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13 Nov 2019
EXAMPLE 4 Suppose the odometer on our car is broken and wo want to ostimato the distance driven over a 30 seoond time and record them in the following table. Time (s) 5 10 15 20 25 30 Velocity (mi/h) 18 20 22 29 33 31 29 In order to have the time and the velocity in consistent units, let's convert the velodity readings to feot per second (1 miht/s). (Round your answers to the nearost whole number.) 3600 Time (s) |0| Velocity (ft/s) 26 |10| |20|25|30| 5 15 X 32 X48 45 43 During the first five seconds the velocity doesnt assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (26 f/s), then we obtain the approximate distance traveled during the first five seconds: aly much, so we can estimate the distance travoled during that time by 26 f/s x 5s Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t 5s. So our estimate for the distance traveled from t= 5 s to t= 10 s is 29ft/s à 5 s = ft. If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (26 x 5) + (29 x 5) + (32 à 5) + (43 à 5) + (48 x 5) + (45 x 5) = 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 5) + (32 x 5) + (43 x 5) + (48 à 5) + (45 x 5) + (43x5).
EXAMPLE 4 Suppose the odometer on our car is broken and wo want to ostimato the distance driven over a 30 seoond time and record them in the following table. Time (s) 5 10 15 20 25 30 Velocity (mi/h) 18 20 22 29 33 31 29 In order to have the time and the velocity in consistent units, let's convert the velodity readings to feot per second (1 miht/s). (Round your answers to the nearost whole number.) 3600 Time (s) |0| Velocity (ft/s) 26 |10| |20|25|30| 5 15 X 32 X48 45 43 During the first five seconds the velocity doesnt assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (26 f/s), then we obtain the approximate distance traveled during the first five seconds: aly much, so we can estimate the distance travoled during that time by 26 f/s x 5s Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t 5s. So our estimate for the distance traveled from t= 5 s to t= 10 s is 29ft/s à 5 s = ft. If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (26 x 5) + (29 x 5) + (32 à 5) + (43 à 5) + (48 x 5) + (45 x 5) = 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 5) + (32 x 5) + (43 x 5) + (48 à 5) + (45 x 5) + (43x5).
Beverley SmithLv2
16 Aug 2019