Fastest-accelerating legal car-speed v. distances.

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Thu 26 Feb 2009 14:06
Fastest-accelerating legal car-speed v. distances. - FotheringtonThomas
It's a "Maxximus G-Force", according to an article in "The Daily Telegraph". 0-60 in just over 2 seconds. 0-100 in 4 1/2. 100-0 in just under 9 seconds.

What are the distances involved?

0-60 in 2 seconds = ? yards
0-100 in 4 1/2 = ?
100-0 in 9 = ?
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Thu 26 Feb 2009 14:54
Fastest-accelerating legal car-speed v. distances. - maz64
Assuming constant acceleration (which it won't be, but have to assume something):
0-60 in 2 seconds = ? yards


Same as 60mph for 1 second. 60mph = 1 mile per minute so 1760 / 60 = 30yds approx.
0-100 in 4 1/2 = ?


Yards per second = 50 * 1760 / (60 * 60) = 24.4
Yards in 4.5 seconds = 110yds
100-0 in 9 = ?


Double previous answer = 220yds
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Thu 26 Feb 2009 15:31
Fastest-accelerating legal car-speed v. distances. - Dwight Van Driver
Just blown the dust off my Accident Invetsigation notes and cranked the rusty brain:

Distance = (1.47 (a constant) x Time x Speed) + (16.1 (a constant) x F (Drag factor of Road taken as .7) x Speed squared.)

So 0.60 in 2 secs =221. 08 ft

0 to a 100 in 4.5 secs = 889.7ft.

100 to 0 would depend on whther free wheeling, gentlr or hard braking?

No boubt a Phyisist will be allow to help...........my brain hurts

dvd

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Thu 26 Feb 2009 15:53
Fastest-accelerating legal car-speed v. distances. - maz64
So 0.60 in 2 secs =221. 08 ft


Doesn't that indicate an average speed > 60mph?

60mph = 1760yds / minute = 30yds/s = 60yds in 2s = 180ft
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Thu 26 Feb 2009 16:03
Fastest-accelerating legal car-speed v. distances. - rtj70
But it's not doing 60mph until it gets to 60mph. It is accelerating. So until up to speed it's moving forward at anywhere between 1mph and 59mph for a while and therefore covering ground. For most of the 2s it will not have been doing 30yds/s.
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Thu 26 Feb 2009 16:10
Fastest-accelerating legal car-speed v. distances. - maz64
But it's not doing 60mph until it gets to 60mph.


Sorry - so which are you saying is more likely to be correct?
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Thu 26 Feb 2009 16:06
Fastest-accelerating legal car-speed v. distances. - craig-pd130

Don't the calculations above assume a linear rate of acceleration over the whole time though?

I believe that at sprint meetings, competitors get a 60ft time and speed as well as the quarter-mile ET and terminal speed. It's a measure of traction AND reaction time.

I've seen figures on an American Audi S5 which did the following:

0-30: 1.45 sec
0-60: 4.47 sec

the 60ft time was 1.8 sec

So the car has already hit around 35mph in just 60 feet (less than 4 car lengths) after starting!
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Thu 26 Feb 2009 16:14
Fastest-accelerating legal car-speed v. distances. - maz64
Don't the calculations above assume a linear rate of acceleration over the whole time though?


Mine do- as I said, assuming constant acceleration. If acceleration is greater to start with, distance will be bigger. If acceleration increases with speed, distance will be lower.

Think of a point (x,y) on a graph, and the area under a line between (0,0) and (x,y). X axis is time, Y axis is speed. Area under line is distance travelled. Changing shape of line corresponds to changing acceleration and area under graph (distance) without changing final time and speed values.
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Thu 26 Feb 2009 19:57
Fastest-accelerating legal car-speed v. distances. - Number_Cruncher
DVD's calculations are braking distance calculations, sensibly, assuming a co-efficient of friction of 0.7 between the tyre and the road, and an initial speed.

In broad agreement with Focus's sums;
% Assuming constant acceleration
% s=t*(v+u)/2
% u=0
% s= 0.5 * v * t
v_mph=60;
% convert to metres per second
v_ms=v_mph*1609.3/3600;
t=2;
s=0.5*v_ms*t


s =

26.8217
a=v_ms/t


a =

13.4108

So, 27 metres or so - The strange thing is that it requires a coefficient of friction between the tyre and the road of nearly 1.4, which means the tyres must be quite grippy, and there would be no chance of acheiving those figures on a typical wet, greasy UK road.
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