Question

The velocity of a car at one second intervals is given in the following table: Time t(s) 0 1 2 3 4 5 6 Speed v(m/s) 0 2.0 4.5 8.00 14.00 21.0 29.0 Determine the distance travelled in 6 seconds (i.e. the area under the v/t graph) using Simpson’s rule.

Answer 1

For a better understanding of the solution please find the attached file which is the graphical representation of the table given.

Simpson's Rule can be applied as:

$\int f(x)dx=\frac{\triangle x}{3}[f(x_{0})+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)]$

Where $\triangle x=\frac{6-0}{6}=1$

Applying the rule we get the distance travelled in 6 seconds (i.e. the area under the v/t graph) using Simpson’s rule as to be:

\frac{1}{3}[0+4\times2+2\times 4.5+4\times 8+2\times 14+4\times 21+29]\approx63.33

Thus 63.33 metres is the correct answer.

Answer 2

Answer: The distance traveled in 6 seconds is 61.182.

$\begin{matrix} Time\quad t(s) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ Speed & 0 & 0.2 & 5.8 & 8 & 14 & 21 & 29 \\ y\quad (m/s) & { y }_{ 0 } & { y }_{ 1 } & { y }_{ 2 } & { y }_{ 3 } & { y }_{ 4 } & { y }_{ 5 } & { y }_{ n } \end{matrix}$

According to the Simpson’s rule,

$\int _{ a }^{ b }{ f(x) } dx\quad =\quad \frac { h }{ 3 } [{ y }_{ 0 }+\quad 4({ y }_{ 1 }+{ y }_{ 3 }+{ y }_{ 5 }+...)+\quad 2\quad ({ y }_{ 2 }+{ y }_{ 4 }+{ y }_{ 6 }+......)\quad +{ y }_{ n }]$

b = 6 , a = 0 , no of intervals (n) = 6

$h =\frac{ b-a}{n} = \frac{6-0} { 6} = 1$

Therefore,

$= \frac{1}{3} [ 0+4(0.2+8+21)+2(5.8+14)+29 ]$

= 0.33 [ 4 x 29.2 + 2 x 19.8 + 29] = 0.33[116.8 + 39.6 + 29]

Distance Traveled = 0.33 x 185.4 = 61.182