Chapter 11.4, Problem 25E

Solution

a.

To find: The average velocity $\frac{\Delta y}{\Delta x}$ on each subinterval of length $0.5$ . Also, create a table showing midpoints and average velocities.

The table that shows the midpoints of subintervals and average velocities is shown in Table (1).

Given information: The table below shows the height of the ball at different time intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

Make a table that shows the midpoints and average velocities at on each subinterval of length $0.5$ .

Interval	Midpoint	$\frac{\Delta y}{\Delta x}$
$\left[0,0.5\right]$	$0.25$	$\frac{495-500}{0.5}=-10$
$\left[0.5,1\right]$	$0.75$	$\frac{485-495}{0.5}=-20$
$\left[1,1.5\right]$	$1.25$	$\frac{465-485}{0.5}=-40$
$\left[1.5,2\right]$	$1.75$	$\frac{435-465}{0.5}=-60$
$\left[2,2.5\right]$	$2.25$	$\frac{400-435}{0.5}=-70$
$\left[2.5,3\right]$	$2.75$	$\frac{355-400}{0.5}=-90$
$\left[3,3.5\right]$	$3.25$	$\frac{305-355}{0.5}=-100$
$\left[3.5,4\right]$	$3.75$	$\frac{245-305}{0.5}=-120$
$\left[4,4.5\right]$	$4.25$	$\frac{175-245}{0.5}=-140$
$\left[4.5,5\right]$	$4.75$	$\frac{100-175}{0.5}=-150$
$\left[5,5.5\right]$	$5.25$	$\frac{15-100}{0.5}=-170$

Table (1)

Thus, the above table shows the midpoints of subintervals of length $0.5$ and average velocities.

b.

To find: The scatter plot of the data showing the numbers in second column as a function of the numbers in first column and the regression model.

The scatter plot of the data is shown in figure (1) and the linear regression model is $y=-32.182x+0.318$ .

Given information: The table below shows the height of the ball at different time intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

The general equation of the linear regression model is $y=ax+b$ .

Use the table form in part (a), and find the values $a=-32.182$ and $b=0.318$ .

The required linear regression model is given as:

  $y=-32.182x+0.318$

Plot all the points from the table in part (a) and draw the linear regression as shown below.

  

Figure (1)

Thus, the scatter plot of the data is shown in figure (1) and the linear regression model is $y=-32.182x+0.318$ .

c.

To find: The approximate velocity of the ball at $t=1.5\sec$ .

The approximate velocity of the ball at $t=1.5\sec$ is $-47.95$ ft/sec.

Given information: The ball is dropped from the roof of a $30$ -story building. The height of the falling ball in feet above the ground is measured at $0.5$ sec intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

From part (b), the linear regression model is $y=-32.182x+0.318$ .

Substitute $1.5$ for $t$ in the above equation.

  $\begin{array}{c}y=-32.182\left(1.5\right)+0.318\\ =-47.95\end{array}$

Thus, the approximate velocity of the ball at $t=1.5\sec$ is $-47.95$ ft/sec.