A driver is traveling at 55 mi/h on a wet road. An object is spotted on the road 450 ft ahead and the driver is able to come to a stop just before hitting the object. Assuming standard perception/reaction time, determine the grade of the road.

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 %