Chapter 11.1, Problem 33E

Solution

a)

To find:the average speed.

The average speed is from 0.5 to 0.6 sec is 9ft/sec

The average speed is from 0.8 to 0.9 sec is 15ft/sec

Given information:

Given the distance data of the Lead Ball.

Given set of points are $\left(0.5,2.3\right)$ and $\left(0.6,3.2\right)$ .

Another set of points are $\left(0.8,5.8\right)$ and $\left(0.9,7.3\right)$ .

Formula used:

The average speed from one point $\left(a,f\left(a\right)\right)$ to another point $\left(b,f\left(b\right)\right)$ is given as

  $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

The average speed of from one point $\left(0.5,2.3\right)$ to another $\left(0.6,3.2\right)$ is

  $\begin{array}{c}\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{3.2-2.3}{0.6-0.5}\\ =\frac{0.9}{0.1}\\ =9\end{array}$

Thus, the average speed of from the point $\left(0.5,2.3\right)$ to another $\left(0.6,3.2\right)$ is 9 ft/sec.

The average speed of from one point $\left(0.8,5.8\right)$ to another $\left(0.9,7.3\right)$ is

  $\begin{array}{c}\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{7.3-5.8}{0.9-0.8}\\ =\frac{1.5}{0.1}\\ =15\end{array}$

Thus, the average speed of from one point $\left(0.8,5.8\right)$ to another $\left(0.9,7.3\right)$ is 15 ft/sec.

b)

To find:the quadratic regression model for the given data

The quadratic regression model for the given data is $f\left(x\right)=8.94x^2+0.05x+0.01$ ,

  $x=$ time in seconds

Given information:

Given the distance data of the Lead Ball.

Formula used:

Calculator for quadratic regression model used.

Calculation:

The scatter plot of the given is shown in Figure (1).

  

Figure (1)

Quadratic regression model fit is shown below.

  

Figure (2)

Using the calculator for quadratic regression model based given data is $f\left(x\right)=8.94x^2+0.05x+0.01$ , $x=$ time in seconds

  $$

c)

To find:to estimate the depth of the lake if the ball hits the bottom after 2 sec.

The depth of the of the lake is 35.9 ft.

Given information:

From part (b), quadratic regression model for given data is $f\left(x\right)=8.94x^2+0.05x+0.01$ , $x=$ time in seconds

Calculation:

Substitute $x=2$ in the equation $f\left(x\right)=8.94x^2+0.05x+0.01$ .

  $\begin{array}{c}f\left(x\right)=8.94x^2+0.05x+0.01\\ =8.94{\left(2\right)}^2+0.05\left(2\right)+0.01\\ =35.87\end{array}$

The depth of the of the lake is 35.9 ft.