is given by equation, $\frac{d v}{d t} = 1 - 2 v$ . The velocity at any time $t$ is:

(1 - e^{t})

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(1 - e^{- t})

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\frac{1}{2} (1 - e^{t})

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\frac{1}{2} (1 - e^{- 2 t})

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Solution

The correct option is D $\frac{1}{2} (1 - e^{- 2 t})$
$\frac{d v}{d t} = 1 - 2 v$
$\frac{d v}{1 - 2 v} = d t$
Integrating
$\frac{- l o g (1 - 2 v)}{2} = t$
$1 - 2 v = e^{- 2 t}$
$v = \frac{1}{2} (1 - e^{- 2 t})$

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The motion of a body falling from rest in a medium is given by equation, dvdt=1−2v. The velocity at any time t is:

The correct option is D 12(1−e−2t)dvdt=1−2vdv1−2v=dtIntegrating −log(1−2v)2=t1−2v=e−2tv=12(1−e−2t)

The motion of a body falling from rest in a medium is given by equation, $\frac{d v}{d t} = 1 - 2 v$ . The velocity at any time $t$ is:

The correct option is D $\frac{1}{2} (1 - e^{- 2 t})$
$\frac{d v}{d t} = 1 - 2 v$
$\frac{d v}{1 - 2 v} = d t$
Integrating
$\frac{- l o g (1 - 2 v)}{2} = t$
$1 - 2 v = e^{- 2 t}$
$v = \frac{1}{2} (1 - e^{- 2 t})$