Chapter 6.5, Problem 11E

To determine

To calculate: The area under the velocity curve in which the time intervals will get vary in length.

Answer to Problem 11E

The distance travelled is 0.633 mile

Explanation of Solution

Given information:

The accompanying table shows time-Io-speed data for a 2011 Scarpellone Panther accelerating from rest to 130 mph.

  

Formula used:

  $\text{distance traveled}=\frac{1}{3600}\left[\left(t_1-t_0\right)\frac{v_0-v_1}{2}+\left(t_2-t_1\right)\frac{v_1-v_2}{2}+\mathrm{.....}+\left(t_{11}-t_0\right)\frac{v_{11}-v_{10}}{2}\right]$

Calculation:

Since, here time intervals vary in length: the graph of f on [a,b] can be approximated by astraight line segment at each subinterval, The region between the curve and the x-axis is thenapproximated by the trapezoids: the area of each trapezoid being the length of its horizontal“altitude" times the average of two vertical “bases“: Since time is given in second so, right sightside must be divided by 3600 to convert the time into hour. That is,

  $\begin{array}{l}\text{distance traveled}=\frac{1}{3600}\left[\left(t_1-t_0\right)\frac{v_0-v_1}{2}+\left(t_2-t_1\right)\frac{v_1-v_2}{2}+\mathrm{.....}+\left(t_{11}-t_0\right)\frac{v_{11}-v_{10}}{2}\right]\\ \text{distance traveled}=\frac{1}{7200}\left(\left(t_1-t_0\right)\left(v_0+v_1\right)+\left(t_2-t_1\right)\left(v_1+v_2+\mathrm{.......}+\left(t_{11}-t_{10}\right)\left(v_{11}+v_{10}\right)\right)\right)\end{array}$

  

Now, substitute the values from given table in the above equation and dince

  $\begin{array}{l}\text{distance traveled}=\frac{1}{7200}\left(\left(t_1-t_0\right)\left(v_0+v_1\right)+\left(t_2-t_1\right)\left(v_1+v_2\right)+\mathrm{......}+\left(t_{11}-t_{10}\right)\left(v_{11}+v_{10}\right)\right)\\ =\frac{1}{7200}\left(\begin{array}{l}\left(2.0-0\right)\left(0+30\right)+\left(3.2-2.0\right)\left(30+40\right)+\left(4.5-3.2\right)\left(40+50\right)\\ +\left(5.8-4.5\right)\left(50+60\right)\\ +\left(7.7-5.8\right)\left(60+70\right)+\left(9.5-7.7\right)\left(70+80\right)+\left(11.6-9.5\right)\left(80+90\right)+\\ \left(14.9-11.6\right)\left(90+100\right)+\left(17.8-14.9\right)\left(100+110\right)\\ +\left(21.7-17.8\right)\left(110+120\right)+\left(26.3-21.7\right)\left(120+130\right)\end{array}\right)\\ \text{distance traveled}=0.633\text{mile}\end{array}$

Conclusion:

The distance travelled is 0.633 mile