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Your team of test engineers is to release the parking brake so an 768.0-kg car will roll down a very long 6.1 percent grade in preparation for a crash test at the bottom of the incline. (On a 6.1 percent grade the change in altitude is 6.1 percent of the horizontal distance traveled.) The total resistive force (air drag plus rolling friction) for this car has been previously established to be Fd = 100 N + (1.2 N · s2/m2)v2, where v is the speed of the car. What is the terminal speed for the car rolling down this grade?

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temdan2001

Answer: 19.98m/s

Explanation: since total resistive force (air drag plus rolling friction) for this car has been established to be

100 + 1.2v²] = Fᵈ

The angle of inclination is:

tanθ = 0.061/1 = tan(3.491°)

terminal(constant) speed is achieved when net force down along the incline = Fᵈ

=> mgsin3.491° = [100 + 1.2v²]

=> 768(9.81)0.061 = 100 + 1.2v²

579.62 = 100 + 1.2v²

1.2v² = 579.62 - 100

v² = 479.62/1.2

=> v² = 399.38

=> v ~= 19.98 m/s

So the car achieve terminal speed of 19.98m/m as it's rolling down the grade.

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