Chapter 8.1, Problem 34E

To determine

Tofind the approximate velocity at $t=15$ using the trapezoidal rule.

Answer to Problem 34E

Approximately the velocity is found to be $125ft/\sec$ option (d).

Explanation of Solution

Given information:

The data for the acceleration of a car from $0$ to $15$ seconds are given by the below table. If the velocity at $t=0$ is $5ft/\sec$ .

  

Formula:

The Trapezoidal rule is

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\approx \frac{1}{2}\Delta x\left[f\left(x_0\right)+2f\left(x_2\right)+2f\left(x_3\right)+\dots +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

Where $\Delta x=\frac{b-a}{n},a=x_0,b=x_n$

From given we know that the distance between each point is given by

  $\begin{array}{l}15-12=3\\ 12-9=3\\ 9-6=3\\ 6-3=3\\ 3-0=3\end{array}$

The Trapezoidal rule is

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\approx \frac{1}{2}\Delta x\left[f\left(x_0\right)+2f\left(x_2\right)+2f\left(x_3\right)+\dots +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ \Rightarrow \frac{1}{2}\left(3\right)\left(4+2\left(8\right)+2\left(6\right)+2\left(9\right)+2\left(10\right)+10\right)\\ =120\end{array}$

The original velocity was $5ft/\sec$ ,

Hence the total velocity is given by

  $\Rightarrow 120+5=125$

Therefore, approximately the velocity is found to be $125ft/\sec$ option (d).