the physics of car safety
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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THE PHYSICS OF CAR SAFETYTHE PHYSICS OF CAR SAFETY
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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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Driving is the number one cause of death and injury for people aged 15 to 44.
.
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Key Equations:
Speed = distance / time s = d / t
Energy and Work
Work done = Energy transferred
Force = mass x acceleration F = m x a
W = F x dWork done = Force x distance
Force
Acceleration = change in speed / time a = (v-u) / t
Kinetic Energy KE = ½ m v2
Motion
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Both cars are moving at constant velocity.No acceleration.No net force.
Observe what happens when the drive force changes to zero, and the brakingforces are unequal…
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Observe what happens when the drive force changes to zero, and the brakingforces are unequal…
What are the mathematicalrelationships between force,stopping distance andstopping time?
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=
If the force increases, whathappens to the distance?
FW
Work done = energy transferred = Force x distance
x dMore force means Less distance.
11
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= v – u t
= m x
If the acceleration is greater, what does
it say aboutthe time?
If F is larger, whatcan you say aboutthe acceleration?
a
a
F
More force means Less time.
12
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How do these ideas apply in crashes?
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New car with crumple zone.
Old car without.
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d
2d
Compare the impacts in slow motion…Compare the impacts in slow motion…
What is the effect of increasing the stopping distance?
Which equation provides the answer? [see slide 7]
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Now consider the duration of the impact…Now consider the duration of the impact…
What is the effect of decreasing the stopping time?
Which equation provides the answer? [see slide 8]
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= FW
Work done = energy transferred = Force x distance
x d
= F x d½ mv2
dv2 assumption…
Braking force is constant.
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The passenger’s head will decelerate
rapidly if it hits the dashboard.
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With an airbag, the time to
decelerate will increase as will the
distance travelled during
deceleration
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How do seatbelts protect the
passenger?
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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
What is the average retarding force acting on the car as it stops?
Force = mass x acceleration = 1000 x 5 = 5000 N
GCSE
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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
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 !
A Level