Chapter 1, Problem 1P

Use calculus to solve Eq. (1.9) for the case where the initial velocity, $v\left(0\right)$ is nonzero.

To determine

To calculate: The solution of the equation $\frac{dv}{dt}=g-\frac{c}{m}v$ by the use of calculus if the initial velocity $v\left(0\right)$ is non-zero.

Answer to Problem 1P

Solution:

The solution of the equation $\frac{dv}{dt}=g-\frac{c}{m}v$ is $v=v\left(0\right)e^{-\left(\frac{c}{m}\right)t}+\frac{mg}{c}\left(1-e^{-\left(\frac{c}{m}\right)t}\right)$.

Explanation of Solution

Given Information:

The equation, $\frac{dv}{dt}=g-\frac{c}{m}v$.

The initial velocity $v\left(0\right)$ is non-zero.

Formula used:

Integration property of polynomial,

${\displaystyle \int \frac{1}{a+bx}dx}=\frac{1}{b}\ln \left(a+bx\right)+C$

Logarithmic property,

$\ln a-\ln b=\ln \left(\frac{a}{b}\right)$

Calculation:

Consider the equation,

$\frac{dv}{dt}=g-\frac{c}{m}v$

Separate the variables as below,

$\frac{dv}{g-\frac{c}{m}v}=dt$

Integrate both the sides of above equation,

${\displaystyle \int \frac{dv}{g-\frac{c}{m}v}}={\displaystyle \int dt}$

Use the integration property of polynomial,

$-\frac{\ln \left(g-\frac{c}{m}v\right)}{\left(\frac{c}{m}\right)}=t+C$ …… (1)

Where, C is the constant of integration.

For $t=0$, the initial velocity is $v\left(0\right)$. Therefore,

$\begin{array}{c}-\frac{\ln \left(g-\frac{c}{m}v\left(0\right)\right)}{\left(\frac{c}{m}\right)}=0+C\\ -\frac{\ln \left(g-\frac{c}{m}v\left(0\right)\right)}{\left(\frac{c}{m}\right)}=C\end{array}$

Substitute the value of C in the equation (1) and simplify as below,

$\begin{array}{c}-\frac{\ln \left(g-\frac{c}{m}v\right)}{\left(\frac{c}{m}\right)}=t-\frac{\ln \left(g-\frac{c}{m}v\left(0\right)\right)}{\left(\frac{c}{m}\right)}\\ -\ln \left(g-\frac{c}{m}v\right)=\left(\frac{c}{m}\right)t-\ln \left(g-\frac{c}{m}v\left(0\right)\right)\\ -\ln \left(g-\frac{c}{m}v\right)+\ln \left(g-\frac{c}{m}v\left(0\right)\right)=\left(\frac{c}{m}\right)t\\ \ln \frac{\left(g-\frac{c}{m}v\left(0\right)\right)}{\left(g-\frac{c}{m}v\right)}=\left(\frac{c}{m}\right)t\end{array}$

Simplify furthermore,

$\begin{array}{c}\frac{\left(g-\frac{c}{m}v\left(0\right)\right)}{\left(g-\frac{c}{m}v\right)}=e^{\left(\frac{c}{m}\right)t}\\ g-\frac{c}{m}v=\frac{g-\frac{c}{m}v\left(0\right)}{e^{\left(\frac{c}{m}\right)t}}\\ g-\frac{c}{m}v=\left(g-\frac{c}{m}v\left(0\right)\right)e^{-\left(\frac{c}{m}\right)t}\\ -\frac{c}{m}v=\left(g-\frac{c}{m}v\left(0\right)\right)e^{-\left(\frac{c}{m}\right)t}-g\end{array}$

Simplify furthermore,

$\begin{array}{c}v=\left(-\frac{m}{c}\right)\left(g-\frac{c}{m}v\left(0\right)\right)e^{-\left(\frac{c}{m}\right)t}-\left(-\frac{m}{c}\right)g\\ =-\frac{mg}{c}{e}^{-\left(\frac{c}{m}\right)t}+v\left(0\right)e^{-\left(\frac{c}{m}\right)t}+\frac{mg}{c}\\ =v\left(0\right)e^{-\left(\frac{c}{m}\right)t}+\frac{mg}{c}\left(1-e^{-\left(\frac{c}{m}\right)t}\right)\end{array}$

Since, $v\left(0\right)$ is non-zero. Hence, the solution of the equation is $v=v\left(0\right)e^{-\left(\frac{c}{m}\right)t}+\frac{mg}{c}\left(1-e^{-\left(\frac{c}{m}\right)t}\right)$.