Consider the following velocity data: t (min)

Consider the following velocity data: t (min) | v m/s --- | --- 1 | 5 2 | 6 3.25 | 5.5 4.5 | 7 6 | 8.5 7 | 8 8 | 6 8.5 | 7 9 | 7 10 | 5 Use the trapezoidal rule to find the distance traveled from the given data. Also, determine the average velocity. (Round the final answers to four decimal places.) The required distance is 60.0224 m and the average velocity is 6.0125 m/s.

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00:01 All i like to think about is the area of a trapezoid.

00:04 Area trap is equal to one half and i always do the height times the sum of the bases.

00:12 So what you have to be careful of on this problem is that the width of each of these values is not the same.

00:19 So i guess what i'm looking at is going from one to two.

00:25 So this approximation has a width of well i need to have a one half in all these.

00:30 And really you could factor out the one half because it's going to be the same in all of them.

00:35 So the first height is one, the width of the two bases.

00:38 Then i have to add together five and six.

00:41 Then i have to add to it the next rectangle or trapezoid which going from two to 3 .25 is 1 .25 with the sum of its two bases which is six and 5 .5.

00:55 And then going from 3 .25 to 4 .5 is another 1 .25 with 5 .5 again plus its other piece of seven.

01:10 Now let me just move down here because now 4 .5 to 6 has a width of 1 .5 with a height of seven and i believe it was 8 .5.

01:20 And then the next width going from six to seven was one, 8 .5 and eight.

01:27 And then the next one seven to eight has a width of one using eight and the second height is six.

01:41 Now we're going to add to it.

01:45 I'm losing track...

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