Question

The stopping sight distance for a design speed of 75 kmph, assuming the coefficient of friction as 0.34 and total reaction time of driver as 2.5 seconds will be

• A. 65.7
• B. 125
• C. 107.3
• D. 117.9

Answer 1

The correct option is D 117.9

Given,
$\text{Design speed}=75kmph$
$f=0.34$
$t=2.5sec$
$\text{Lag distance}=0.278Vt=0.278\times 75\times 2.5$
$=52.125m$
$\text{Braking distance}=\frac{V^2}{2gf}=\frac{(0.278\times 75{)}^2}{2\times 9.81\times 0.34}=65168m$
$\text{Stopping sight distance = Lag distance + Braking distance}$
$=52.125+65.68$
$=117.29m$