Question

A force $F=B e^{-C t}$ acts on a particle whose mass is $m$ and whose velocity is 0 at $t=0$. It's terminal velocity is :

• A. $\frac{C}{m B}$
• B. $\frac{B}{m C}$
• C. $\frac{B C}{m}$
• D. $-\frac{B}{m C}$

Answer 1

Answer: (B)

$F =m a $
$ m a =B e^{-c t} $
$a =\frac{B e^{-C t}}{m} $
$ \frac{d v}{d t} =\frac{B e^{-C t}}{m} $
$ \int d v =\int \frac{B e^{-C t}}{m} d t $
$v(t)=-\frac{B e^{-C t}}{m C}+k$
at $t=0, v=0$
$k=\frac{B}{m C} \Rightarrow$
$\Rightarrow v(t)=\frac{B}{m C}\left(1-e^{-c t}\right)$
When $t$ is very large
$e^{-c t} \rightarrow 0$
$\Rightarrow$ Terminal velocity, $v=\frac{B}{m C}$