Chapter 6.5, Problem 12E

To determine

To find: The approximation the distance the bobsled travels.

Answer to Problem 12E

The distance traveled is $0.045\text{mile}$ .

Explanation of Solution

Given information:

  

Calculation:

In the trapezoidal rule, if $\left[a,b\right]$ is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of f on $\left[a,b\right]$ can be approximated by a straight line segment at each sub-interval. The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal "attitude" times the average of two vertical "bases".

That is,

  ${{\displaystyle \int }}_a^{b}f(x)dx\approx h\frac{y_0+y_1}{2}+h\frac{y_1+y_2}{2}+\dots .+h\frac{y_{n-1}+y_n}{2}$

  $\begin{array}{l}=h\left(\frac{y_0}{2}+y_1+y_2+\dots .+y_{n-1}+\frac{y_n}{2}\right)\hfill \\ =\frac{h}{2}\left(y_0+2y_1+2y_2+\dots .+2y_{n-1}+y_n\right)\hfill \end{array}$

Where,

  $\begin{array}{cc}{y}_0& =f(a)\\ y_1& =f\left(x_1\right)\\ & \cdot \\ & \cdot \\ y_{n-1}& =f\left(x_{n-1}\right)\\ y_n& =f(b)\end{array}$

Now, apply the trapezoidal rule with

  $h=1$

$time\left(sec\right)$		$speed\left(mph\right)=v\left(t\right)$
$t_0$	0	$v_0$	0
$t_1$	1	$v_1$	3
$t_2$	2	$v_2$	7
$t_3$	3	$v_3$	12
$t_4$	4	$v_4$	17
$t_5$	5	$v_5$	25
$t_6$	6	$v_6$	33
t7	7	$v_7$	41
$t_8$	8	$v_8$	48

Now, substitute the values from given table in this equation, equation has factor of $\frac{1}{3600}$ because time is given in second then it must converted in hour.

  $\text{distance traveled}=\frac{1}{3600}\times \frac{h}{2}\left(v_0+2v_1+2v_2+2v_3+2v_4+2v_5+2v_6+2v_7+v_8\right)$

  $\begin{array}{l}=\frac{1}{7200}(0+2(3)+2(7)+2(12)+2(17)+2(25)+2(33)+2(41)+48)\hfill \\ =\frac{1}{7200}(0+6+14+24+34+50+66+82+48)\hfill \\ =0.045\text{mile}\hfill \end{array}$

  $$

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