Chapter 7.4, Problem 54E

a.

To determine

Show that, $s\left(t\right)=\frac{v_{0}m}{k}\left(1-e^{-\left(k/m\right)t}\right)$ .

Explanation of Solution

Given information: The resistance encountered by a moving object is proportional to the object’s velocity so that its velocity is $v=v_0{e}^{-\left(k/m\right)t}$ .

Proof: The required proved is obtained as,

  $v=v_0{e}^{-\left(k/m\right)t}$

Since velocity is an instantaneous rate of change of displacement, then $v=\frac{ds}{dt}$ .

  $\begin{array}{l}v=v_0{e}^{-\left(k/m\right)t}\\ \frac{ds}{dt}=v_0{e}^{-\frac{k}{m}t}\\ ds=v_0{e}^{-\frac{k}{m}t}dt\\ {\displaystyle \int ds={\displaystyle \int v_0{e}^{-\frac{k}{m}t}dt}}\\ s=v_0{\displaystyle \int e^{-\frac{k}{m}t}dt}\\ s=v_0\left(\frac{e^{-\frac{k}{m}t}}{-\frac{k}{m}}\right)+C\\ =v_0\left(-\frac{m}{k}\right)e^{-\frac{k}{m}t}+C\\ =-\frac{v_{0}m}{k}{e}^{-\frac{k}{m}t}+C\end{array}$

Substitute $s=0$ and $t=0$ to find C.

  $\begin{array}{l}s=-\frac{v_{0}m}{k}{e}^{-\frac{k}{m}t}+C\\ 0=-\frac{v_{0}m}{k}{e}^{-\frac{k}{m}\left(0\right)}+C\\ 0=-\frac{v_{0}m}{k}{e}^{\left(0\right)}+C\\ 0=-\frac{v_{0}m}{k}+C\\ C=\frac{v_{0}m}{k}\end{array}$

Substitute $C=\frac{v_{0}m}{k}$ into the equation for s.

  $\begin{array}{l}s=-\frac{v_{0}m}{k}{e}^{-\frac{k}{m}t}+\frac{v_{0}m}{k}\left(-e^{-\frac{k}{m}t}+1\right)\\ =\frac{v_{0}m}{k}\left(1-e^{-\frac{k}{m}t}\right)\end{array}$

b.

To determine

To show that the total coasting distance travelled by the object as it coast to a complete stop is $\frac{v_{0}m}{k}$ .

Explanation of Solution

Given information: The resistance encountered by a moving object is proportional to the object’s velocity so that its velocity is $v=v_0{e}^{-\left(k/m\right)t}$ .

Proof:

The required proved is obtained as,

The velocity of the object is $v=v_0{e}^{-\left(k/m\right)t}$ . Which never equal to 0 but gets infinitesimally close to 0 since the function has a horizontal asymptote at $v=0$ . The total coasting distance is then the limit of the distance function at $t\to \infty$ .Find the limit using $e^{-\infty }\to 0$ .

  $\underset{t\to \infty }{\lim }s\left(t\right)=\underset{t\to \infty }{\lim }\frac{v_{0}m}{k}\left(1-e^{-\frac{k}{m}t}\right)=\frac{v_{0}m}{k}\left(1-e^{-\infty }\right)=\frac{v_{0}m}{k}\left(1-0\right)=\frac{v_{0}m}{k}$ .