Multiple choice

The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and coefficient of friction of pavement surface as 0.35, the required stopping distance for two-way traffic on a single lane road is

  1. 82.1 m

  2. 102.4 m

  3. 164.2 m

  4. 186.4 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stopping distance = vt + v^2 / (2gf). v = 60 kmph = 16.67 m/s. SSD = (16.67 * 2.5) + (16.67^2) / (2 * 9.81 * 0.35) = 41.67 + 277.89 / 6.867 = 41.67 + 40.47 = 82.14 m. For two-way traffic on a single lane, OSD/SSD is doubled, so 82.14 * 2 = 164.28 m.

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