SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Question : 141 of 200

Marks: +1, -0

Calculate the lag distance for design speed of 47 km/h for two-way traffic on a single-lane road (assume coefficient of friction as 0.38 and reaction time of driver as 2.5 seconds)

32.64 m
111.04 m
55.52 m
65.28 m

Validate

Solution:

Concept:
Lag distance = 0.278 × V × $\t_\R$
Where,
V = Speed in Kmph

\t_\R

= Reaction time in sec
Calculation:
Lag distance = 0.278 × 47 × 2.5 = 32.665
For two way traffic on a single lane lag distance = 2 × 32.665 = 65.33 m

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