Speed, Force and Energy in a Road Accident
A police team investigates a collision on a straight road. A car of mass 1,200 kg was travelling at 54 km/h when the driver braked. Skid marks show the car stopped over 22.5 m. The road was dry, and the team assumes constant braking force. A second car of mass 1,500 kg was moving at 36 km/h in the same lane. Investigators want to know the braking force, the stopping distance if the speed had been higher, and how the two cars' kinetic energies compare. Describe the masses, speeds, skid distance and the assumption of constant force.
1. Convert the speeds to SI units and calculate the deceleration and braking force for the first car.
Read the worked answer
Convert speeds.
- 54 km/h = 54 × (1000/3600) = 15 m/s
- 36 km/h = 10 m/s
Deceleration. Using v² = u² + 2as with v = 0, u = 15 m/s, s = 22.5 m:
0 = 15² + 2a(22.5) → 0 = 225 + 45a → a = −225 ÷ 45 = −5 m/s².
So deceleration is 5 m/s².
Braking force. F = ma = 1,200 × 5 = 6,000 N (opposite to motion).
Judgement. The constant-force assumption lets us use the simple equations; real braking varies, so 6,000 N is an average. The calculation is still the correct exam approach.
2. If the same braking force acted on the second car, what would its stopping distance be from 36 km/h?
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Deceleration of the second car. F = ma → a = F ÷ m = 6,000 ÷ 1,500 = 4 m/s².
Stopping distance. v² = u² + 2as with u = 10 m/s, v = 0:
0 = 100 + 2(−4)s → 100 = 8s → s = 12.5 m.
Check by energy. KE = ½mv² = ½ × 1,500 × 100 = 75,000 J. Work done = F × s = 6,000 × 12.5 = 75,000 J. ✔
Judgement. The heavier car stops in a shorter distance here only because the same force gives it lower deceleration but it also started slower. The key insight is that stopping distance depends on speed squared, so the faster car needs far more road even though it is lighter.
3. Compare the kinetic energies of the two cars at their stated speeds and explain the investigative significance.
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Car 1. KE = ½ × 1,200 × 15² = ½ × 1,200 × 225 = 135,000 J.
Car 2. KE = ½ × 1,500 × 10² = ½ × 1,500 × 100 = 75,000 J.
Ratio. 135,000 ÷ 75,000 = 1.8. Car 1 carries 1.8 times the kinetic energy of Car 2 despite being lighter.
Significance. The lighter car is more dangerous because energy rises with the square of speed. A 50% increase in speed (10 → 15 m/s) raises energy by 2.25× for the same mass; the mass difference (1,200 vs 1,500) only partly offsets it.
Judgement for the report. Speed, not mass, dominates crash severity. This supports the conclusion that the first car's higher speed was the main factor in the longer skid and greater collision energy.