Question

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) 18 20 22 29 31 31 26 5280 In order to have the time and the velocity in consistent units, let's convert the velocity readings to feet per second (1 mi/h = ft/s). (Round your answers 3600 to the nearest whole number.) Time (5) 0 5 10 15 20 25 30 Velocity (t/s) 26 32 45 45 38 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 (26 ft/s), then we obtain the approximate distance traveled during the first five seconds: 26 / 5S it. Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when - So our estimate for the distance traveled from 1-5 to 10 is 29 Ft/55 it. If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: (26 * 5) + (29 * 5) + (32 x 5) + (435) + (45 x 5) + (455) - 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 * 5) + (32 * 5) + (435) + (45 * 5) + (455) + (385) . If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second.

Answer 1

Givn: & mifn= 5200 feface then go mish = 5200 x 20 = 29.3å 229 3607 29 milh-598 42.53 ~42 3600 Honce Timerno 0 15 30 velocity left/l) 26 129 32/(43| 451 45 38 B6Hfred »(5) (51) - 145ft And (2635) +12945) +Bax 5)+(43,065)+(4545) -(4515)= 3100.ft (2905) + (325) + (435) + (45%65) + (4545) + (3805)= 1160 fr (99/8) and