SUVAT v² Equation — Braking Distance (Worked Example)

Medium

Problem

A car travelling at 25 m/s brakes hard with a constant deceleration of 5 m/s² until it stops. What is its braking distance?

Given

u = 25 m/s
v = 0 m/s
a = −5 m/s²

Find

The stopping distance s.

Principle

When time is unknown, v² = u² + 2as links velocities directly to distance. Deceleration is negative acceleration.

Formula

v^2 = u^2 + 2as

v² = u² + 2as

Solution

Write the equation: $v^2 = u^2 + 2as$
Substitute: $0 = 25^2 + 2(-5)s$
Rearrange: $10s = 625$
Solve: $s = 62.5\ \mathrm{m}$

62.5 m

Unit check

(m/s)² ÷ (m/s²) = m. A distance — correct.

What this means physically

62.5 m is over half a football field — at 90 km/h even hard braking needs serious road. This is why speed limits matter: braking distance grows with the square of speed.

Common mistake

Using a = +5 m/s². The sign matters: deceleration opposes motion, so a is negative and the equation yields a positive distance.