Question

The stopping sight distance of a vehicle moving with a speed of 50 kmph in a two lane road is _______ m if the reaction time of driver is 2.3 sec and coefficient of longitudinal friction is 0.38.

1. 57.87
2. 35.65
3. 48.32
4. 12.43

Answer 1

Correct Answer - Option 1 : 57.87

Concept:

Stopping sight distance is given by:

= Lag distance + Breaking distance

\(SSD = 0.278 \times {V\times t_{R}} + \frac{{{V^2}}}{{254\;f}}\)

Calculation:

SSD = 95 m

V = 50 kmph

tR = 2.3 seconds

\(SSD\; = \;0.278\; \times \;50\; \times \;2.3 + \frac{{{{50}^2}}}{{254 \times 0.38}}\)

SSD = 57.87