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THE PHYSICS OF CAR SAFETY

THE PHYSICS OF CAR SAFETY. Road traffic accidents kill more than one million people a year (one person every 30 seconds) , injuring another thirty-eight million ( about one per heart beat ). Driving is the number one cause of death and injury for people aged 15 to 44.

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THE PHYSICS OF CAR SAFETY

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  1. THE PHYSICS OF CAR SAFETY

  2. Road traffic accidents kill more than one million people a year(one person every 30 seconds), injuring another thirty-eight million (about one per heart beat). Driving is the number one cause of death and injury for people aged 15 to 44.

  3. Driving is the number one cause of death and injury for people aged 15 to 44. .

  4. Key Equations: Motion s = d / t Speed = distance / time a = (v-u) / t Acceleration = change in speed / time Energy and Work Work done = Energy transferred W = F x d Work done = Force x distance KE = ½ m v2 Kinetic Energy Force F = m x a Force = mass x acceleration

  5. Observe what happens when the drive force changes to zero, and the braking forces are unequal… Both cars are moving at constant velocity. No acceleration. No net force.

  6. Observe what happens when the drive force changes to zero, and the braking forces are unequal… What are the mathematical relationships between force, stopping distance and stopping time?

  7. Work done = energy transferred = Force x distance If the force increases, what happens to the distance? = d F W x More force means Less distance.  11

  8. If F is larger, what can you say about the acceleration? = m x F a a = v – u t If the acceleration is greater, what does it say about the time? More force means Less time.  12

  9. How do these ideas apply in crashes?

  10. New car with crumple zone. Old car without.

  11. Compare the impacts in slow motion… 2d What is the effect of increasing the stopping distance? Which equation provides the answer? [see slide 7] d

  12. Now consider the duration of the impact… What is the effect of decreasing the stopping time? Which equation provides the answer? [see slide 8]

  13. Work done = energy transferred = Force x distance = d F W x ½ mv2 = d F x d v2  assumption… Braking force is constant.

  14. The passenger’s head will decelerate rapidly if it hits the dashboard.

  15. With an airbag, the time to decelerate will increase as will the distance travelledduring deceleration

  16. How do seatbelts protect the passenger?

  17. A car travels a distance of 1000m in 40s (constant speed). What is the speed of the car? s = d/t = 1000/40 = 25 m/s What is the acceleration if the car stops in a further 5s? a = (v-u)/t = (0 -25)/5 = - 5 m/s2 If the car has a mass of 1000kg, how much kinetic energy did it have during the 40s? KE = ½ mv2 = ½ x 1000 x 252 = 312500 J = 312.5 kJ How much work is done by the brakes in stopping the car? Work Done = Energy Transferred = 312.5 kJ GCSE What is the average retarding force acting on the car as it stops? Force = mass x acceleration = 1000 x 5 = 5000 N

  18. In a crash, passenger ‘A’ hits an airbag and decelerates to a stop in 0.1s. The initial speed of ‘A’ is 15 m/s. The mass of ‘A’ is 60kg. What is the acceleration of ‘A’? a = (v – u)/t = (0 – 15)/0.1 = - 150 m/s2 What is the retarding force on ‘A’? F = ma = 60 x 150 = 9 kN If the area of contact between face / torso and airbag is 0.3m2, what is the pressure on ‘A’? Pressure = Force / Area = 9000 / 0.3 = 30000 Pa How far into the airbag does passenger ‘A’ travel? WD = E = ½ mv2 = ½ x 60 x 225 = 6750J = Fd d = 6750 / 9000 = 0.75m A Level Passenger ‘B’, also 60kg, has a similar crash (15m/s) but with no air bag. The deceleration occurs over a distance of just 2cm (0.02m). The area of contact (now between forehead and dashboard) is 25cm2 (0.0025m2). What is the pressure on ‘B’? P = F/A = W/dA = 6750/(0.02x0.0025) = 135000000Pa !

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