Stopping distance with brake efficiency and reaction time If V is the speed in km/h, t is the driver's brake-reaction time in seconds, η is the brake efficiency (0 to 1), and f is the longitudinal friction coefficient, the total stopping distance S on level road is:

Difficulty: Medium

Correct Answer: S = 0.278 * V * t + V^2 / (254 * η * f)

Explanation:


Introduction / Context:
Stopping sight distance combines perception–reaction distance and braking distance. Real vehicles do not always exploit full tire–road friction due to brake condition and balance, so an efficiency factor η reduces the effective deceleration.



Given Data / Assumptions:

  • V in km/h; t in seconds.
  • f = coefficient of longitudinal friction.
  • η = brake efficiency (0 < η ≤ 1).
  • Level road (no grade term included).


Concept / Approach:
Total distance S = perception–reaction distance + braking distance. The first term is v * t, with v = 0.278 * V. The second term derives from uniform deceleration a = η * f * g. Using standard highway units introduces the constant 254 when V is in km/h and distances are in metres.



Step-by-Step Solution:

Perception–reaction distance = (V / 3.6) * t = 0.278 * V * t.Braking distance = V^2 / (254 * effective friction).Effective friction = η * f ⇒ braking distance = V^2 / (254 * η * f).Total S = 0.278 * V * t + V^2 / (254 * η * f).


Verification / Alternative check:
When η = 1, the expression reduces to the familiar SSD formula on level grade: S = 0.278 * V * t + V^2 / (254 * f).



Why Other Options Are Wrong:
Option B multiplies friction by η in the numerator, which unrealistically shortens distance. Option C divides f by η inside the denominator, effectively increasing deceleration instead of reducing it. Option D omits braking distance entirely.



Common Pitfalls:
Using η > 1; forgetting that adverse grades require replacing f by (f ± G).



Final Answer:
S = 0.278 * V * t + V^2 / (254 * η * f)

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