A trucking company pays drivers an hourly wage to drive trailer-loads of cargo to distant destinations. A fully-loaded tractor-trailer rig itself also costs the company an average of $1.25/mile to operate (for fuel, maintenance, insurance, etc.). To monitor the safety and whereabouts of drivers and cargo, each truck is equipped with a GPS tracker, which transmits a continuous (second-by-second) record of the truck’s position; and a speedometer recorder, which transmits a continuous record of its speed (in mph). And the GPS has a “trip-meter” odometer (measuring miles) that is reset to zero at the start of each trip. A certain driver delivers a full load to some distant town, driving non-stop, costing the company a total of $930 (including the driver’s pay and benefits) for this one-way trip. As it turns out, the speedometer record for this trip best fits the following curve:

800s = (t/2)^(5/4) – (t/275)^(5/2)

where: t is the time in seconds (t= 0 when the driver first puts the truck in motion); and s is the speed in miles per hour. Suppose, at the above halfway point (that’s halfway in time), the driver had decided to slow the truck at a constant rate so that his total travel time for the day would have been the same as in the actual record above.

(i) How many miles from his intended destination would he have stopped under this alternate scenario?

(ii) Now, modeling the direction of travel to be always the positive x-direction (and still assuming that t = 0 at the start of the entire trip), write three equations of motion (for position, velocity, acceleration) as functions of time for this alternate second half of the trip. (Thus the curve given above would be the correct velocity function for the first half of the travel time, but your own velocity equation will be correct for the other half.) Express each function with coefficients, as in item (1) above.