Chapter 5.3, Problem 70E

(a)

To determine

To find: The third degree polynomial model for the velocity with the help of graphing calculator.

Answer to Problem 70E

The third degree polynomial model for the velocity is $y=0.001x^3-0.115x^2+24.981x-21.268$ .

Explanation of Solution

Given information: The below table gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters:

Event	Time $\left(\text{s}\right)$	Velocity $\left(\text{ft}/\text{s}\right)$
Launch	$0$	$0$
Begin roll maneuver	$10$	$185$
End roll maneuver	$15$	$319$
Throttle to $89\%$	$20$	$447$
Throttle to $67\%$	$32$	$742$
Throttle to $104\%$	$59$	$1325$
Maximum dynamic pressure	$62$	$1445$
Solid rocket booster separation	$125$	$4151$

Calculation:

For the cubic model of data follow the steps below:

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now, enter the keystrokes $\boxed{\text{STAT}}\boxed{\to }\boxed{6}$ and press the $\boxed{\text{ENTER}}$ key for the regression.

  

Figure (1)

Therefore, the third degree polynomial model for the velocity is $y=0.001x^3-0.115x^2+24.981x-21.268$ .

(b)

To determine

To find: The height reached by the Endeavour $125$ seconds after the liftoff.

Answer to Problem 70E

The height reached by the Endeavour $125$ seconds after the liftoff is equal to $179314.156\text{ft}$ .

Explanation of Solution

Given information: The below table gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters:

Event	Time $\left(\text{s}\right)$	Velocity $\left(\text{ft}/\text{s}\right)$
Launch	$0$	$0$
Begin roll maneuver	$10$	$185$
End roll maneuver	$15$	$319$
Throttle to $89\%$	$20$	$447$
Throttle to $67\%$	$32$	$742$
Throttle to $104\%$	$59$	$1325$
Maximum dynamic pressure	$62$	$1445$
Solid rocket booster separation	$125$	$4151$

Calculation:

The height reached by the Endeavour after $125$ seconds of liftoff is the integration of the velocity function with limits $0$ to $125$ that is ${\displaystyle \underset{0}{\overset{125}{\int }}y\left(x\right)dx}$ .

Simplify the integral.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{125}{\int }}y\left(x\right)dx}={\displaystyle \underset{0}{\overset{125}{\int }}\left(0.001x^3-0.115x^2+24.981x-21.268\right)dx}\\ ={\left(0.001\frac{x^4}{4}-0.115\frac{x^3}{3}+24.98\frac{x^2}{2}-21.268x\right)|}_0^{125}\\ ={\left(0.00025x^4-0.038x^3+12.49x^2-21.268x\right)|}_0^{125}\\ =\left(0.00025{\left(125\right)}^4-0.038{\left(125\right)}^3+12.49{\left(125\right)}^2-21.268\left(125\right)\right)\\ -\left(0.00025{\left(0\right)}^4-0.038{\left(0\right)}^3+12.49{\left(0\right)}^2-21.268\left(0\right)\right)\end{array}$

Further simplify.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{125}{\int }}y\left(x\right)dx}=\left(0.00025\left(244140625\right)-0.038\left(1953125\right)+12.49\left(15625\right)-21.268\left(125\right)\right)\\ =\left(61035.156-74218.75+195156.25-2658.5\right)\\ =179314.156\end{array}$

Therefore, the height reached by the Endeavour $125$ seconds after the liftoff is equal to $179314.156\text{ft}$ .