Difficulty: Easy
Correct Answer: 12 metres
Explanation:
Introduction / Context:Braking distance depends primarily on initial speed and available longitudinal friction. For quick checks, standard values are tabulated for typical dry conditions and effective brakes.
Given Data / Assumptions:
Concept / Approach:Using uniform deceleration, braking distance s = v^2 / (2 * g * f). Adopting f ≈ 0.3–0.35 gives conservative distances commonly cited in design aids for low speeds.
Step-by-Step Solution:
Compute v = 8.33 m/s; v^2 ≈ 69.4.With f = 0.30, s = 69.4 / (2 * 9.81 * 0.30) ≈ 11.8 m (≈ 12 m).With f = 0.35, s ≈ 10.1 m. The 12 m choice matches a conservative tabulated value.Verification / Alternative check:Using highway units: s ≈ V^2 / (254 * f) = 30^2 / (254 * 0.30) ≈ 900 / 76.2 ≈ 11.8 m, consistent with the above.
Why Other Options Are Wrong:10 m is feasible but less conservative; 15 m and 18 m are excessive for the assumed conditions.
Common Pitfalls:Forgetting to include reaction distance in full SSD; using wet or icy friction without adjusting the estimate.
Final Answer:12 metres
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