Question

A driver is traveling at 52 mi/hr on a wet road.An object is spotted on the road 415 feet ahead and the driver is able to come to a stop just before hitting the object. Assuming standard perception/reaction time and practical stopping distance, determine the grade of the road.

Answer 1

Grade of road (G) = -0.58 •/•

Step-by-step explanation:

Vd = 52mi/h = 83.868 Km/hr

Vd is approximately 84Km/hr

Let's take >>>>> D = d = 135m

Perception Reaction Time (PRT) = tp = 2.5 sec

`a= 3.5m/ sec^2`

To find grade of road (G), we have,

`D= 0.278Vd * tp + Vd^2 / (254 [(a/9.81) + 0.01G]` )

`135 = 0.278 * 84 * 2.5 + (84^2 / 254 [(3.5 / 9.81) - 0.01G] )`

76.62 = 7056 / ( 90.622 - 2.54G )

6493.457 - 194.6148G = 7056

-112.542 = 194.61G

G = -112.542 / 194.6148 = -0.578 •/•

G = 0.578 = 0.58