Calculus

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The police department must determine a safe speed limit on a bridge so that the flow rate of cars is at a maximum per unit time. The greater the speed limit, the farther apart the cars
must be expected to be in order to allow for a safe stopping distance. The total distance needed for a car to stop, if a car in front of it stops suddenly, depends on the speed of the car through two factors: the time needed to react and the braking distance. Experimental data on the braking distance d (in feet), on the bridge surface, for various speeds s (in miles per hour) is given in the
following table. The table also provides an estimate for reaction distance r (in feet); this is the
distance the car will travel before the driver reacts.
s (in mph) 5 10 20 30 40 50 60
d (in feet) 4 11 33 62 100 149 203
r (in feet) 5 10 20 30 40 50 60
The police department has also identified the lengths of the 5 most common types of vehicles that are expected to use the bridge:
Model Length (in inches)
Fiat 500 142.0
Ford Fiesta 153.1
Dodge Caliber 173.8
Honda Civic 176.5
Dodge Grand Caravan 202.5
The bridge will also be used occasionally by tractor-trailers with an average length of 75 feet and stopping distances that are about 40% greater than the stopping distances for an automobile.
1. Find a function of the form d(s) = as2 +bs+c that models the breaking distance in terms of speed s. What should d(0) equal? What is a reasonable estimate for dŒ(0)? Select the constants a, b and c that best fits the data and produces the value of d(0) and dŒ(0) that
you identified. Also find a function r(s) that models the reaction distance in terms of the
speed.
Please Help!!!!!

  • Calculus -

    That's a lot of spurious data for answering one simple question. Something tells me there are followup questions coming. For now, all we are interested in is speed and BRAKing distance. (If the braking distance is too large, then we will get into the breaking distance!)

    s (in mph) 5 10 20 30 40 50 60
    d (in feet) 4 11 33 62 100 149 203

    Letting d = 1.5(s/5)^2 gets us


    s/5 (in mph) 1 2 4 6 8 10 12
    d (in feet) 1.5 6 24 54 96 150 216

    It underestimates for small s, and starts to overestimate at about s=50

    You can play around with other terms, or use quadratic regression or divided differences.

    Others may offer further insights.

    You might notice that the slope increases by about 10 in each interval of 10. That is, the slope is just about the value of s.

    d' = s

    and then integrate that twice to get a quadratic and evaluate a,b,c.

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