Chapter 3.4, Problem 85E

a.

To determine

Complete the table using given model.

Answer to Problem 85E

The complete table is

$x$	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$
$y$	$162.6$	$78.5$	$52.5$	$40.5$	$33.94$

Explanation of Solution

Given:

The givenmodel equation is $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$ and the given table for $y\left(g\text{'}\text{s}\text{value}\right)$ is

  

Calculation:

  $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$

Substitute all the $x$ -values in the equation and find corresponding $y$ -values.

Hence the complete table is given below.

$x$	$y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$	$y$
$\begin{array}{l}\\ 0.2\\ \\ 0.4\\ \\ 0.6\\ \\ 0.8\\ \\ 1.0\end{array}$	$y=-3.00+11.88\ln \left(0.2\right)+\left(\frac{36.94}{0.2}\right)$   $y=-3.00+11.88\ln \left(0.4\right)+\left(\frac{36.94}{0.4}\right)$   $y=-3.00+11.88\ln \left(0.6\right)+\left(\frac{36.94}{0.6}\right)$   $y=-3.00+11.88\ln \left(0.8\right)+\left(\frac{36.94}{0.8}\right)$   $y=-3.00+11.88\ln \left(1\right)+\left(\frac{36.94}{1}\right)$	$\begin{array}{l}\\ 162.6\\ \\ 78.5\\ \\ 52.5\\ \\ 40.5\\ \\ 33.94\end{array}$   $78.5$

b.

To determine

Draw a graph of data points and the model in the same viewing window .

Explanation of Solution

Given:

The given model equation is $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$ and the given table for $y\left(g\text{'}\text{s}\text{value}\right)$ is

  

Calculation:

  $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$

Substitute all the $x$ -values in the equation and find corresponding $y$ -values.

The complete table is given below.

$x$	$y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$	$y$
$\begin{array}{l}\\ 0.2\\ \\ 0.4\\ \\ 0.6\\ \\ 0.8\\ \\ 1.0\end{array}$	$y=-3.00+11.88\ln \left(0.2\right)+\left(\frac{36.94}{0.2}\right)$   $y=-3.00+11.88\ln \left(0.4\right)+\left(\frac{36.94}{0.4}\right)$   $y=-3.00+11.88\ln \left(0.6\right)+\left(\frac{36.94}{0.6}\right)$   $y=-3.00+11.88\ln \left(0.8\right)+\left(\frac{36.94}{0.8}\right)$   $y=-3.00+11.88\ln \left(1\right)+\left(\frac{36.94}{1}\right)$	$\begin{array}{l}\\ 162.6\\ \\ 78.5\\ \\ 52.5\\ \\ 40.5\\ \\ 33.94\end{array}$   $78.5$

Hence the graph of the given model is

  

c.

To determine

Find the distance traveled during impact.

Answer to Problem 85E

The passenger declaration does not exceed $30$ g’s when $x\approx 1.2$ .

Explanation of Solution

Given:

The given model equation is $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$ and the given table for $y\left(g\text{'}\text{s}\text{value}\right)$ is

  

Calculation:

  $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$

Substitute all the $x$ -values in the equation and find corresponding $y$ -values.

Hence the complete table is given below.

$x$	$y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$	$y$
$\begin{array}{l}\\ 0.2\\ \\ 0.4\\ \\ 0.6\\ \\ 0.8\\ \\ 1.0\end{array}$	$y=-3.00+11.88\ln \left(0.2\right)+\left(\frac{36.94}{0.2}\right)$   $y=-3.00+11.88\ln \left(0.4\right)+\left(\frac{36.94}{0.4}\right)$   $y=-3.00+11.88\ln \left(0.6\right)+\left(\frac{36.94}{0.6}\right)$   $y=-3.00+11.88\ln \left(0.8\right)+\left(\frac{36.94}{0.8}\right)$   $y=-3.00+11.88\ln \left(1\right)+\left(\frac{36.94}{1}\right)$	$\begin{array}{l}\\ 162.6\\ \\ 78.5\\ \\ 52.5\\ \\ 40.5\\ \\ 33.94\end{array}$   $78.5$

Graph of the given model with data points is

  

From the table and graph, the passenger declaration does not exceed $30$ g’s when $x\approx 1.2$ .

d.

To determine

Give the explanation for the given statement.

Answer to Problem 85E

These values of $x$ do not make any sense.

Explanation of Solution

Given:

The given model equation is $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$ and the given table for $y\left(g\text{'}\text{s}\text{value}\right)$ is

  

The given statement is that to lower the number of g’s experienced during impact to fewer than $23$ .

Calculation:

  $y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$

Substitute all the $x$ -values in the equation and find corresponding $y$ -values.

Hence the complete table is given below.

$x$	$y=-3.00+11.88\ln x+\left(\frac{36.94}{x}\right)$	$y$
$\begin{array}{l}\\ 0.2\\ \\ 0.4\\ \\ 0.6\\ \\ 0.8\\ \\ 1.0\end{array}$	$y=-3.00+11.88\ln \left(0.2\right)+\left(\frac{36.94}{0.2}\right)$   $y=-3.00+11.88\ln \left(0.4\right)+\left(\frac{36.94}{0.4}\right)$   $y=-3.00+11.88\ln \left(0.6\right)+\left(\frac{36.94}{0.6}\right)$   $y=-3.00+11.88\ln \left(0.8\right)+\left(\frac{36.94}{0.8}\right)$   $y=-3.00+11.88\ln \left(1\right)+\left(\frac{36.94}{1}\right)$	$\begin{array}{l}\\ 162.6\\ \\ 78.5\\ \\ 52.5\\ \\ 40.5\\ \\ 33.94\end{array}$   $78.5$

Graph of the given model with data points is

  

From the table and graph, the value of $x$ will be between $2.28$ m and $4.40$ m when the number of g’s is less than $23$ .

Hence, these values of $x$ do not make any sense.