Drivers must slow down from 60 mi/hr to 40 mi/hr to negotiate a severe curve on a rural highway. A warning sign for the curve is clearly visible for a distance of 120 ft. How far in advance of the curve must the sign be located to ensure that vehicles have sufficient distance to decelerate safely. Use the standard reaction time and deceleration rate recommended by AASHTO for basic braking maneuvers.

the distance in advance of the curve the sign must be located to ensure that vehicles have sufficient distance to decelerate safely is 292.07 ft

Explanation:

first we calculate the distance to place sign board in advance to curve for decelerating the speed

d = 1.47Si t + [ (Si² - Sf²) / 30(0.348 ± 0.01G)

where d is the safe stopping distance, initial speed is Si (60 mi/hr ), time reaction is t (2.5s), Sf is the final speed (40 mi/hr), G is the grade (0).

so we substitute

d = (1.47 × 60 × 2.5) + [ (60² - 40²) / 30(0.348 ± 0)

= 220.5 + ( 2000/10.44)

= 220.5 + 191.57

= 412.07 ft

Now giving that a warning sign is clearly visible at a distance of 120 ft

optimum safe distance will be

d = minimum distance - sign visible distance

d = 412.07 - 120

d = 292.07 ft

therefore the distance in advance of the curve the sign must be located to ensure that vehicles have sufficient distance to decelerate safely is 292.07 ft

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answer:

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