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Question: On a rainy day, a raindrop falls from very high clouds and faces retardation due to air. This retard...

On a rainy day, a raindrop falls from very high clouds and faces retardation due to air. This retardation is directly proportional to the instantaneous speed of the drop. An expression for the distance traveled by the drop in time is

A

s=gα2(e−αt−1)+gαts = \frac{g}{\alpha^{2}}\left( e^{- \alpha t} - 1 \right) + \frac{g}{\alpha}t

B

s = gtα\frac{gt}{\alpha}

C

s=gα2(e−αt−1)s = \frac{g}{\alpha^{2}}\left( e^{- \alpha t} - 1 \right)

D

s=gtα2+gα(e−αt−1)s = \frac{gt}{\alpha^{2}} + \frac{g}{\alpha}\left( e^{- \alpha t} - 1 \right)

Answer

s=gα2(e−αt−1)+gαts = \frac{g}{\alpha^{2}}\left( e^{- \alpha t} - 1 \right) + \frac{g}{\alpha}t

Explanation

Solution

Let the retardation produced by instantaneous opposition

= αv (where α is a constant)

Net instantaneous acceleration = g - αv

i.e., dv/dt = (g - αv)

Integrating, ∫0vdv(g−αv)=∫0tdt\int_{0}^{v}\frac{dv}{(g - \alpha v)} = \int_{0}^{t}{dt}

In g−αvg−αti.e.,g−αvg=e−αt\frac{g - \alpha v}{g} - \alpha ti.e.,\frac{g - \alpha v}{g} = e^{- \alpha t}

i.e., v = gα\frac{g}{\alpha} (1 – e-αt)

i.e., dSdt=gα(1−e−αt)\frac{dS}{dt} = \frac{g}{\alpha}\left( 1 - e^{- \alpha t} \right)

i.e., dS = gα(1−e−αt)\frac{g}{\alpha}\left( 1 - e^{- \alpha t} \right)dt

Integrating, ∫0sdS=gα∫0t(1−e−αt)dt\int_{0}^{s}{dS} = \frac{g}{\alpha}\int_{0}^{t}{\left( 1 - e^{- \alpha t} \right)dt}

= gα∫0tdt−gα∫0te−αtdt\frac{g}{\alpha}\int_{0}^{t}{dt} - \frac{g}{\alpha}\int_{0}^{t}{e^{- \alpha t}dt}

or S = gαt+gα2e−αt−gα2\frac{g}{\alpha}t + \frac{g}{\alpha^{2}}e^{- \alpha t} - \frac{g}{\alpha^{2}}

or S = gα2\frac{g}{\alpha^{2}} (e-αt – 1 ) + gα\frac{g}{\alpha}t