A car is traveling 50 mi/h when the driver applies the brakes. The car is traveling on good wet pavement and is going down a 3% grade. The antilock braking system works intermittently during the stop. When the antilock brakes system is functioning, the braking efficiency is 100%. When the antilock system is not functioning the wheels lock and the braking efficiency is 70%. The antilock braking system turns off/on every 60 ft while the brakes are being applied (it is on for the first 60 ft of the stop, off for the second 60 ft, on for the third 60 ft of the stop, and so on). Under these conditions, what is the total distance needed for the car to stop after the brakes are applied? (Assume theoretical stopping distance, ignore aerodynamic resistance, and let frl = 0.011.)

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

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Eq. 2.43 can be applied to solve this problem,

( ) ( )

2 2 1 2

2 sinbb rl g

V V Sg fγη μ θ− =+ ±

but it will have to be applied in 60-ft distance increments to account for the intermittent antilock brake failure. For the first 60 ft with the antilock brakes functioning, in Eq. 2.43 S will be 60 ft, frl and sin g are given in the problem, b is a known constant (see text), μ can be obtained from Table 2.5 and will alternate between maximum and slide values depending on whether or not the antilock braking system is functioning, and b will alternate between 1.0 and 0.7 depending on whether the antilock braking system is functioning or not. We wish to determine the vehicle speed at the end of the first 60-ft interval (an interval where the antilock brakes are working), so rearranging Eq. 2.43 we have,

( ) ( )× × × + − = × − =22

60 2 0.9 1.467 45.91 ft/s

V The final vehicle speed of 45.91 ft/s after the first 60-ft interval will become the initial speed of the second 60-ft interval, an interval in which the antilock braking system does not function. For the second 60-ft interval, μ and b will change to their non-antilock brake values and Eq. 2.43 is applied again to give,