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
