Chapter 6.4, Problem 55E

To determine

To find: Find the model for the position function.

Answer to Problem 55E

The model for the distance is $s=1.32\left(1-e^{-0.75t}\right)$ .

Explanation of Solution

Given information:

Given that the initial velocity was $v_0=1 \text{m}/\text{sec}$ , mass was $m=49.90 \text{kg}(110\text{lb})$ and total costing distance is $s=1.32m$ that is $\frac{v_{0}m}{k}=1.32$

The model for the velocity is given as $v=v_0{e}^{\left(\frac{-k}{m}\right)t}$ .

The table for the distance and time is given below:

t	0	0.1	0.3	0.5	0.7	0.9	1.1
s	0	0.07	0.22	0.36	0.49	0.60	0.71
t	1.5	1.7	1.9	2.1	2.3	2.5	2.7
s	0.89	0.97	1.05	1.11	1.17	1.22	1.25
t	2.9	3.1	3.3	3.5	3.7
s	1.28	1.30	1.31	1.32	1.32

Calculation:

In order to find the position, function the given velocity function can be written as:

  $\begin{array}{l}\frac{ds}{dt}=v_0{e}^{\left(\frac{-k}{m}\right)t}\\ ds=v_0{e}^{\left(\frac{-k}{m}\right)t}dt\end{array}$

Integrate the above equation:

  $\begin{array}{l}s=\frac{v_0{e}^{\left(\frac{-k}{m}\right)t}}{\frac{-k}{m}}+c\\ s=\frac{-m{v}_0}{k}{e}^{\left(\frac{-k}{m}\right)t}+c\mathrm{....}(1)\end{array}$

Apply the initial conditions $t=0,s=0$ in equation (1):

  $\begin{array}{l}0=\frac{-m{v}_0}{k}{e}^{\left(\frac{-k}{m}\right)\cdot 0}+c\\ 0=\frac{-m{v}_0}{k}+c\\ c=\frac{m{v}_0}{k}\end{array}$

Substitute the value of $c$ in equation (1) gives:

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

This is the required model.

Given that the total coasting distance is 1.32 m.

  $\frac{v_{0}m}{k}=1.32$

Substitute $v_0=1 \text{m}/\text{sec}$ and $m=49.9$ in $\frac{v_{0}m}{k}=1.32$ gives:

  $\begin{array}{l}\frac{49.9}{k}=1.32\\ k=\frac{49.9}{1.32}\\ k=37.80\end{array}$

Substitute the value of $k$ and $\frac{v_{0}m}{k}=1.32$ in $s=\frac{m{v}_0}{k}\left(1-e^{\left(\frac{-k}{m}\right)t}\right)$ :

  $\begin{array}{l}s=1.32\left(1-e^{\left(\frac{-37.80}{49.9}\right)t}\right)\\ s=1.32\left(1-e^{-0.75t}\right)\end{array}$

The scatter plot of the above data is given below: