SOLUTION: A radar gun was used to record the speed of a car (in feet per minute) during selected times in the first 2 minutes of a race. Use a trapezoidal sum with 4 intervals to estimate th

Question 1119297: A radar gun was used to record the speed of a car (in feet per minute) during selected times in the first 2 minutes of a race. Use a trapezoidal sum with 4 intervals to estimate the distance the car covered during those 2 minutes.
t 0 0.3 1.0 1.6 2
v(t) 0 24.5 27.8 28.3 29.0
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Those are extremely low speeds in feet per minute for a race car; they are even slow if they were feet per second.

But how reasonable the given data is is not important; the important thing is how to use a trapezoidal rule to estimate the total distance.

Using a trapezoidal rule essentially means that in each time interval you assume the acceleration is constant. That means the average speed over each interval is the average of the speeds at the beginning and end of each period. So

(1) for 0.3 minutes, between t=0 and t=0.3, the average speed was (0+24.5)/2 = 12.25 ft/min; the total distance is (0.3)(12.25) = 3.675 ft.
(2) for 0.7 minutes, between t=0.3 and t=1.0, the average speed was (24.5+27.8)/2 = 26.15 ft/min; the total distance is (0.7)(26.15) = 18.305 ft.
(3) for 0.6 minutes, between t=1.0 and 7=1.6, the average speed was (27.8+28.3)/2 = 28.05 ft/min; the total distance is (0.6)(28.05) = 16.83 ft.
(4) for 0.4 minutes, between t=1.6 and t=2, the average speed was (28.3+29.0)/2 = 28.65 ft/min; the total distance is (0.4)(28.65) = 11.46 ft.

The total estimated distance using a trapezoidal rule is
3.675+18.305+16.83+11.46 = 50.27 feet

....sounds like a crawling insect -- not a race car.

Change the "feet per minute" to "meters per second" and you get reasonable speeds, as long as it is a small race track.
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