Stopping sight distance at 50 kmph

Assumed data (IRC 66 / NRS 2070)

Design speed $V=50$ km/h, total reaction time $t=2.5$ s, coefficient of longitudinal friction $f=0.37$ (IRC value for 50 km/h), level road.

Basic stopping sight distance

Lag distance $=0.278Vt=0.278(50)(2.5)=34.75$ m. Braking distance $=\dfrac{V^2}{254f}=\dfrac{50^2}{254(0.37)}=26.60$ m.

$$SSD=0.278Vt+\frac{V^2}{254f}=61.35\ \text{m}$$

(i) Two-way traffic on a two-lane road

Each vehicle keeps to its own lane and only has to stop for an obstruction in its path, so SSD ≈ 61.4 m ≈ 65 m (IRC gives 60 m for 50 km/h).

(ii) Two-way traffic on a single-lane road

Vehicles from opposite directions share the same lane and may approach head-on, so each must be able to stop within half of the sight distance. The sight distance is twice the SSD:

$$SD=2\times SSD=2(61.35)=122.70\ \text{m}$$

≈ 125 m (IRC gives 120 m).