Stopping distance for a loaded delivery van
An empty delivery van travels at 14 m/s on a straight, level road. The driver's reaction time is 0.70 s, and the brakes produce a constant deceleration of 6.0 m/s². The driver sees a stationary truck 30 m ahead, applies the brakes, and stops safely. On a later trip the van is fully loaded, so its total mass is twice the empty mass, but the maximum braking force the brakes can apply is unchanged. The driver again travels at 14 m/s, and again a hazard appears 30 m ahead; reaction time is unchanged. The transport officer says the van will now take twice as long to stop because the braking force is unchanged. The mechanic disagrees: he says the tyres, not the brakes, limited the empty van's deceleration, so the loaded van will still stop in about the same distance. The two claims cannot both be right.
1. Calculate the reaction distance, braking distance and total stopping distance for the empty van, and decide whether it stops in time.
Read the worked answer
Reaction distance = v × t = 14 × 0.70 = 9.8 m. No braking occurs during this time, so the van still travels at 14 m/s when the brakes are applied.
Braking distance from v² = u² + 2as with final speed 0 and a = −6.0 m/s²: s = u²/(2a) = 14²/(2 × 6.0) = 196/12 = 16.3 m.
Total stopping distance = 9.8 + 16.3 = 26.1 m, which is less than 30 m, so the van stops with about 3.9 m to spare. This is consistent with the account.
Note the split: the reaction distance is about 38 % of the total. At 14 m/s, every extra 0.1 s of reaction time adds 1.4 m, so reaction time is not a minor detail. The calculation assumes level ground, dry tyres and no delay in the brakes actually biting.
2. Assume the brakes limit the deceleration. Calculate the speed at which the loaded van would reach the hazard 30 m ahead.
Read the worked answer
Constant braking force with double the mass gives half the deceleration: a = 3.0 m/s².
The reaction phase is unchanged, so 9.8 m is covered before braking starts. Distance left for braking = 30 − 9.8 = 20.2 m.
Using v² = u² + 2as with a = −3.0 m/s²: v² = 196 − 2(3.0)(20.2) = 196 − 121.2 = 74.8 v = 8.6 m/s (about 31 km/h).
The van hits the obstacle at roughly 8.6 m/s, not at zero. Equivalently, the loaded van would need 9.8 + 32.7 = 42.5 m to stop, well beyond the 30 m available.
If instead the tyres limit the braking, the friction force is μmg, so a = μg is independent of mass and the deceleration stays at 6.0 m/s²; the van then stops in the original 26.1 m. The two assumptions give completely different outcomes, so the next task is to decide which one applies.
3. Which claim, the officer's or the mechanic's, is better supported, and what test would settle it?
Read the worked answer
The critical question is whether the empty van's 6.0 m/s² was limited by the brakes or by the tyres.
6.0 m/s² corresponds to μ ≈ a/g = 0.60, a realistic tyre–road coefficient on dry asphalt, and close to the usual limit (about 0.7–0.8 for good tyres). That makes it plausible that the tyres were already at their limit when empty. If so, the friction force on the loaded van is twice as large, because it equals μ(2m)g, exactly the force needed to give the same 6.0 m/s². The mechanic is then right.
If the brakes could have produced more than 6.0 m/s² and the tyres had reserve grip, the brakes were the limit, and doubling the mass halves the deceleration. The officer is then right.
A practical test: measure the deceleration of the loaded van directly, or carry out a skid test to find μ. Another indicator is whether the empty van could lock its wheels under hard braking: if it could, the tyres were at their limit.