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 %

b. check your rearview mirror

explanation: