Reaction Time Versus Speed on a Wet Road
A car travels at 72 km/h on a dry road where its brakes can decelerate it at 5.0 m/s². The driver's reaction time before touching the brake pedal is 0.80 s. A pedestrian steps into the road 40 m ahead at the moment the hazard becomes visible. On the same road in rain, the usable deceleration drops to 3.0 m/s², and the driver's reaction time is unchanged. The driver cannot change the 40 m distance to the pedestrian.
1. On the dry road, does the car stop before reaching the pedestrian? Support the answer with the reaction distance and braking distance.
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Convert units first: 72 km/h ÷ 3.6 = 20 m/s.
Reaction distance (no braking yet, constant speed): d = v·t = 20 × 0.80 = 16 m.
Braking distance (uniform deceleration to rest): d = v²/(2a) = 400 / (2 × 5.0) = 40 m.
Total stopping distance = 16 + 40 = 56 m.
Verdict: 56 m > 40 m, so the car is still moving when it reaches the pedestrian. The shortfall is 16 m - exactly the distance covered before the brakes engage. The driver does not stop in time.
2. What speed would allow the car to stop exactly at the pedestrian on the dry road? Express the result in both m/s and km/h.
Read the worked answer
Budget the 40 m. The reaction phase consumes part of it at whatever speed the car is doing, and both phases shrink together as v falls - so this must be solved as one equation, not by scaling the old answer.
Let v be the unknown speed:
- reaction: 0.80v
- braking: v²/(2 × 5.0) = v²/10
Set 0.80v + v²/10 = 40. Multiply by 10: v² + 8v − 400 = 0.
v = [−8 + √(64 + 1600)] / 2 = (−8 + √1664)/2 = (−8 + 40.79)/2 = 16.4 m/s.
In km/h: 16.4 × 3.6 ≈ 59 km/h.
Check: 0.80 × 16.4 = 13.1 m reaction; 16.4²/10 = 26.9 m braking; total 40.0 m. ✓ The car must travel roughly 13 km/h slower than it does.
3. In rain the car keeps its 72 km/h and 0.80 s reaction time. A safety officer argues that "cutting the reaction time to zero would fix it." Test that claim, and say which single change actually matters.
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Wet-road stopping distance at 20 m/s: reaction 20 × 0.80 = 16 m; braking 400/(2 × 3.0) = 66.7 m; total 82.7 m - more than double the 40 m available.
Test the reaction-time claim on the dry road (the officer's best case). With reaction time reduced to zero, the car still needs 40 m of braking distance to stop from 20 m/s at 5.0 m/s². That is exactly the 40 m gap - no margin at all - and on the wet road it needs 66.7 m.
Conclusion: eliminating reaction time entirely still leaves a collision on the wet road and a zero-margin stop on the dry one. Reaction time is the smaller of the two terms at this speed (16 m versus 40–67 m), and it cannot compensate for kinetic energy, which grows with v².
Which change matters: reducing speed. Halving the speed roughly quarters the braking distance, whereas reaction time can at best remove 16 m.