How do I estimate a car's braking distance from speed and tyre friction, and where does the energy go?

With all wheels at the friction limit the deceleration is µ × g = 0.8 × 9.81 = 7.85 m/s², independent of mass. Braking distance is d = v² / (2 × µ × g). 100 km/h is 27.8 m/s, so d = 772 / 15.7 = 49 m. 50 km/h is 13.9 m/s, so d = 193 / 15.7 = 12.3 m.

The ratio is four because the distance goes with v squared: the brakes have to remove kinetic energy, ½mv², and they do it with a roughly constant force, so the distance is proportional to the energy.

That is only the braking phase. Total stopping distance adds the distance covered during the driver's reaction time at full speed. With a reaction time of 1 s, often used for estimates, that is 27.8 m at 100 km/h and 13.9 m at 50 km/h, so the totals are about 77 m and 26 m. At low speeds the reaction distance is the larger part; at high speeds the braking distance takes over.

Friction also varies widely: wet asphalt is typically 0.4 to 0.6 and ice around 0.1, and the braking distance grows in inverse proportion.

The energy goes into the brakes as heat. At 100 km/h the car has ½ × 1500 × 27.8² = 579 kJ. If that ends up in four cast-iron discs with a combined mass of about 25 kg and a specific heat of about 460 J/kg·K, the average temperature rise is 579,000 / (25 × 460) = about 50 K for one stop, ignoring cooling during the stop and the share taken by the pads.

One stop is easy. The problem is repeated stops or a long descent, where heat goes in faster than the discs can shed it. The temperature climbs by a step each time, and eventually the pad friction falls off (fade) or the fluid boils.

Front brakes are larger because of load transfer. Deceleration acts at the centre of gravity, above the road, and that shifts load from the rear axle to the front: ΔW = m × a × h / L. With a = 7.85 m/s², a centre of gravity height of 0.55 m and a wheelbase of 2.6 m, ΔW = 1500 × 7.85 × 0.55 / 2.6 = 2490 N, about 17 percent of the car's weight.

A car with a 55/45 static split is then at roughly 72/28 during hard braking. Friction force is proportional to load, so the front tyres can do, and the front brakes must do, most of the work, and the front discs absorb most of the heat calculated above.

The formula assumes the tyres stay at peak friction, which is what anti-lock braking is for. A tyre produces its highest braking force at around 10 to 20 percent slip; a fully locked wheel slides with somewhat lower friction on most surfaces and, more importantly, cannot generate side force, so the car cannot be steered. ABS modulates pressure to hold each wheel near the peak.

For the assignment, state your assumptions: constant µ, no aerodynamic drag, level road. On a slope, replace µ × g with g × (µ × cos θ + sin θ) uphill and g × (µ × cos θ - sin θ) downhill.