Question

A driver is traveling at 52 mi/h on a wet road. an object is spotted on the road 415 ft 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

Answer:

G = 0.424

Explanation:

Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))

Where Ds = stopping sight distance = 415miles = 126.5m

G = absolute grade road

V = velocity of vehicle = 52miles/hr

f = friction = 0 because the road is wet

tr = standard perception / reaction time = 2.5s

So therefore:

Substituting to get G

We have

2479.4G = 705.6G + 751.72

1773.8G = 751.72

G = 751.72/1773.8

G = 0.424

Answer 2

Answer: 12.6°

Explanation: