Question

The safe stopping distance for a design speed of 50 kmph for a two lane road with coefficient of friction of 0.37 in m is

• A. 61.3
• B. 81.7
• C. 123.7
• D. 161.6

Answer 1

The correct option is A 61.3

We know that
$\text{SSD}=vt+\frac{v^2}{2gf}$

As per IRC
$t=2.5sec$

$\text{SSD}=50\times \frac{5}{18}\times 2.5+\frac{(50\times \frac{5}{18}{)}^2}{2\times 9.81\times 0.37}$

$=61.3m$