SightDistanceSolutions

# SightDistanceSolutions - CEE 4600 Sight Distance Homework...

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CEE 4600 Sight Distance Homework Problems Solutions 1. A driver takes 3.0 seconds to react to a complex situation while traveling at a speed of 55 mph. How far does the vehicle travel before the driver initiates a physical response (i.e., putting their foot on the brake)? How does this compare to the standard reaction time of 2.5 seconds and compared to an attentive, younger driver with a reaction time of 1.25 seconds? Brake Reaction Distance d R = 1.47 Vt = 1.47(55)(3) = 242.55 ft = 1.47(55)(2.5) = 202.13 ft = 1.47(55)(1.25) = 101.06 ft A reduction in reaction time provides a significant decrease in the brake reaction distance. 2. Use Table 7-6 in your textbook to determine the stopping sight distance for a typical car (deceleration rate = 11.2 ft/s 2 ) on a level road with a design speed of 40 mph assuming the driver has a standard reaction time of 2.5 seconds. Brake Reaction Distance = 147.0 ft Braking Distance = 153.6 ft Stopping Sight Distance = 147 + 153.6 = 300.6 ft Design = 305 ft 3. Assume the same driver is on the same road as in Problem 2 (design speed of 40 mph) and a dog runs into the street 300’ in front of them. Will they be able to stop in time? Will they be able

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## This note was uploaded on 06/07/2009 for the course CEE 4600 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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SightDistanceSolutions - CEE 4600 Sight Distance Homework...

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