Answer:
74.7 m
Explanation:
You want the minimum radius of curvature of a vertical circular arc so that vehicles traveling 95 km/h making the transition from horizontal to an 18° down grade will not leave the road.
The acceleration due to gravity on the vehicle at the point it is traveling 18° down is ...
g·cos(18°) ≈ (9.8 m/s²)·0.95106 ≈ 9.32 m/s²
The radius of an arc and the speed of travel are related to the normal acceleration by ...
r = v²/a
where v is the speed along the arc, and 'a' is the normal acceleration required to make an object follow the circular curve.
The speed of 95 km/h can be translated to m/s by ...
(95 km/h) · (1000 m/(1 km)) · (1 h)/(3600 s) = 95/3.6 m/s ≈ 26.389 m/s
Using the above relation to find the radius, we get ...
r = (26.389 m/s)²/(9.32 m/s²) ≈ 74.72 m
The minimum radius of the vertical circular arc should be 74.7 meters to prevent vehicles from becoming airborne.
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Additional comment
In actual roadway design, factors of sight distance and stopping distance come into play. Sight distance over this crest would be about 25 m. It would need to be more than 6 times that distance for the roadway to be safe at 95 km/h.
An 18° down grade is about 3 times the usual maximum for roadway design, which will severely increase stopping distance. The vertical arc described here would leave the car with no normal force on the roadway at the end of the arc, so braking would be impossible. Passengers would be feeling weightless, too.
Roadway curves are generally parabolic (not circular), so they have a constant rate of change of slope with horizontal distance.
