A car is traveling up a 2% grade at 70 mi/h on good, wet pavement. The driver brakes to try to avoid hitting stopped traffic that is 250 ft ahead. The driver's reaction time is 0.5 s. At first, when the driver applies the brakes, a software flaw causes the antilock braking system to fail (brakes work in non-antilock mode with 80% efficiency), leaving 80 ft skid marks. After the 80 ft skid, the antilock brakes work with 100% efficiency. How fast will the driver be going when the stopped traffic is hit if the coefficient of rolling resistance is constant at 0.013? (Assume minimum theoretical stopping distance and ignore aerodynamic resistance.)

Question

Nolan Mccann

Engineering

1 July, 20:15

A car is traveling up a 2% grade at 70 mi/h on good, wet pavement. The driver brakes to try to avoid hitting stopped traffic that is 250 ft ahead. The driver's reaction time is 0.5 s. At first, when the driver applies the brakes, a software flaw causes the antilock braking system to fail (brakes work in non-antilock mode with 80% efficiency), leaving 80 ft skid marks. After the 80 ft skid, the antilock brakes work with 100% efficiency. How fast will the driver be going when the stopped traffic is hit if the coefficient of rolling resistance is constant at 0.013? (Assume minimum theoretical stopping distance and ignore aerodynamic resistance.)

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Answers (1)

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Carlee Todd · Answer 1

Given data;

time = 0.5 sec; velocity, v = 70 m/hr = 102.667 ft/sec; skid distance = 80 ft, efficiency = 100 %

Step 1: calculate distance traveled, dt = velocity * time = 102.667 * 0.5 = 51.33 ft

Step 2: Calculate distance traveled before applying brake = distance traveled + skid distance = 51.33 ft + 80 = 131.33 ft

Thus, required distance = 250-131.33 = 118.67 ft

Step 3: Calculate velocity, using this formula;

V = √[Required distance * reaction time (coefficient of resistance + percentage grade) ]

V = √[118.67 * 30 (0.013 + 0.02) ]

V = √118.67 * 0.99

V = √117.4833

V = 10.8 ft/sec