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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.

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Chrisnando

Answer:

Grade of road (G) = -0.58 •/•

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

[tex] a= 3.5m/ sec^2 [/tex]

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

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

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

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

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