Question

The solution of $\frac{dv}{dt} +\frac{k}{m}v = -g$ is

• A. $v = ce^{-^{\frac{k}{m}t}} - \frac{mg}{k}$
• B. $v = c- \frac{mg}{k} e^{-^{\frac{k}{m}t}}$
• C. $v e^{-^{\frac{k}{m}t}} = c- \frac{mg}{k}$
• D. $v e^{^{\frac{k}{m}t}} = c- \frac{mg}{k}$

Answer 1

Answer: (A)

$\frac{dv}{dt}+\frac{k}{m}v = -g \Rightarrow \frac{dv}{dt} = -\frac{k}{m}\left(v+\frac{mg}{k}\right)$
$\Rightarrow \frac{dv}{v+mg/k} = -\frac{k}{m}dt \Rightarrow log \left(v+\frac{mg}{k}\right)$
$= -\frac{k}{m}t+log\,C$
$\Rightarrow v+\frac{mg}{k}= Ce^{-kt/m} \Rightarrow v = Ce^{-kt/m} -\frac{mg}{k}$