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13 Nov 2019
EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take the speedometer readings every five seconds and record them in the following table Time (s) 0 5 10 15 20 25 30 Velocity (mi/h) 17 21 |24 27 32 31 27 In order to have the time and the velocity in consistent units, let's convert the velocity readings to feet per second 5280 3600 (1 mi/h = ft/s). (Round your answers to the nearest whole number.) Time (s) 0 10 15 20 25 30 Velocity (ft/s) 25 35 47 45 40 During the first five seconds the velocity doesn't change very much, so we can estimate the distance traveled during that time by assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (25 ft/s), then we obtain the approximate distance traveled during the first five seconds 25 ft/s à 5 s- ft Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t = 5 s. So our estimate for the distance traveled from t = 5 s to t = 10 s is 31ft/s à 5 s- ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled (25 à 5) + (31 à 5) + (35 à 5) + (40 à 5) + (47 à 5) + (45 à 5) = ft. We could just as wel have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity. Then our estimate becomes (31 à 5) + (35 à 5) + (40 à 5) + (47 à 5) + (45 à 5) + (40 à 5) = ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second
EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take the speedometer readings every five seconds and record them in the following table Time (s) 0 5 10 15 20 25 30 Velocity (mi/h) 17 21 |24 27 32 31 27 In order to have the time and the velocity in consistent units, let's convert the velocity readings to feet per second 5280 3600 (1 mi/h = ft/s). (Round your answers to the nearest whole number.) Time (s) 0 10 15 20 25 30 Velocity (ft/s) 25 35 47 45 40 During the first five seconds the velocity doesn't change very much, so we can estimate the distance traveled during that time by assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (25 ft/s), then we obtain the approximate distance traveled during the first five seconds 25 ft/s à 5 s- ft Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t = 5 s. So our estimate for the distance traveled from t = 5 s to t = 10 s is 31ft/s à 5 s- ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled (25 à 5) + (31 à 5) + (35 à 5) + (40 à 5) + (47 à 5) + (45 à 5) = ft. We could just as wel have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity. Then our estimate becomes (31 à 5) + (35 à 5) + (40 à 5) + (47 à 5) + (45 à 5) + (40 à 5) = ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second
Lelia LubowitzLv2
20 Jul 2019