Stopping sight distance on a highway

SSD = lag distance + braking distance = 0.278Vt + V²/(254f), with V in km/h.

Worked answer.

Stopping sight distance (SSD) is the minimum distance a driver needs to see ahead to stop safely before hitting a stationary obstacle. It has two parts.

Step 1: Lag distance (vehicle travels at constant speed while the driver perceives and reacts).
Converting 80 km/h to m/s: 80/3.6 ≈ 22.22 m/s.
Lag = v × t = 22.22 × 2.5 ≈ 55.6 m (IRC's form: 0.278 × 80 × 2.5 = 55.6 m).

Step 2: Braking distance on level ground.
Braking = v² / (2gf). In km/h units this becomes V² / (254f).
= 80² / (254 × 0.35) = 6400 / 88.9 ≈ 72.0 m.

Step 3: SSD.
SSD = 55.6 + 72.0 ≈ 127.6 m, which you would round up to about 128 m for design. Always state the reaction time and friction coefficient you assumed, because the answer depends directly on them.

Step 4: On a 3% downgrade. Gravity helps the vehicle keep moving, so the effective friction is reduced:
Braking = V² / [254(f − n)] = 6400 / [254 × (0.35 − 0.03)] = 6400 / 81.28 ≈ 78.7 m.
SSD ≈ 55.6 + 78.7 ≈ 134.3 m.
On a 3% upgrade it would be 6400 / (254 × 0.38) ≈ 66.3 m braking, giving SSD ≈ 121.9 m.

Check. Using SI directly: 22.22² / (2 × 9.81 × 0.35) = 493.8 / 6.867 ≈ 71.9 m, which agrees with 72.0 m within rounding.

Why f decreases with speed. IRC uses lower friction coefficients at higher speeds: about 0.40 at low speeds down to 0.35 at 80 km/h and above. Tyre-road friction decreases as speed rises.

Common mistake. Forgetting the lag distance, or using V in km/h in v²/(2gf) without converting.

Likely follow-up. "What is intermediate sight distance?" On undivided two-lane roads where overtaking sight cannot be provided, IRC recommends twice the SSD as the intermediate sight distance.