THE PHYSICS OF CAR SAFETY

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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 Presentation

Transcript of THE PHYSICS OF CAR SAFETY

THE PHYSICS OF CAR SAFETYTHE 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.

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

.

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

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…

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?

=

If the force increases, whathappens to the distance?

FW

Work done = energy transferred = Force x distance

x dMore force means Less distance.

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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.

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How do these ideas apply in crashes?

New car with crumple zone.

Old car without.

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]

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]

= FW

Work done = energy transferred = Force x distance

x d

= F x d½ mv2

dv2 assumption…

Braking force is constant.

The passenger’s head will decelerate

rapidly if it hits the dashboard.

With an airbag, the time to

decelerate will increase as will the

distance travelled during

deceleration

How do seatbelts protect the

passenger?

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

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