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13 Nov 2019
The velocity graph of a braking car is shown. Use it to estimate the distance traveled by the car while the brakes are applied. (Use Ms to get the most precise estimate.) 155 x ft v (ft/s) 90 60 30 t (seconds) 4
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Related questions
The velocity graph of a car accelerating from rest to a speed of 120 km/h over a period of 30 seconds is shown. Use the Midpoint Rule with {it n} = 6 to estimate the distance (in km) traveled during this period. Distance traveled
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The velocity graph of a car accelerating from rest to a speed of 120 km/h over a period of 30 seconds is shown. Use the Midpoint Rule with {it n} = 6 to estimate the distance (in km) traveled during this period. Distance traveled
ivorycicada928
A ball is dropped from a state of rest at time t = 0. The distance traveled after t seconds is s(t) = 16t2 ft. How far does the ball travel during the time interval [4, 4.5] ? Delta s = ft Compute the average velocity over [4, 4.5] . Delta s / Delta t = ft / sec Compute the average velocity over time intervals [4, 4.01] , [4, 4.001] , [4, 4.0001] , [3.9999, 4] , [3.999, 4] , [3.99, 4] . Use this to estimate the object's instantaneous velocity at t = 4 . V(4) = ft / sec
Show transcribed image text
A ball is dropped from a state of rest at time t = 0. The distance traveled after t seconds is s(t) = 16t2 ft. How far does the ball travel during the time interval [4, 4.5] ? Delta s = ft Compute the average velocity over [4, 4.5] . Delta s / Delta t = ft / sec Compute the average velocity over time intervals [4, 4.01] , [4, 4.001] , [4, 4.0001] , [3.9999, 4] , [3.999, 4] , [3.99, 4] . Use this to estimate the object's instantaneous velocity at t = 4 . V(4) = ft / sec
sangriahippopotamus492
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
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