Question

): drivers must slow down from 60 to 40 mi/hr to negotiate a severe curve. A warning sign is visible for a distance of 120 ft. How far in advance of the curve must the sign be located.

Answer 1

Answer:

Explanation:

Total Stopping distance is the sum of the reaction distance and the braking distance, such that ;

d=dr + db =1.47St +[ ( Si 2- Sf2 ) / 30(F/0.01G) ]

In this case, the reaction time, t, is the AASHTO standard, or 2. 5 s. The friction factor, F, is based upon the standard AASHTO deceleration rate of 11.2 ft/s2 (F = 11.2/32.2 = 0.348) and the speed is given as 40 mph.

d = 1.47x60x2.5+ [ (60^2-40^2) / 30x 0.348 ]

d = 220.5 + ( (3600-1600) / 10.44 )

d = 220.5 + 191.57

d = 412.07ft.

Since the sign can be seen clearly at 120 ft.

Then the position of the sign should be,

= 412.07 - 120

= 292.07 ft