What this answers
From the moment a driver sees a hazard to the moment the car stands still, it covers two distances: the one travelled while the driver reacts, and the one travelled while the brakes work. The first grows with the speed, the second with the square of it, which is why speed matters more than it feels like.
Formula
Reaction distance = v × t
Braking distance = v² / (2 × g × (μ cos θ + sin θ))
Stopping distance = reaction distance + braking distance
- v — speed in m/s (km/h divided by 3.6); t — reaction time in seconds
- μ — coefficient of friction between tyre and road; θ — the slope of the road
- g = 9.80665 m/s², standard gravity
Worked examples
50 km/h, 1.5 s, μ 0.7. 13.89 m/s × 1.5 s = 20.8 m of reaction, and 13.89² / (2 × 9.80665 × 0.7) = 14.1 m of braking: 34.9 m in all.
The same at μ 0.1. Braking grows to 98.4 m. With these example coefficients, braking on ice takes seven times the distance it takes on dry asphalt.
What this doesn’t cover
- Reaction time and friction are examples you change, not figures from a road authority. Real values depend on the driver, the tyres and the road.
- Brake systems, ABS, tyre condition and load. The car is taken to slow at a constant rate.
Common questions
Why does doubling the speed more than double the distance?
The reaction distance doubles, but the braking distance has the speed squared in it, so it
quadruples.