Answer:
G = +/- 0.0583 = +/- 5.83 %
Explanation:
Given:
- Speed of the vehicle V = 55 mi/h
- The spotting distance D = 450 ft
- Assume standard perception/reaction time t = 2.5 s
Find:
Determine the grade of the road. When driver stops just before hitting the object.
Solution:
Step 1:
Determine the distance d traveled for the perception reaction time t:
d = V*t
d = 55*1.467*2.5 = 201.7125 ft
Step 2:
The remaining distance left is the practical stopping distance SD:
SD = D - d
SD = 450 - 201.7125
SD = 248.2875 ft
Step 3:
Calculate the grade G using practical stopping distance SD formula:
SD = V^2 / [ 30(a/g +/-G) ]
(a/g +/-G) = V^2 / 30*SD
G = +/- ( V^2 / 30*SD - a/g )
Where, a = 11.2 ft/s^2 and g = 32.2 ft/s^2 and V = 55 mph
Plug in the values:
G = +/- ( 55^2 / 30*248.2875 - 11.2/32.2 )
G = +/- 0.0583 = +/- 5.83 %
