Chapter 7.3, Problem 50E

a.

To determine

$v$ and $s$ at any time $t$ and find the total distance the object travels from time $t=0$ .

Answer to Problem 50E

  $v=v_0{e}^{\frac{-kt}{m}}$

  $s\left(t\right)=s_0+\frac{m{v}_0}{k}\left(1-e^{-\frac{kt}{m}}\right)$

The total distance that the object travels from time $t=0$ is $\frac{m{v}_0}{k}$

Explanation of Solution

Given information :

  $m\frac{d^{2}s}{d{t}^2}=m\frac{dv}{dt}=f\left(v\right)$

  $f\left(v\right)=-kv$

  $v\left(0\right)=v_0$ ; $s\left(0\right)=s_0$

Formula used :

  ${\displaystyle \int \frac{1}{x}}dx=\ln \left|x\right|+C$

  ${\displaystyle \int e^{x}dx=}{e}^x+C$

Calculation:

  $m\frac{d^{2}s}{d{t}^2}=m\frac{dv}{dt}=f\left(v\right)$

  $f\left(v\right)=-kv$

From above two equations we can write, $m\frac{dv}{dt}=-kv$

  $\begin{array}{l}m\frac{dv}{dt}=-kv\\ \Rightarrow \frac{dv}{v}=-\frac{k}{m}dt\end{array}$

Integrating, we get,

  $\begin{array}{l}{\displaystyle \int \frac{dv}{v}}={\displaystyle \int -\frac{k}{m}dt}\\ \Rightarrow \ln \left|v\right|=-\frac{k}{m}t+C\left[C\text{ is integrating constant}\right]\\ \Rightarrow \left|v\right|=e^C\cdot e^{-\frac{k}{m}t}\\ \Rightarrow v=\pm e^C\cdot e^{-\frac{k}{m}t}\\ \Rightarrow v=A{e}^{-\frac{k}{m}t}\left[A=\pm e^C\right]\end{array}$

Now, $v\left(0\right)=v_0$

i.e. $v_0=A$

So, $v=v_0{e}^{\frac{-kt}{m}}$

Now, $\frac{ds}{dt}=v=v_0{e}^{\frac{-kt}{m}}$

  $\begin{array}{l}\Rightarrow \frac{ds}{dt}=v_0{e}^{\frac{-kt}{m}}\\ \Rightarrow ds=v_0{e}^{\frac{-kt}{m}}dt\end{array}$

Integrating both sides, we get,

  $\begin{array}{l}{\displaystyle \int ds}={\displaystyle \int v_0{e}^{\frac{-kt}{m}}dt}\\ \Rightarrow {\displaystyle \int ds}=v_0{\displaystyle \int e^{\frac{-kt}{m}}dt}\end{array}$

  $\left[\text{Let, }-\frac{k}{m}t=q\Rightarrow -\frac{k}{m}dt=dq\Rightarrow dt=-\frac{m}{k}dq\right]$

  $\begin{array}{l}\Rightarrow {\displaystyle \int ds}=v_0\left(-\frac{m}{k}\right){\displaystyle \int e^{q}dq}\\ \Rightarrow s=-\frac{m{v}_0}{k}{e}^q+D\end{array}$

  $\Rightarrow s=-\frac{m{v}_0}{k}{e}^{\frac{-kt}{m}}+D$

Now, $s\left(0\right)=s_0$

  $\begin{array}{l}\Rightarrow s_0=-\frac{m{v}_0}{k}+D\\ \Rightarrow D=s_0+\frac{m{v}_0}{k}\end{array}$

So,

  $\begin{array}{l}s=-\frac{m{v}_0}{k}{e}^{-\frac{kt}{m}}+s_0+\frac{m{v}_0}{k}\\ \Rightarrow s=s_0+\frac{m{v}_0}{k}\left(1-e^{-\frac{kt}{m}}\right)\end{array}$

So, $s\left(t\right)=s_0+\frac{m{v}_0}{k}\left(1-e^{-\frac{kt}{m}}\right)$

The total distance that the object travels from time $t=0$ is

  $\begin{array}{l}\underset{t\to \infty }{\lim }s\left(t\right)-s\left(0\right)\\ =\underset{t\to \infty }{\lim }\left[s_0+\frac{m{v}_0}{k}\left(1-e^{-\frac{kt}{m}}\right)\right]-s_0\\ =s_0+\underset{t\to \infty }{\lim }\left[\frac{m{v}_0}{k}\left(1-e^{-\frac{kt}{m}}\right)\right]-s_0\\ =\frac{m{v}_0}{k}\end{array}$

b.

To determine

$v$ and $s$ at any time $t$ and find the total distance the object travels from time $t=0$ .

Answer to Problem 50E

  $v\left(t\right)=\frac{m{v}_0}{k{v}_{0}t+m}$

  $s\left(t\right)=s_0+\frac{m}{k}\ln \left|\frac{k{v}_{0}t+m}{m}\right|$

The total distance that the object travels from time $t=0$ is $\infty$

Explanation of Solution

Given information :

  $m\frac{d^{2}s}{d{t}^2}=m\frac{dv}{dt}=f\left(v\right)$

  $f\left(v\right)=-k{v}^2$ ............. $k>0$

  $v\left(0\right)=v_0$ ; $s\left(0\right)=s_0$

Formula used :

  ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}}+C$

  ${\displaystyle \int \frac{dx}{ax+b}}=\frac{1}{a}\ln \left|ax+b\right|+C$

Calculation:

  $m\frac{d^{2}s}{d{t}^2}=m\frac{dv}{dt}=f\left(v\right)$

  $f\left(v\right)=-k{v}^2$

From above two equations we can write, $m\frac{dv}{dt}=-k{v}^2$

  $\begin{array}{l}m\frac{dv}{dt}=-k{v}^2\\ \Rightarrow \frac{dv}{dt}=-\frac{k}{m}{v}^2\\ \Rightarrow \frac{dv}{v^2}=-\frac{k}{m}dt\end{array}$

Integrating both sides, we get,

  $\begin{array}{l}{\displaystyle \int \frac{dv}{v^2}}={\displaystyle \int -\frac{k}{m}dt}\\ \Rightarrow -\frac{1}{v}=-\frac{k}{m}t+C\\ \Rightarrow \frac{1}{v}=\frac{k}{m}t+D\left[D=-C\right]\end{array}$

Now, $v\left(0\right)=v_0$

  $\frac{1}{v_0}=D$

So, $\frac{1}{v}=\frac{k}{m}t+\frac{1}{v_0}$

  $\begin{array}{l}\Rightarrow \frac{1}{v}=\frac{k{v}_{0}t+m}{m{v}_0}\\ \Rightarrow v=\frac{m{v}_0}{k{v}_{0}t+m}\end{array}$

  $v\left(t\right)=\frac{m{v}_0}{k{v}_{0}t+m}$

Now, $\frac{ds}{dt}=v\left(t\right)=\frac{m{v}_0}{k{v}_{0}t+m}$

  $\begin{array}{l}\frac{ds}{dt}=\frac{m{v}_0}{k{v}_{0}t+m}\\ \Rightarrow ds=\frac{m{v}_0}{k{v}_{0}t+m}dt\end{array}$

Integrating both sides, we get,

  $\begin{array}{l}{\displaystyle \int ds}={\displaystyle \int \frac{m{v}_0}{k{v}_{0}t+m}dt}\\ \Rightarrow {\displaystyle \int ds}=m{v}_0{\displaystyle \int \frac{dt}{k{v}_{0}t+m}}\\ \Rightarrow s=m{v}_0\times \frac{1}{k{v}_0}\ln \left|k{v}_{0}t+m\right|+E\\ \Rightarrow s=\frac{m}{k}\ln \left|k{v}_{0}t+m\right|+E\end{array}$

Now, $s\left(0\right)=s_0$

i.e. $s_0=\frac{m}{k}\ln \left|m\right|+E$

  $\begin{array}{l}{s}_0=\frac{m}{k}\ln \left|m\right|+E\\ \Rightarrow E=s_0-\frac{m}{k}\ln \left|m\right|\end{array}$

So,

  $\begin{array}{l}s=\frac{m}{k}\ln \left|k{v}_{0}t+m\right|+s_0-\frac{m}{k}\ln \left|m\right|\\ \Rightarrow s=s_0+\frac{m}{k}\ln \left|\frac{k{v}_{0}t+m}{m}\right|\end{array}$

  $s\left(t\right)=s_0+\frac{m}{k}\ln \left|\frac{k{v}_{0}t+m}{m}\right|$

The total distance that the object travels from time $t=0$ is

  $\begin{array}{l}\underset{t\to \infty }{\lim }s\left(t\right)-s\left(0\right)\\ =\underset{t\to \infty }{\lim }\left[s_0+\frac{m}{k}\ln \left|\frac{k{v}_{0}t+m}{m}\right|\right]-s_0\\ =s_0+\underset{t\to \infty }{\lim }\left[\frac{m}{k}\ln \left|\frac{k{v}_{0}t+m}{m}\right|\right]-s_0\\ =\infty \end{array}$