Minibus on a Wet Road: Stopping Distance and the Unbanked Bend
A school minibus of mass 2 000 kg is travelling along a straight, level road at 15 m/s. The driver's reaction time is measured as 0.80 s. Rounding a corner the driver suddenly sees a stationary lorry blocking the road 30 m ahead. The road is wet, the tyres give a coefficient of kinetic friction of 0.50 against the road, and the brakes apply a constant retarding force. Later on the same journey the minibus has to go round a flat, unbanked bend of radius 60 m at 15 m/s on the same wet surface. Take g = 9.8 m/s² and treat the minibus as a particle. The police note that the tyres were near the legal wear limit.
Afterwards the transport officer argues that the driver 'was not speeding, so the crash was unavoidable', while a second driver insists that 'a bus at 15 m/s needs only about 11 m to stop, so the driver could easily have pulled up in time'.
1. Using the particle model, calculate the total stopping distance at 15 m/s on the wet road. If the minibus cannot stop in 30 m, find its speed at the moment it reaches the lorry.
Read the worked answer
Step 1 - thinking distance. No braking force acts during the reaction time, so the minibus runs at a constant 15 m/s: s_th = u t = 15 × 0.80 = 12.0 m.
Step 2 - braking distance. The only horizontal force is kinetic friction F = μmg, so a = μg = 0.50 × 9.8 = 4.9 m/s². s_b = u²/(2a) = 15²/(2 × 4.9) = 225/9.8 = 23.0 m.
Step 3 - total. s = 12.0 + 23.0 = 35.0 m > 30 m, so a collision does occur. The single formula is s = u t_r + u²/(2μg).
Step 4 - impact speed. Only 30 − 12.0 = 18.0 m of braking is available: v² = u² − 2as = 225 − 2(4.9)(18.0) = 225 − 176.4 = 48.6, v = 7.0 m/s (about 25 km/h). The lorry is hit at roughly 7 m/s, not at 15 m/s: the reaction time and the wet surface between them remove most of the kinetic energy before impact.
2. Find the greatest speed at which the minibus can take the 60 m bend on the wet road, and use this with the stopping calculation to judge both drivers' claims. Also find the initial speed that would just have allowed the minibus to stop in 30 m.
Read the worked answer
Bend. Friction must supply the centripetal force: μmg ≥ mv²/r, so v_max = √(μgr) = √(0.50 × 9.8 × 60) = √294 = 17.1 m/s. At 15 m/s the required friction is mv²/r = 2 000 × 225/60 = 7 500 N, against a limit μmg = 9 800 N - 77 % of the limit. The bend is passable, but with only 2.1 m/s in hand. If diesel or leaves drop μ to 0.30, v_max = √(0.30 × 9.8 × 60) = 13.3 m/s and 15 m/s would skid. So 'not speeding' holds only for the straight road; it does not make the crash unavoidable.
Just-stopping speed. Put 0.80v + v²/(2 × 4.9) = 30: v² + 7.84v − 294 = 0 v = (−7.84 + √(61.5 + 1 176))/2 = (−7.84 + 35.2)/2 = 13.7 m/s. A reduction from 15 to about 13.7 m/s (9 %) would have avoided the crash.
Judging the claims. The officer ignores the wet surface and the reaction distance. The second driver's 11 m corresponds to v²/(2a) with a ≈ 10 m/s², i.e. a dry road with μ ≈ 1, and it ignores reaction time entirely; the true figure is more than three times that, so both statements are wrong for opposite reasons.
3. Explain why the constant-friction calculation may not match the real stopping distance. State assumptions in the calculation that could fail and say which way each would move the answer.
Read the worked answer
The calculation assumes a constant μ, a constant braking force applied instantly at t = 0.80 s, a particle with no weight transfer, a level road, and no air resistance.
In reality:
- Brake build-up: the pads take a few tenths of a second to reach full force, adding distance.
- Weight transfer: under braking the front tyres carry extra load while the rear tyres unload and may lock; once a tyre locks, sliding friction is lower than the peak (static) friction, so distance increases - ABS is designed to keep the tyre near the peak.
- Speed-dependent μ: μ falls as speed rises and with a water film, so the effective deceleration early in the stop is smaller than 4.9 m/s².
- Camber and gradient: a downhill or adverse camber adds a component of weight along the road, again lengthening the stop.
- Reaction time: 0.80 s is a laboratory average; a distracted driver may take 1.5 s or more, adding about 10 m. All of these make the real stopping distance longer than 35 m, so the real impact speed would be higher than 7 m/s, not lower - the calculation is an optimistic lower bound on the severity of the crash.