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Video Example EXAMPLE4 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) 05 10 15 20 2530 Velocity (mi/h) 16 20 23 30 32 31 29 In order to have the time and the velocity in consistent units, let's convert the velocity readings to 5280 ft/s) (Round your answers to the nearest whole number.) 3600 feet per second (1 mi/h Time (s) 20 25 30 0 5 10 15 Velocity (ft/s) 23 47 45 43 34 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 (23 ft/s), then we obtain the approximate distance traveled during the first five seconds: ft 23 ft/s x 5 s = 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 29 ft/s x 5 s = ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (23 x 5)(29 x 5) + (34 x 5) (44 x 5) + (47 x 5) (45 x 5) = ft. We could just as well 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 (29 x 5)(34 x 5) (44 x 5) (47 x 5) + (45 x 5) + (43 x 5) ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second.

Video Example EXAMPLE4 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) 05 10 15 20 2530 Velocity (mi/h) 16 20 23 30 32 31 29 In order to have the time and the velocity in consistent units, let's convert the velocity readings to 5280 ft/s) (Round your answers to the nearest whole number.) 3600 feet per second (1 mi/h Time (s) 20 25 30 0 5 10 15 Velocity (ft/s) 23 47 45 43 34 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 (23 ft/s), then we obtain the approximate distance traveled during the first five seconds: ft 23 ft/s x 5 s = 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 29 ft/s x 5 s = ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (23 x 5)(29 x 5) + (34 x 5) (44 x 5) + (47 x 5) (45 x 5) = ft. We could just as well 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 (29 x 5)(34 x 5) (44 x 5) (47 x 5) + (45 x 5) + (43 x 5) ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter3: Straight Lines And Linear Functions
Section3.3: Modeling Data With Linear Functions
Problem 23E: When Date Are Unevenly speed. If data are evenly spaced, we need only calculate differences to see...
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Video Example EXAMPLE4 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) 05 10 15 20 2530 Velocity (mi/h) 16 20 23 30 32 31 29 In order to have the time and the velocity in consistent units, let's convert the velocity readings to 5280 ft/s) (Round your answers to the nearest whole number.) 3600 feet per second (1 mi/h Time (s) 20 25 30 0 5 10 15 Velocity (ft/s) 23 47 45 43 34 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 (23 ft/s), then we obtain the approximate distance traveled during the first five seconds: ft 23 ft/s x 5 s = 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 29 ft/s x 5 s = ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (23 x 5)(29 x 5) + (34 x 5) (44 x 5) + (47 x 5) (45 x 5) = ft. We could just as well 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 (29 x 5)(34 x 5) (44 x 5) (47 x 5) + (45 x 5) + (43 x 5) ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second.
Transcribed Image Text:Video Example EXAMPLE4 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) 05 10 15 20 2530 Velocity (mi/h) 16 20 23 30 32 31 29 In order to have the time and the velocity in consistent units, let's convert the velocity readings to 5280 ft/s) (Round your answers to the nearest whole number.) 3600 feet per second (1 mi/h Time (s) 20 25 30 0 5 10 15 Velocity (ft/s) 23 47 45 43 34 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 (23 ft/s), then we obtain the approximate distance traveled during the first five seconds: ft 23 ft/s x 5 s = 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 29 ft/s x 5 s = ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (23 x 5)(29 x 5) + (34 x 5) (44 x 5) + (47 x 5) (45 x 5) = ft. We could just as well 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 (29 x 5)(34 x 5) (44 x 5) (47 x 5) + (45 x 5) + (43 x 5) ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second.
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