Question

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

• A. $v=c{e}^{{-}^{\frac{k}{m}t}}-\frac{mg}{k}$
• 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)

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