Question

On one way traffic road, the speeds of overtaking and overtaken vehicles are 80 kmph and 50 kmph respectively. If the acceleration of the overtaking vehicle is 2.5 kmph per second, calculate the overtaking sight distance (Assume spacing between vehicles is 16 m, reaction time of driver is 2 seconds)

• A. 191.05

Answer 1

The correct option is A 191.05
Given one way traffic road.



Let A be overtaking vehicle and B be the overtaken vehicle

$V_A=80kmph=22.22m/s$

$V_B=50kmph=13.88m/s$

$a=2.5kmph/s=0.694m/s^2$

$s_1=s_2=16m$

Reaction time, t = 2 sec

Overtaking sight distance,
$\text{OSD}=d_1+d_2$

Where $d_1$ is distance travelled during reaction time

$d_2$ is distance travelled during overtaking operation

$d_1=V_a\times T=13.88\times 2m$

$\Rightarrow d_1=27.76m$

$d_2=V_{B}T+2s$

$T=\sqrt{\frac{4s}{a}}sec$

Where,
$s=s_1=s_2=16m$

$\Rightarrow T=\sqrt{\frac{4\times 16}{0.694}}sec$

$T=9.603sec$

$\Rightarrow d_2=(13.8\times 9.603)+(2\times 15)$

$=163.289m$

$\Rightarrow d=d_1+d_2$

$=27.76+163.29$

$=191.05m$