Chapter 14, Problem 74RE

a.

To determine

Find the average speed from $t=1$ to $t=4$ seconds.

Answer to Problem 74RE

  $\boxed{80feet/\sec ond}$

Explanation of Solution

Given information:

The following data represent the distances $s$ (in feet) that a parachutist has fallen over time $t$ (in seconds)

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

Find the average speed from $t=1$ to $t=4$ seconds.

Calculation:

The data in this table represents the distance $x$ in feet that a parachutist has fallen over the time $t$ in seconds.

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

To find the average speed from $t=1$ to $t=4$ seconds, consider the change in distance $o$ the change in time, $\frac{\Delta s}{\Delta t}$ . This is called the average speed. We need to know the initial distance and as well as the final distance and time.

The initial time is $t=1$ , the initial distance is $s\left(1\right)=16$ , the final time is $t=4$ , and the final distance is $s(4)=256$ . Plug these values into the average rate of change formula to find 1 average speed from $t=1$ to $t=4$ seconds

  $\frac{\Delta s}{\Delta t}=\frac{s(4)-s(1)}{4-1}$

  $\begin{array}{l}=\frac{256-16}{3}\\ =\frac{240}{3}\end{array}$

Hence, the average speed from $t=1$ to $t=4$ seconds is $80feet/\sec ond$

b.

To determine

Find the average speed from $t=1$ to $t=3$ seconds.

Answer to Problem 74RE

  $\boxed{64feet/\sec ond}$

Explanation of Solution

Given information:

The following data represent the distances $s$ (in feet) that a parachutist has fallen over time $t$ (in seconds)

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

Find the average speed from $t=1$ to $t=3$ seconds.

Calculation:

The data in this table represents the distance $x$ in feet that a parachutist has fallen over the time $t$ in seconds.

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

To find the average speed from $t=1$ to $t=3$ seconds, consider the change in distance $o$ the change in time, $\frac{\Delta s}{\Delta t}$ . This is called the average speed. We need to know the initial distance and as well as the final distance and time.

The initial time is $t=1$ , the initial distance is $s\left(1\right)=16$ , the final time is $t=3$ , and the final distance is $s(3)=144$ . Plug these values into the average rate of change formula to find 1 average speed from $t=1$ to $t=3$ seconds

  $\frac{\Delta s}{\Delta t}=\frac{s(3)-s(1)}{3-1}$

  $\begin{array}{l}=\frac{144-16}{2}\\ =\frac{128}{2}\end{array}$

Hence, the average speed from $t=1$ to $t=3$ seconds is $64feet/\sec ond$ .

c.

To determine

Find the average speed from $t=1$ to $t=2$ seconds.

Answer to Problem 74RE

  $\boxed{48feet/\sec ond}$

Explanation of Solution

Given information:

The following data represent the distances $s$ (in feet) that a parachutist has fallen over time $t$ (in seconds)

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

Find the average speed from $t=1$ to $t=2$ seconds.

Calculation:

The data in this table represents the distance $x$ in feet that a parachutist has fallen over the time $t$ in seconds.

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

To find the average speed from $t=1$ to $t=2$ seconds, consider the change in distance $o$ the change in time, $\frac{\Delta s}{\Delta t}$ . This is called the average speed. We need to know the initial distance and as well as the final distance and time.

The initial time is $t=1$ , the initial distance is $s\left(1\right)=16$ , the final time is $t=2$ , and the final distance is $s(2)=64$ . Plug these values into the average rate of change formula to find 1 average speed from $t=1$ to $t=2$ seconds

  $\frac{\Delta s}{\Delta t}=\frac{s(2)-s(1)}{2-1}$

  $\begin{array}{l}=\frac{64-16}{1}\\ =48\end{array}$

Hence, the average speed from $t=1$ to $t=2$ seconds is $48feet/\sec ond$ .

d.

To determine

Find the power function of best fit.

Answer to Problem 74RE

  $\boxed{s\left(t\right)=\mathrm{10.39.}(2.19)}$

Explanation of Solution

Given information:

The following data represent the distances $s$ (in feet) that a parachutist has fallen over time $t$ (in seconds)

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

Using a graphing utility, find the power function of best fit.

Calculation: Use a best fit function on a graphing utility to find the quadratic function the best fits the data in the table.

On a graphing utility, enter the data points found in the table by hitting STAT, 1: Edit, and then entering the $x$ values into L1 and the $s\left(t\right)$ values into L2. Then hit STAT, CLAC, $0$ : ExpReg, VARS, Y-VARS, 1: Function, 1: $Y_1$ and then hit enter. Then go back to the Y= screen, and the quadratic function of best fit should appear in the $Y_1$ spot.

Hence, the power function that fits the data in the table the best is

  $s\left(t\right)=\mathrm{10.39.}(2.19)$

e.

To determine

Determine the instantaneous at $t=1$ second.

Answer to Problem 74RE

  $\boxed{17.77feet/\sec ond}$

Explanation of Solution

Given information:

The following data represent the distances $s$ (in feet) that a parachutist has fallen over time $t$ (in seconds)

  $\begin{array}{cc}\begin{array}{l}Time,t\\ (in-\sec onds)\end{array}& \begin{array}{l}Dis\tan ce,s\\ (in-feet)\end{array}\\ 1& 16\\ 2& 64\\ 3& 144\\ 4& 256\\ 5& 400\end{array}$

Using the function found in part (d) , determine the instantaneous at $t=1$ second.

Calculation:

To find the instantaneous speed at $t=1$ second, use the derivative function on a graphing utility. Make sure that position function $s\left(t\right)=\mathrm{10.39.}{(2.19)}^t$ from part (d) is still input as $Y_1$ .

Go to MATH, 8:nDeriv(, VARS, Y-VARS, 1:Function, 1: $Y_1$ , type a comma, X, comma,1, and then a closing parenthesis.

Hence, after hitting enter, the calculator displays that the instantaneous speed at $t=1$ second is about $17.77feet/\sec ond$