Engineering
Stopping a car
A 1,200 kg car travelling at 30 m/s brakes steadily and stops in 60 m. What is the average braking force? What happens to the stopping distance if the car is going twice as fast?
- Hint 1Use v² = u² + 2as to find the deceleration.
- Hint 2Or use energy: the brakes' work equals the car's kinetic energy.
- 0 = 30² − 2a × 60, so a = 900 ÷ 120 = 7.5 m/s².
- Force = ma = 1,200 × 7.5 = 9,000 N.
- Check with energy: ½ × 1,200 × 30² = 540,000 J; 9,000 N × 60 m = 540,000 J.
- With the same braking force, distance scales with speed squared: twice the speed needs four times the distance, 240 m.
Where it lands: 9,000 N; at twice the speed, four times the distance.
The trap: Assuming stopping distance doubles when speed doubles.
What a tutor might ask next: Why do real stopping distances include a 'thinking distance', and how does that scale with speed?
Now do one out loud. In the interview you think aloud with a tutor. Try an Engineering problem with hints, follow-ups and Mia's feedback on how you think.