Chapter 4.2, Problem 64E

(a)

To determine

To find: The cubic polynomial that best models the velocity of the shuttle.

Answer to Problem 64E

The cubic polynomial that best models the velocity of the shuttle for the time interval is $t\in \left[0,125\right]$ is $V=0\cdot 00146t^3-0\cdot 1153t^2+24\cdot 98169t-21\cdot 26872$ .

Explanation of Solution

Given information:

The given time interval is $t\in \left[0,125\right]$ .

The given data is shown in table (1).

  

Table (1)

Calculation:

Graph the given data by the computer and get the cubic polynomial that best model the velocity of the shuttle for the time interval $t\in \left[0,125\right]$ .

The cubic polynomial is.

  $V=0\cdot 00146t^3-0\cdot 1153t^2+24\cdot 98169t-21\cdot 26872$

  

Therefore, the cubic polynomial that best models the velocity of the shuttle for the time interval is $t\in \left[0,125\right]$ is $V=0\cdot 00146t^3-0\cdot 1153t^2+24\cdot 98169t-21\cdot 26872$ .

(b)

To determine

To find: The minimum and maximum value for the acceleration of the shuttle.

Answer to Problem 64E

The minimum and maximum value for the acceleration of the shuttle are $21\cdot 9{\text{ft}}^2\text{/s}$ and $64\cdot 53{\text{ft}}^2\text{/s}$ respectively.

Explanation of Solution

Given information:

The cubic polynomial is $V=0\cdot 00146t^3-0\cdot 1153t^2+24\cdot 98169t-21\cdot 26872$

Calculation:

The acceleration of the shuttle is equal to $\frac{dV}{dt}=a$ .

Calculate the acceleration.

  $a\left(t\right)=V\text{'}=0\cdot 00438t^2-0\cdot 23106t+24\cdot 98169$

Differentiate $a\left(t\right)$ with respect to $t$ for getting critical values.

  $\frac{da}{dt}=0\cdot 00876t-0\cdot 23106$

For $\frac{da}{dt}=0$ when

  $\begin{array}{c}0\cdot 00876t-0\cdot 23106=0\\ t=\frac{0\cdot 23106}{0\cdot 00876}\\ t\approx 26\cdot 37\text{seconds}\end{array}$

Calculate the value of $a\left(t\right)$ at $t\approx 26\cdot 37\text{seconds}$ .

  $\begin{array}{c}a\left(26\cdot 37\right)=0\cdot 00438{\left(26\cdot 37\right)}^2-0\cdot 23106\left(26\cdot 37\right)+24\cdot 98169\\ a\approx 21\cdot 9{\text{ft}}^2\text{/s}\end{array}$

Calculate the value of $a\left(t\right)$ at the end points of the interval $\left[0125\right]$ .

At $t=0$

  $\begin{array}{c}a\left(0\right)=0\cdot 00438{\left(0\right)}^2-0\cdot 23106\left(0\right)+24\cdot 98169\\ a=24\cdot 981669{\text{ft}}^2\text{/s}\end{array}$

At $t=125$

  $\begin{array}{c}a\left(125\right)=0\cdot 00438{\left(125\right)}^2-0\cdot 23106\left(125\right)+24\cdot 98169\\ a=64\cdot 53{\text{ft}}^2\text{/s}\end{array}$

The smallest number is $21\cdot 9{\text{ft}}^2\text{/s}$ .

The minimum value of the acceleration is about $21\cdot 9{\text{ft}}^2\text{/s}$ .

The greatest number is $64\cdot 53{\text{ft}}^2\text{/s}$ .

The maximum value of the acceleration is about $64\cdot 53{\text{ft}}^2\text{/s}$ .

Therefore, the minimum and maximum value for the acceleration of the shuttle are $21\cdot 9{\text{ft}}^2\text{/s}$ and $64\cdot 53{\text{ft}}^2\text{/s}$ respectively.