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Buster fin said..Ian K said..Crusoe said.
S - alway drives 10ks under the speed limit
They're the ones who paid attention in yr 12 physics.
( Vf^2 - Vi^2 = 2as )
Legion. If you please.
Sorry. I was assuming those who paid attention didn't need one and those that didn't wouldn't be interested anyway.
Vf is the final velocity. ie your collision velocity after you've jammed the brakes on when a car unexpectedly comes through a red light say 20 metres in front of you. Or a drunk or a toddler.
Vi is your initial velocity say 65 kph (18 m/sec) in a 70kph zone or 75 kph (20.8) in the same 70 kph zone.
a is the deceleration -0.7 g
~ -7 m/sec^2 might be typical of a modern car.
s is the distance of deceleration (let's try 20 metres)
In the first case Vf ^2 - 18^2 = 2 X -7 X20 ... Vf^2 = 324 - 280 = 44 Vf = 6.6 m/sec = 23.8 kph
In the case of the fast driver the collision speed is
Vf^2 - 20.833^2 = 2 X -7 X 20 ... Vf^2 = 434.0 - 280 = 12.41 m/sec = 44.68 kph
So the collision speed is doubled for only an extra 10 kph on board. This is once the brakes are applied so the distance covered in the reaction time is extra. This will make it even worse for the faster driver. He closes down more of the remaining gap so would have to have identified the red light runner sooner. 1.5 seconds is a typical driver reaction time. ie 27 metres for the driver doing 18 m/sec and 31 metres for the faster driver. Do the calculations again with 4 metre less stopping distance ie. 16 metres for the faster driver and you get
Vf^2 = 434.0 - 224 ...... Vf = 14.49 m/sec or 52.16 kph. More than double the collision speed of the person 5 kph under the limit.
These simple equations were the basis of the Victorian TAC campaign. But it was discontinued. I think they found that unless you go through the equations, and most people don't, you'd just say "Bull **** "