FruitBooTeR
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Registered: 18th Jan 07
Location: Wolverhampton Drives: S15
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Ok people with intelligence help me come up with a calculation.
Is there a way to work out how far a car has travelled over a period of time from just the 0-60 time.
Ok say a car does 0-60 in 6.5 seconds,
How far would it of travelled in the 6.5 seconds, 3.25 seconds or even 1 second?
Can you work it out just from these details would there need to be more info?
Cheers
Oh and i know from school about the S=D/T but i dont think you can use that as its only doing 60mph at the very end of the time period were testing, its the speed,time and distance before that.
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John
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Registered: 30th Jun 03
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Calculus
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FruitBooTeR
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Registered: 18th Jan 07
Location: Wolverhampton Drives: S15
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Being a C student at GCSE and having not really done maths since then (3 years) think thats beyond me
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Planty02
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Registered: 5th Mar 05
Location: Burslem, Stoke-on-Trent
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Wouldnt have thought you could work it out from just those details.
theres no telling how fast the car was going at 3.25 secs, so no way of telling what distance it would cover
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ed
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Registered: 10th Sep 03
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You can work out a very good estimation as the acceleration is linier. The estimation and the real world distances would differ because in the real world the acceleration would fluctuate... I'll get my notes out and tell you how its done if I can find them.
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RichR
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Registered: 17th Oct 01
Location: Waterhouses, Staffordshire
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Assuming that the acceleration is linear [which it won't be] and that at 0 seconds it is 0mph and at 6.5 seconds it is exactly 60mph
the average speed over the duration would be 30mph.
30mph is 0.5miles per minute or 0.00833miles per second.
if the car travels at an average of 30mph for 6.5 seconds it would cover 0.0542miles or 87.23metres
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John
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Registered: 30th Jun 03
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http://www.stvincent.ac.uk/Resources/Physics/Speed/speed/formulae.html
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ed
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Registered: 10th Sep 03
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Mmmmm, velocity time graphs. Bring back many happy memorys of studying mechanical engineering
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RichR
Premium Member
Registered: 17th Oct 01
Location: Waterhouses, Staffordshire
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quote: Originally posted by ed
Mmmmm, velocity time graphs. Bring back many happy memorys of studying mechanical engineering
and hydro & fluid dynamics
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Fonz
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Registered: 12th May 06
Location: Newbury, Berks
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although correct to say no, as acceleration is not linear, an equation can be formed
a= acceleration
u = initial speed (in this case 0mph)
v = final speed (in this case 60mph)
t = time(in this case 6.5s)
s = displacement
using a = (v+u)/t we can find acceleration
a = (v+u)/t
= 60/6.5
= 9.23m/s^2
using s = ut + 1/2at^2
s = ut + 1/2at^2
= (0*6.5) + 0.5*9.23*t^2
= (0) + 4.61t^2
s = 4.61t^2
when t = 6, s = 166m
t = 3.5, s = 56m
t = 1, s = 4.61m
hope that helps?
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Skylined
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Registered: 27th Sep 05
Location: Sideways, Surrey
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v^2=u^2+2as
using acceleration as worked out above
s = 195m
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RichR
Premium Member
Registered: 17th Oct 01
Location: Waterhouses, Staffordshire
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Fonz; thats wrong - you need to convert 60mph to m/s to get the correct answer.
Youve done a = (v+u)/t
where
a = m/s^2
v = miles/hr
t = s
your calc actually gives a = miles/hour/second
you require consistency with your units.
ie.
a = m/s^2
v = m/s
t = s
Therefore:
60mph = 26.822m/s
a = (v+u)/t
= 26.822/6.5
= 2.91m/s^2
using s = ut + 1/2at^2
s = ut + 1/2at^2
= (0*6.5) + 0.5*2.91*t^2
= (0) + 1.455t^2
s = 1.4551t^2
when t = 6.5, s = 61.47m
t = 3.5, s = 17.7625m
t = 1, s = 1.455m
hope that helps?
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Fonz
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Registered: 12th May 06
Location: Newbury, Berks
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ah yes very true.....it was a quick calc off the top of my head. my initial thoughts were of given the final distance covers but waved it off!
the method is correct, just the figure wrong!
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Paul_J
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Registered: 6th Jun 02
Location: London
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haha this reminds me of A-Level Mechanics Maths
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Kurt
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Registered: 23rd Oct 05
Location: Hi
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geek day?
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BarnshaW
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Registered: 25th Oct 06
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wtf with this thread lol. isnt there a button on a calculator which does it
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loo_goblin
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Registered: 14th Jul 04
Location: Horsham, West Sussex
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Argh its like being back at uni
acceleration is much easier when gravity is involved
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