Question

A highway is designed for a design speed of 80 kmph. However a driver is travelling at a speed of 120 kmph on this highway at a descending gradient of 2.5%. The increase in the stopping sight distance required for this driver will be_________. Given coefficient of friction, f = 0.35 and reaction time as 2.5 sec.

Answer 1

Given; Design speed = 80 kmph
$\Rightarrow V_1=\frac{80}{3.6}=22.22m/s$
Driver speed = 120 kmph
$\Rightarrow V_2=\frac{120}{3.6}=33.33m/s$
Reaction time = 2.5 sec and f = 0.35,
x = -2.5% = - 0.025

Stopping sight distance for design speed,
$S_1=V_1\times t+\frac{V_1^2}{2\times g\times (f\pm x)}$
$\Rightarrow S_1=22.22\times 2.5+\frac{{22.22}^2}{2\times 9.81\times (0.35-0.025)}$
$\Rightarrow S_1=132.98m$

Stopping sight distance for driver's speed,
$S_2=33.33\times 2.5+\frac{{33.33}^2}{2\times 9.81\times (0.35-0.025)}$
$S_2=257.54m$

Increase in sight distance,
$\Delta S=S_2=S_1$
$=257.54-132.98$
$=124.6m$