Chapter 7.4, Problem 56E

To determine

To find a model for her position in the form given in exercise 54(a) and superimpose its graph on a scatter plot of the data.

Answer to Problem 56E

The model of her position is $s\left(t\right)=0.97\left(1-e^{-0.887t}\right)$ .

Graph,

  

Explanation of Solution

Given information: Initial velocity was $v_0=0.86m/\sec$ , her mass $m=30.84kg$ and her total coasting distance was 0.97m.

  

Calculation: The required solution is obtained as,

First find the model for her position using $s\left(t\right)=\frac{v_{0}m}{k}\left(1-e^{-\frac{k}{m}t}\right)$ and coasting distance $=\frac{v_{0}m}{k}$ .

It is given that her coasting distance is 0.97m so $s\left(t\right)=0.97\left(1-e^{-\frac{k}{m}t}\right)$ .

Let c be the coasting distance and use its equation to find an expression for $\frac{k}{m}$ .

  $\begin{array}{l}\frac{v_{0}m}{k}=c\\ \frac{m}{k}=\frac{c}{v_0}\\ \frac{k}{m}=\frac{v_0}{c}\end{array}$

Substituting in the given values of $v_0=0.86m/\sec$ and $c=0.97$ then,

  $\frac{k}{m}=\frac{0.86}{0.97}\approx 0.887$

Hence, the model of her position is $s\left(t\right)=0.97\left(1-e^{-0.887t}\right)$ .

Graph the points from the table (shown in blue ) and the model $s\left(t\right)=0.97\left(1-e^{-0.887t}\right)$ (shown in green).