Chapter 7.3, Problem 49E

To determine

To Find: The expression for the terminal velocity in terms of g and k .

Answer to Problem 49E

The expression for the terminal velocity is $\frac{g}{k}$ .

Explanation of Solution

Given:

The mass as the function of the t is $m\left(t\right)$ .

The rate of the growth of the mass is k.

The given expression is ${\left(mv\right)}^{\prime }=gm$ .

The terminal velocity of the raindrop is $\underset{t\to \infty }{\lim }v\left(t\right)$ .

Calculation:

Consider the given expression is,

  ${\left(mv\right)}^{\prime }=gm$

Then,

  $m^{\prime }v+v^{\prime }m=gm$

Consider the rate of the growth of the mass is $m^{\prime }=km\left(t\right)$ .

Then,

  $\begin{array}{l}kmv+v^{\prime }m=gm\\ v^{\prime }=g-kv\\ \frac{dv}{dt}=g-kv\\ \left(\frac{1}{g-kv}\right)dv=dt\end{array}$

Then, integrate both the sides as,

  $\begin{array}{l}{\displaystyle \int \left(\frac{1}{g-kv}\right)}=t+C\\ -\ln \left(g-kv\right)=kt+kC\end{array}$

Consider the case when the drop falls the initial velocity at time $t=0$ is $v=0$ is,

  $\begin{array}{l}-\ln \left(g-0\right)=0+kC\\ C=-\frac{1}{k}\ln g\end{array}$

Therefore,

  $\begin{array}{l}-\ln \left(g-kv\right)=kt-\ln g\\ g-kv=e^{\ln g-kt}\\ g-kv=e^{\ln g}{e}^{-kt}\\ v\left(t\right)=\frac{g}{k}\left(1-e^{-kt}\right)\end{array}$

Then, consider the limit $t\to \infty$ to find the terminal velocity is,

  $\begin{array}{c}\underset{t\to \infty }{\lim }v\left(t\right)=\underset{t\to \infty }{\lim }\frac{g}{k}\left(1-e^{-kt}\right)\\ =\frac{g}{k}\left(1-0\right)\\ =\frac{g}{k}\end{array}$