Braking Distance on a Wet Road
A 1200 kg car travels at 20 m/s on a dry asphalt road. The driver sees a hazard and applies the brakes. On dry asphalt the friction coefficient is 0.70; on the wet surface ahead it drops to 0.45. The driver's reaction time is 0.80 s before the brakes engage. A pedestrian steps out 45 m ahead of the car at the moment the driver first sees the hazard. The car's ABS keeps the wheels rolling so kinetic friction acts throughout braking. Take g = 9.8 m/s². During the reaction time the car continues at constant speed.
1. Determine whether the car stops before reaching the pedestrian, distinguishing the dry and wet cases.
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Reaction distance: during 0.80 s at 20 m/s the car covers $d_r = vt = 20 \times 0.80 = 16$ m. This is the same in both cases.
Braking distance from $\tfrac12 mv^2 = \mu mg, d_b$ gives $d_b = \dfrac{v^2}{2\mu g}$.
- Dry: $d_b = \dfrac{400}{2 \times 0.70 \times 9.8} = \dfrac{400}{13.72} \approx 29.2$ m.
- Wet: $d_b = \dfrac{400}{2 \times 0.45 \times 9.8} = \dfrac{400}{8.82} \approx 45.4$ m.
Total stopping distance: dry $= 16 + 29.2 = 45.2$ m; wet $= 16 + 45.4 = 61.4$ m.
Against the 45 m gap: on dry asphalt the car just barely stops (45.2 m > 45 m by a small margin - it stops essentially at the pedestrian, marginally short). On the wet surface it clearly does not stop; it would still be moving fast at the pedestrian's position. The mass cancels, so the 1200 kg is irrelevant to the distance.
2. A student claims that doubling the speed only doubles the stopping distance. Test this claim using the wet-road case.
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The braking distance goes as $v^2$: $d_b = v^2/(2\mu g)$. Doubling speed from 20 to 40 m/s quadruples the braking distance, not doubles it.
Wet road at 40 m/s: $d_b = \dfrac{1600}{8.82} \approx 181$ m, versus 45.4 m at 20 m/s - a factor of 4.
The reaction distance, however, is linear in $v$ ($d_r = vt$), so it only doubles (16 m → 32 m). The total stopping distance is therefore a mix: it is neither purely linear nor purely quadratic, but at high speed the quadratic braking term dominates. The student's claim is wrong; the correct statement is that braking distance scales with the square of speed.