Chapter 24, Problem 45P

Employ the multiple-application Simpson's rule to evaluate the vertical distance traveled by a rocket if the vertical velocity is given by

$\begin{array}{l}v=11t^2-5t\\ v=1100-5t\\ v=50t+2{\left(t-20\right)}^2\end{array}$ $\begin{array}{l}0\le t\le 10\\ 10\le t\le 20\\ 20\le t\le 30\end{array}$

In addition, use numerical differentiation to develop graphs of the acceleration $\left(dv/dt\right)$ and the jerk $\left(d^{2}v/d{t}^2\right)$ versus time for $t=0$ to 30. Note that the jerk is very important because it is highly correlated with injuries such as whiplash.

To determine

To calculate: The vertical distance travelled by the rocket if the vertical velocity in different time interval mentioned in the problem by using the multiple-application Simpson’s rule, and plot the graph between acceleration $\left(\frac{dv}{dt}\right)$ versus time $\left(t\right)$ and also plot jerk $\left(\frac{d^{2}v}{d{t}^2}\right)$ versus time $\left(t\right)$.

Answer to Problem 45P

Solution:

The vertical distance travelled by the rocket is $26833.3309$.

The graph between acceleration $\left(\frac{dv}{dt}\right)$ versus time $\left(t\right)$ is,

The graph between jerk $\left(\frac{d^{2}v}{d{t}^2}\right)$ versus time $\left(t\right)$ is,

Explanation of Solution

Given Information:

The vertical velocity of the rocket is,

$v=\{\begin{array}{cc}11t^2-5t& 0\le t\le 10\\ 1100-5t& 10\le t\le 20\\ 50t+2{\left(t-20\right)}^2& 20\le t\le 30\end{array}$

The acceleration is,

$a=\frac{dv}{dt}$

The jerk is,

$\text{Jerk}=\frac{d^{2}v}{d{t}^2}$

Formula Used:

The expression for Simpson’s 1/3 rule is,

${\displaystyle {\int }_{t_1}^{t_2}y\left(t\right)dt=\frac{h}{3}\left[y\left(t_1\right)+4y\left(\frac{t_1+t_2}{2}\right)+y\left(t_2\right)\right]}$

The expression for the distance is,

$y={\displaystyle {\int }_{t_1}^{t_2}v\left(t\right)dt}$

Write the first derivative forward difference formula.

$f^{\prime }\left(x_i\right)=\frac{-f\left(x_{i+2}\right)+4f\left(x_{i+1}\right)-3f\left(x_i\right)}{2h}$

Write the first derivative two-point central difference formula.

$f^{\prime }\left(x_i\right)=\frac{f\left(x_{i+1}\right)-f\left(x_{i-1}\right)}{2h}$

Write the first derivative backward difference formula.

$f^{\prime }\left(x_i\right)=\frac{f\left(x_{i-2}\right)-4f\left(x_{i-1}\right)+3f\left(x_i\right)}{2h}$

Write the second derivative forward difference formula.

$f^{″}\left(x_i\right)=\frac{f\left(x_{i+2}\right)-2f\left(x_{i+1}\right)+f\left(x_i\right)}{h^2}$

Write the second derivative two-point central difference formula.

$f^{″}\left(x_i\right)=\frac{f\left(x_{i+1}\right)-2f\left(x_i\right)+f\left(x_{i-1}\right)}{h^2}$

Write the second derivative backward difference formula.

$f^{″}\left(x_i\right)=\frac{f\left(x_{i-2}\right)-2f\left(x_{i-1}\right)+f\left(x_i\right)}{h^2}$

Calculation:

Recall the expression of the velocity.

$v=\{\begin{array}{cc}11t^2-5t& 0\le t\le 10\\ 1100-5t& 10\le t\le 20\\ 50t+2{\left(t-20\right)}^2& 20\le t\le 30\end{array}$

Use the Excel to find the velocity at a different time.

For the range of $0\le t\le 10$.

$v=11t^2-5t$

Open the Excel and substitute the respective time. Apply the formula in column B.

For the range of $10\le t\le 20$.

$v=1100-5t$

Open the Excel and substitute the respective time. Apply the formula in column B.

For the range of $20\le t\le 30$.

$v=50t+2{\left(t-20\right)}^2$

Open the Excel and substitute the respective time. Apply the formula in column B.

Thus, the velocity at every time is,

time, t	Velocity
0	0
1	6
2	34
3	84
4	156
5	250
6	366
7	504
8	664
9	846
10	1050
11	1045
12	1040
13	1035
14	1030
15	1025
16	1020
17	1015
18	1010
19	1005
20	1000
21	1052
22	1108
23	1168
24	1232
25	1300
26	1372
27	1448
28	1528
29	1612
30	1700

Recall the expression for the distance,

The expression for the distance is,

$y={\displaystyle {\int }_{t_1}^{t_2}v\left(t\right)dt}$

Use Simpson’s 1/3 rule tointegrate the above data.

Recall the expression for Simpson’s 1/3 rule.

${\displaystyle {\int }_{t_1}^{t_2}y\left(t\right)dt=\frac{h}{3}\left[y\left(t_1\right)+4y\left(\frac{t_1+t_2}{2}\right)+y\left(t_2\right)\right]}$

Take the integral for the even space time, $h=1$.

Substitute all the value from the above table.

$\begin{array}{c}{\displaystyle {\int }_0^{2}y\left(t\right)dt}=\frac{1}{3}\left[y\left(0\right)+4y\left(1\right)+y\left(2\right)\right]\\ =\frac{1}{3}\left[0+4\left(6\right)+34\right]\\ =19.33333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_2^{4}y\left(t\right)dt}=\frac{1}{3}\left[y\left(2\right)+4y\left(3\right)+y\left(4\right)\right]\\ =\frac{1}{3}\left[34+4\left(84\right)+156\right]\\ =175.33333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_4^{6}y\left(t\right)dt}=\frac{1}{3}\left[y\left(4\right)+4y\left(5\right)+y\left(6\right)\right]\\ =\frac{1}{3}\left[156+4\left(250\right)+366\right]\\ =507.3333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_6^{8}y\left(t\right)dt}=\frac{1}{3}\left[y\left(6\right)+4y\left(7\right)+y\left(8\right)\right]\\ =\frac{1}{3}\left[366+4\left(504\right)+664\right]\\ =1015.333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_8^{10}y\left(t\right)dt}=\frac{1}{3}\left[y\left(8\right)+4y\left(9\right)+y\left(10\right)\right]\\ =\frac{1}{3}\left[664+4\left(846\right)+1050\right]\\ =1699.333\end{array}$

Solve further,

$\begin{array}{c}{\displaystyle {\int }_{10}^{12}y\left(t\right)dt}=\frac{1}{3}\left[y\left(10\right)+4y\left(11\right)+y\left(12\right)\right]\\ =\frac{1}{3}\left[1050+4\left(1045\right)+1040\right]\\ =2090\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{12}^{14}y\left(t\right)dt}=\frac{1}{3}\left[y\left(12\right)+4y\left(13\right)+y\left(14\right)\right]\\ =\frac{1}{3}\left[1040+4\left(1035\right)+1030\right]\\ =2070\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{14}^{16}y\left(t\right)dt}=\frac{1}{3}\left[y\left(14\right)+4y\left(15\right)+y\left(16\right)\right]\\ =\frac{1}{3}\left[1030+4\left(1025\right)+1020\right]\\ =2050\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{16}^{18}y\left(t\right)dt}=\frac{1}{3}\left[y\left(16\right)+4y\left(17\right)+y\left(18\right)\right]\\ =\frac{1}{3}\left[1020+4\left(1015\right)+1010\right]\\ =2030\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{20}^{22}y\left(t\right)dt}=\frac{1}{3}\left[y\left(20\right)+4y\left(21\right)+y\left(22\right)\right]\\ =\frac{1}{3}\left[1000+4\left(1052\right)+1108\right]\\ =2105.333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{22}^{24}y\left(t\right)dt}=\frac{1}{3}\left[y\left(22\right)+4y\left(23\right)+y\left(24\right)\right]\\ =\frac{1}{3}\left[1108+4\left(1168\right)+1232\right]\\ =2337.333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{24}^{26}y\left(t\right)dt}=\frac{1}{3}\left[y\left(24\right)+4y\left(25\right)+y\left(26\right)\right]\\ =\frac{1}{3}\left[1232+4\left(1300\right)+1372\right]\\ =2601.333\end{array}$

Solve further,

$\begin{array}{c}{\displaystyle {\int }_{26}^{28}y\left(t\right)dt}=\frac{1}{3}\left[y\left(26\right)+4y\left(27\right)+y\left(28\right)\right]\\ =\frac{1}{3}\left[1372+4\left(1448\right)+1528\right]\\ =2897.333\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{28}^{30}y\left(t\right)dt}=\frac{1}{3}\left[y\left(28\right)+4y\left(29\right)+y\left(30\right)\right]\\ =\frac{1}{3}\left[1528+4\left(1612\right)+1700\right]\\ =3225.333\end{array}$

The total integral is equal to the sum of all individual integral.

$\begin{array}{c}I=\left(\begin{array}{c}19.3333+175.3333+507.3333+1015.333+1699.333+2090+2070+2050\\ +2030+2010+2105.333+2337.333+2601.333+2897.333+3225.333\end{array}\right)\\ =26833.3309\end{array}$

Thus, the vertical distance travelled by the rocket is $26833.3309$.

Calculate the acceleration by numerical difference of the velocity.

$a\left(0\right)=\frac{dv}{dt}\left(0\right)$

Use the first derivative forward difference formula.

$\frac{dv}{dt}\left(0\right)=\frac{-f\left(x_{i+2}\right)+4f\left(x_{i+1}\right)-3f\left(x_i\right)}{2h}$

Substitute all the value from the time velocity table.

$\begin{array}{c}\frac{dv}{dt}\left(0\right)=\frac{-34+4\left(6\right)-3\left(6\right)}{2\left(1\right)}\\ =-5\end{array}$

Calculate the acceleration for time $\text{1 to 29}$ by using the first order central difference formula.

$\begin{array}{c}a\left(1\right)=\frac{dv}{dt}\left(1\right)\\ =\frac{f\left(x_{i+1}\right)-f\left(x_{i-1}\right)}{2h}\end{array}$

Substitute all the value from the time velocity table.

$\begin{array}{c}a\left(1\right)=\frac{\left(34\right)-0}{2\left(1\right)}\\ =17\end{array}$

Similarly calculate all the value by using the Excel.

Use the first derivative backward difference formula to calculate acceleration for $t=\text{30}$.

Recall the first derivative backward difference formula.

$\frac{dv}{dt}\left(t\right)=\frac{f\left(x_{i-2}\right)-4f\left(x_{i-1}\right)+3f\left(x_i\right)}{2h}$

Solve for $t=\text{30}$, Substitute all the value from the time velocity table.

$\begin{array}{c}\frac{dv}{dt}\left(30\right)=\frac{1528-4\left(1612\right)+3\left(1700\right)}{2\left(1\right)}\\ =90\end{array}$

The below plot shows the relation between the Acceleration and time.

Write the expression for the Jerk.

$\text{Jerk}=\frac{d^{2}v}{d{t}^2}$

Calculate the jerk by using the second order finite difference formula.

$\text{Jerk}\left(0\right)=\frac{d^{2}v}{d{t}^2}\left(0\right)$

Write the second derivative forward difference formula.

$\frac{d^{2}v}{d{t}^2}\left(0\right)=\frac{f\left(x_{i+2}\right)-2f\left(x_{i+1}\right)+f\left(x_i\right)}{h^2}$

Substitute all the value from the time velocity table.

$\begin{array}{c}\frac{d^{2}v}{d{t}^2}\left(0\right)=\frac{\left(34\right)-2\left(6\right)+\left(0\right)}{1^2}\\ =22\end{array}$

Calculate the jerk for time $\text{1 to 29}$ by using the second order central difference formula.

$\begin{array}{c}Jerk\left(1\right)=\frac{d^{2}v}{d{t}^2}\left(1\right)\\ =\frac{f\left(x_{i+1}\right)-2f\left(x_i\right)+f\left(x_{i-1}\right)}{h^2}\end{array}$

Substitute all the value from the time velocity table.

$\begin{array}{c}\frac{d^{2}v}{d{t}^2}\left(1\right)=\frac{\left(34\right)-2\left(6\right)+0}{1^2}\\ =22\end{array}$

Similarly calculate all the value by using the Excel.

Use the second derivative backward difference formula to calculate acceleration for $t=\text{30}$.

Write the second derivative backward difference formula.

$f^{″}\left(x_i\right)=\frac{f\left(x_{i-2}\right)-2f\left(x_{i-1}\right)+f\left(x_i\right)}{h^2}$

Solve for $t=\text{30}$, Substitute all the value from the time velocity table.

$\begin{array}{c}\frac{d^{2}v}{d{t}^2}\left(30\right)=\frac{1528-2\left(1612\right)+\left(1700\right)}{{\left(1\right)}^2}\\ =4\end{array}$

The below plot shows the relation between the jerk and time.