The delivery rider, the child, and eighteen metres
A delivery rider on a straight, level road travels at 36 km/h. He suddenly sees a child run onto the road 18 m ahead. His reaction time - the interval between seeing the child and the brakes actually biting - is about 0.6 s, during which the scooter continues at the same speed. Once the brakes bite on the dry road, the scooter slows at about 5 m/s². The day before, in light rain, the same scooter with the same brakes managed only about 3 m/s² before the wheels began to skid. The scooter can carry a pillion, and the rider's 12-year-old brother often rides along without a helmet; when the rider brakes hard, the boy slides forward into his back. The rider insists the scooter is safe because 'the brakes are new'.
1. Calculate the total stopping distance on the dry road: the distance travelled during the reaction time plus the braking distance. Does he stop before reaching the child?
Read the worked answer
Convert the speed: 36 km/h = 36 × 1000/3600 = 10 m/s.
Distance during reaction time - the scooter moves at constant speed because the brakes are not yet applied: s = v × t = 10 × 0.6 = 6 m.
Braking distance - use v² = u² − 2as with final velocity zero: 0 = 10² − 2 × 5 × s s = 100/10 = 10 m.
Total stopping distance = 6 + 10 = 16 m.
The child is 18 m ahead, so he stops with about 2 m to spare. It works - but only just, and only because the road is dry and his reaction time is short. The result should be read as a warning rather than reassurance: the margin is smaller than the length of the scooter. Note how the reaction-time distance (6 m) is over a third of the total; a tired rider with a 1.0 s reaction time would need 10 + 10 = 20 m and would hit the child.
2. Repeat the calculation for the wet road with deceleration 3 m/s². What conclusion should the rider draw, and about how fast would he still be moving at the child's position?
Read the worked answer
Reaction distance is unchanged: 10 × 0.6 = 6 m, because reaction time does not depend on the road surface.
Wet-road braking distance: 0 = 10² − 2 × 3 × s s = 100/6 = 16.7 m.
Total = 6 + 16.7 = 22.7 m, which is greater than 18 m - he does not stop in time.
To find the speed at the child's position, the braking distance available is 18 − 6 = 12 m of braking: v² = u² − 2as = 100 − 2 × 3 × 12 = 100 − 72 = 28 v ≈ 5.3 m/s ≈ 19 km/h.
So he would strike the child at roughly 19 km/h - enough to cause serious injury.
The conclusion: stopping distance depends on the square of the speed and on the road surface, not on how new the brake pads are. New brakes do not shorten the reaction-time distance, and on a wet road the limiting factor is the friction between tyre and road. The safe response is to reduce speed in rain (at 18 km/h, i.e. 5 m/s, the total wet stopping distance is only 3 + 4.2 = 7.2 m) and to keep a larger gap.
3. Explain why the pillion rider slides forward when the brakes are applied, and why carrying him makes the stopping distance longer even with the same brakes.
Read the worked answer
The boy slides forward because of inertia of motion (Newton's first law). While the scooter moves at constant velocity, the boy moves with it. When the brakes are applied, the scooter decelerates because of the friction force on it, but no matching backward force acts on the boy - the only horizontal force on him is a little friction from the seat and whatever he holds. He therefore tends to continue at the original velocity, and in the frame of the scooter he appears to lurch forward. The same effect is what makes a standing passenger in a braking bus fall forward.
Carrying a pillion increases the total mass in motion. For the same speed, the momentum is larger, and the brakes can supply only a limited retarding force (friction between tyres and road, which does not increase in proportion). Since a = F/m, the same braking force on a larger mass gives a smaller deceleration, so the braking distance increases. If deceleration falls from 5 to, say, 4 m/s², the dry braking distance rises from 10 m to 100/8 = 12.5 m - nearly 3 m more.
Practical points that follow from the physics: the boy should wear a helmet and hold a proper grab rail or footrest, the rider should brake earlier and more gently with a pillion, and speed must be reduced. A pillion is not 'free' - it changes the stopping distance and the balance of the vehicle.