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Stopping Distance Under Friction Braking

A car travelling at $25\ \text{m/s}$ brakes to a complete stop. The coefficient of friction between the tyres and the road is $\mu = 0.70$ , and the wheels do not lock.

Take $g = 9.81\ \text{m/s}^2$ .

Find the minimum stopping distance.

Given

v0	25 m/s	Initial speed
μ	0.7 -	Coefficient of friction

Hint 1

Find the deceleration first. What force is doing the stopping?

Hint 2

Friction force is μmg, so a = μg — the mass cancels out entirely.

Hint 3

With no time given, use v² = v₀² − 2ad.

Worked solution — try the problem first

Deceleration from friction. The friction force is $\mu mg$ , so by Newton's second law:

$ma = \mu m g \quad \Longrightarrow \quad a = \mu g$

The mass cancels — stopping distance does not depend on how heavy the car is. That is worth internalising; it surprises people.

$a = 0.70 \times 9.81 = 6.867\ \text{m/s}^2$

Kinematics (no time given, so use the velocity-displacement relation):

$v^2 = v_0^2 - 2ad \quad \Longrightarrow \quad 0 = 25^2 - 2(6.867)d$

$d = \frac{625}{13.734} = 45.5\ \text{m}$

$\boxed{d = 45.5\ \text{m}}$

Equivalent energy argument: the kinetic energy $\tfrac12 mv_0^2$ is dissipated by friction work $\mu mgd$ ; cancelling $m$ gives $d = v_0^2/(2\mu g)$ , the same result.

Concepts:Newton's second lawFrictionKinematicsWork-energy theorem

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