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:
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:
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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