Discussion Forum



1)

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


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

B) $v = c-\frac{mg}{k}\bar{e}^{\frac{k}{m}t}$

C) $v^{-\frac{k}{m}t} = c-\frac{mg}{k}$

D) $v^{\frac{k}{m}t} = c-\frac{mg}{k}$

Answer:

Option A

Explanation:

$\frac{dv}{dt}+\frac{k}{m}v$ = -g → $\frac{dv}{dt}=-\frac{k}{m}\left(v+\frac{mg}{k}\right)$

$\frac{dv}{v+mg/k}=-\frac{k}{m}dt$

$\log_{}{}\left(v+\frac{mg}{k}\right)$ = $-\frac{k}{m}t$+log c

$\left(v+\frac{mg}{k}\right) = Ce^{-\frac{kt}{m}}$

= $v = C\bar{e}^{\frac{k}{m}t}-\frac{mg}{k}$