Highway design sight distance: for a design speed of 80 km/h, reaction time 2.5 s, and coefficient of friction f = 0.35, what is the stopping sight distance (SSD)?

Introduction / Context:Stopping sight distance (SSD) is the minimum distance required for a driver to perceive, react, and bring a vehicle to a stop. It combines perception–reaction distance and braking distance, both of which depend on speed and friction.

Given Data / Assumptions:

• Design speed v = 80 km/h.
• Reaction time t = 2.5 s.
• Coefficient of longitudinal friction f = 0.35.
• Acceleration due to gravity g ≈ 9.81 m/s^2; level grade assumed.

Concept / Approach:SSD = perception–reaction distance + braking distance. Convert speed to m/s and use the kinematic relation for braking on level roads with friction.

Step-by-Step Solution:Convert speed: v = 80 * 1000 / 3600 ≈ 22.222 m/s.Perception–reaction distance: v * t = 22.222 * 2.5 ≈ 55.56 m.Braking distance: v^2 / (2 * g * f) = 22.222^2 / (2 * 9.81 * 0.35).Compute v^2 ≈ 493.83; denominator ≈ 6.867 → braking ≈ 71.94 m.SSD ≈ 55.56 + 71.94 ≈ 127.5 m → closest option 127 m.

Verification / Alternative check:Using rounded g = 9.8 gives essentially the same result, confirming 127 m.

Why Other Options Are Wrong:

• 132 m is slightly high given the parameters.
• 76 m or 56 m ignore either reaction distance or realistic braking distances at 80 km/h.

Common Pitfalls:

• Forgetting to convert km/h to m/s.
• Using grade-adjusted formulas when the road is level.

Final Answer:127 m