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Author Maths People
FruitBooTeR
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Registered: 18th Jan 07
Location: Wolverhampton Drives: S15
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6th May 08 at 11:09   View User's Profile U2U Member Reply With Quote

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.
John
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Registered: 30th Jun 03
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6th May 08 at 11:13   View User's Profile U2U Member Reply With Quote

Calculus
FruitBooTeR
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Registered: 18th Jan 07
Location: Wolverhampton Drives: S15
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6th May 08 at 11:18   View User's Profile U2U Member Reply With Quote

Being a C student at GCSE and having not really done maths since then (3 years) think thats beyond me
Planty02
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Registered: 5th Mar 05
Location: Burslem, Stoke-on-Trent
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6th May 08 at 11:19   View User's Profile U2U Member Reply With Quote

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
ed
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Registered: 10th Sep 03
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6th May 08 at 11:21   View User's Profile U2U Member Reply With Quote

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.
RichR
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Registered: 17th Oct 01
Location: Waterhouses, Staffordshire
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6th May 08 at 11:22   View Garage View User's Profile U2U Member Reply With Quote

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
John
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6th May 08 at 11:24   View User's Profile U2U Member Reply With Quote

http://www.stvincent.ac.uk/Resources/Physics/Speed/speed/formulae.html
ed
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Registered: 10th Sep 03
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6th May 08 at 11:27   View User's Profile U2U Member Reply With Quote

Mmmmm, velocity time graphs. Bring back many happy memorys of studying mechanical engineering
RichR
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6th May 08 at 11:35   View Garage View User's Profile U2U Member Reply With Quote

quote:
Originally posted by ed
Mmmmm, velocity time graphs. Bring back many happy memorys of studying mechanical engineering


and hydro & fluid dynamics
Fonz
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Registered: 12th May 06
Location: Newbury, Berks
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6th May 08 at 12:21   View Garage View User's Profile U2U Member Reply With Quote

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?



Skylined
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Registered: 27th Sep 05
Location: Sideways, Surrey
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6th May 08 at 12:37   View User's Profile U2U Member Reply With Quote

v^2=u^2+2as

using acceleration as worked out above
s = 195m
RichR
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Registered: 17th Oct 01
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6th May 08 at 12:43   View Garage View User's Profile U2U Member Reply With Quote

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?
Fonz
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Registered: 12th May 06
Location: Newbury, Berks
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6th May 08 at 13:14   View Garage View User's Profile U2U Member Reply With Quote

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!
Paul_J
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Registered: 6th Jun 02
Location: London
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6th May 08 at 13:16   View User's Profile U2U Member Reply With Quote

haha this reminds me of A-Level Mechanics Maths
Kurt
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Registered: 23rd Oct 05
Location: Hi
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6th May 08 at 13:17   View User's Profile U2U Member Reply With Quote

geek day?
BarnshaW
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Registered: 25th Oct 06
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6th May 08 at 13:18   View User's Profile U2U Member Reply With Quote

wtf with this thread lol. isnt there a button on a calculator which does it
loo_goblin
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Registered: 14th Jul 04
Location: Horsham, West Sussex
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6th May 08 at 15:09   View User's Profile U2U Member Reply With Quote

Argh its like being back at uni

acceleration is much easier when gravity is involved

 
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