Complete question is;
The design speed of a multilane highway is 60 mi/h. Determine (a) the minimum stopping sight distance that should be provided for a level roadway, and (b) the minimum stopping sight distance that should be provided for a roadway with a maximum grade of 7 percent. Note: The term a/g in the appropriate equation is typically rounded to 0.35 in calculations. Assume perception reaction time of 2.5 sec.
Answer:
A) SSD = 563.357 ft
B) SSD = 506.214 ft
Step-by-step explanation:
A) Formula for sight stopping distance is;
SSD = 1.47ut + [u²/(30((a/g) ± G)]
We are given;
Speed; u = 60 mi/hr
Perception time; t = 2.5 s
Ratio of deceleration to gravitational acceleration; a/g = 0.35
Grade of road will be 0 because we are told that it is a level roadway. Thus, G = 0
Plugging in the relevant values;
SSD = (1.47 × 60 × 2.5) + (60²/30(0.35 ± 0))
SSD = 220.5 + 342.857
SSD = 563.357 ft
B) The road now has a grade of 7%. So, G = 0.07.
Now, this can either be a downgrade or an upgrade.
For minimum sight stopping distance, it will have to be a upgrade as the denominator of [u²/(30((a/g) ± G)] will be larger because we will use positive instead of negative before G.
Thus;
SSD = (1.47 × 60 × 2.5) + (60²/30(0.35 + 0.07))
SSD = 220.5 + 285.714
SSD = 506.214 ft