Article

Braking

From 50 km/h to zero: deceleration a, distance s = v²/(2a), and with reaction time an extra v·tr stretch.

Part 3 of Physics without mysteries. Previous: accelerated motion. Next: vertical throw.

How it works

In the school model, braking is uniformly decelerated motion: enter positive a and v₀, get stopping distance s = v₀²/(2a) and time t = v₀/a. Classic numbers: about 50 km/h is 13.9 m/s; at a = 6 m/s² you get s ≈ 16.1 m.

On the road the driver reaction time t_r matters: before the pedal moves, the car still covers v·t_r. Total stopping distance is that sum — a separate stopping-distance calculator.

From momentum: p = mv must be removed by the brake impulse. From work: W = F·s along the braking stretch. Same numbers, three page intents.

Which formula when

Goal	Formula	Tool
Distance at known a	s = v₀²/(2a)	Uniformly decelerated motion
Time to stop	t = v₀/a	Same calculator
With reaction time	s = v·tᵣ + v²/(2a)	Stopping distance
Momentum to remove	p = mv	Momentum p = mv
Brake work	W = F·s	Work force–distance

Solved problems

City 50 → 0

A car brakes from v = 13.9 m/s (about 50 km/h) to rest at constant deceleration a = 6 m/s². What are the braking distance and time? (no reaction time)

Steps

Model: uniformly decelerated motion, vₖ = 0.
Distance: s = v₀²/(2a) = 13.9² / (2·6).
13.9² = 193.21; divide by 12 → s ≈ 16.10 m.
Time: t = v₀/a = 13.9/6 ≈ 2.32 s.
Fill the decelerated-motion calculator: v₀ = 13.9, a = 6.

Answer: s ≈ 16.1 m, t ≈ 2.32 s

Fill in the calculator

With reaction time

Same v = 13.9 m/s and a = 6 m/s², but reaction time tᵣ = 1 s. What is the total stopping distance?