Distance 1 sec makes 60-100..?
13 November 2006, 04:48 PM
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Distance 1 sec makes 60-100..?
One for the mathematicians:
What sort of distance would there be between two cars 60-100mph if one was one second quicker (ie 60-100mph 7sec v's 6sec)..?
I know 1 second at 100mph = aprox 150 feet, but this isn't as simple as that..
What sort of distance would there be between two cars 60-100mph if one was one second quicker (ie 60-100mph 7sec v's 6sec)..?
I know 1 second at 100mph = aprox 150 feet, but this isn't as simple as that..
13 November 2006, 04:57 PM
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speed = distance over time so revert it
average speed (ish!) is 100 + 60 / 2 = 80
80mph = distance / 1 second
80 / 60 (minutes in an hour) = 1.333333 miles per minutes
1.333333 / 60 (seconds in a minute) = 0.0222222 miles per second
0.0222222 miles I think? 35.76 metres... (about 117.32 feet)
average speed (ish!) is 100 + 60 / 2 = 80
80mph = distance / 1 second
80 / 60 (minutes in an hour) = 1.333333 miles per minutes
1.333333 / 60 (seconds in a minute) = 0.0222222 miles per second
0.0222222 miles I think? 35.76 metres... (about 117.32 feet)
Last edited by Tilly; 13 November 2006 at 05:00 PM.
13 November 2006, 05:22 PM
#3
Same answer by another route, using school physics from rather a long time ago 
v=u+at where v=final velocity, u=initial velocity, a=acceleration and t=time.
Using this formula we can work out accelaration for both cars.
v^2=U^2 + 2as, so we can use this to work out distance travelled and subtract the two to get the same answer (35.76m).
Tilly's way is easier tho!!
v=u+at where v=final velocity, u=initial velocity, a=acceleration and t=time.
Using this formula we can work out accelaration for both cars.
v^2=U^2 + 2as, so we can use this to work out distance travelled and subtract the two to get the same answer (35.76m).
Tilly's way is easier tho!!
13 November 2006, 10:04 PM
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Assuming both cars are doing 60mph at start and have linear acceleration (not the case in real life) then after 6 secs:
Faster car is doing 100mph and has traveled 215m (to nearest m)
Slower car is doing 94.29mph and has traveled 207m (to nearest m)
So distance between is about 8m.
Faster car is doing 100mph and has traveled 215m (to nearest m)
Slower car is doing 94.29mph and has traveled 207m (to nearest m)
So distance between is about 8m.
13 November 2006, 11:21 PM
#6
Darn - you're right. And I got an A in A-level Physics and a B in maths 
I shall go to the bottom of the class immediately.
I calculated the distance car A had travelled at 6 seconds, and the distance car B had travelled at 7 seconds, which is not a lot of use. That's why I failed my S-levels
Having re-calculated it, the difference is 7.66m assuming, as you say, the acceleration is linear.
Not quite 2 car lengths.
I shall go to the bottom of the class immediately. I calculated the distance car A had travelled at 6 seconds, and the distance car B had travelled at 7 seconds, which is not a lot of use. That's why I failed my S-levels
Having re-calculated it, the difference is 7.66m assuming, as you say, the acceleration is linear.
Not quite 2 car lengths.
13 November 2006, 11:22 PM
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1.) Average Speed of the fast car over the 6 seconds is 80mph or 60+((100-60) / 2) so distance of fast car in 6 seconds = 214.58 meters.
2.) If everything is Linear, then the speed of the slow car at 6 seconds is 94.28571 or ((100 / 7) * 6) and therefore the average speed over 6 seconds is 77.142855 or 60+((94.28571-60)/2) so distance of slow car in 6 seconds = 206.92
3.) 214.58 - 206.92 = 7.66 meters to be precise.
This is the right answer, the others are wrong. (appart from the 8m version which I agree with (although mine is more accurate. lol
)
Just remember that this is linear accelleration and as in normal life there would be slight curves, then the answer would most probably be closer to 7 meters.
2.) If everything is Linear, then the speed of the slow car at 6 seconds is 94.28571 or ((100 / 7) * 6) and therefore the average speed over 6 seconds is 77.142855 or 60+((94.28571-60)/2) so distance of slow car in 6 seconds = 206.92
3.) 214.58 - 206.92 = 7.66 meters to be precise.
This is the right answer, the others are wrong. (appart from the 8m version which I agree with (although mine is more accurate. lol
)Just remember that this is linear accelleration and as in normal life there would be slight curves, then the answer would most probably be closer to 7 meters.
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13 November 2006, 11:23 PM
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Originally Posted by tim hardisty
Darn - you're right. And I got an A in A-level Physics and a B in maths I shall go to the bottom of the class immediately.
I calculated the distance car A had travelled at 6 seconds, and the distance car B had travelled at 7 seconds, which is not a lot of use. That's why I failed my S-levels
Having re-calculated it, the difference is 7.66m assuming, as you say, the acceleration is linear.
Not quite 2 car lengths.
I shall go to the bottom of the class immediately. I calculated the distance car A had travelled at 6 seconds, and the distance car B had travelled at 7 seconds, which is not a lot of use. That's why I failed my S-levels
Having re-calculated it, the difference is 7.66m assuming, as you say, the acceleration is linear.
Not quite 2 car lengths.
13 November 2006, 11:30 PM
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Originally Posted by tim hardisty
And, what's worse, we're both sad gits to be working it out at 11.30 at night
13 November 2006, 11:35 PM
#12
Originally Posted by scooter(WRX03 PPP)
I wasn't going to do it, but it started to bug me as I knew it wasn't 35meters. Done enough sprints to that speed to know.
14 November 2006, 02:44 PM
#13
You're all for slightly mistaken here.
Completely missed out that if either of the cars are a clio cup, or have neons (worse if both) that the laws of physics no longer apply.
Dave
Completely missed out that if either of the cars are a clio cup, or have neons (worse if both) that the laws of physics no longer apply.
Dave
14 November 2006, 02:50 PM
#14
Originally Posted by RB5_245
You're all for slightly mistaken here.
Completely missed out that if either of the cars are a clio cup, or have neons (worse if both) that the laws of physics no longer apply.
Dave
Completely missed out that if either of the cars are a clio cup, or have neons (worse if both) that the laws of physics no longer apply.
Dave
Darn - you're right. I didn't even think to ask if either of the vehicles were fitted with a vent-to-air dump valve. How stupid can I get!
14 November 2006, 03:07 PM
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Dont really want to be picky here but the post does not say in which direction the cars were travelling in or if they started at the same time, but if we assume they are travelling in the same direction(parallel) and they started at the same time it would be:
A=v+/- t-vector B(2.33x80mph)[lat,<e=mc"> ]¬ (sxvrr#srs;2)
so Yeah about two car lengths.
A=v+/- t-vector B(2.33x80mph)[lat,<e=mc"> ]¬ (sxvrr#srs;2)
so Yeah about two car lengths.
14 November 2006, 03:13 PM
#16
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FFS you've lost me now..!
How about this one;
If two trains leave.....?
On a serious note, would this distance be double for say a two second difference and so on..?
How about this one;
If two trains leave.....?
On a serious note, would this distance be double for say a two second difference and so on..?
14 November 2006, 03:28 PM
#17
Originally Posted by jasonius
On a serious note, would this distance be double for say a two second difference and so on..?
1s difference: 7.6m
2s difference: 13.4m
3s difference: 17.9m
4s difference: 21.5m
I might regret asking this, but why do you want to know?
14 November 2006, 03:38 PM
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A question for all you maths/physics wizards here.
If a train travelling at 100mph and weighing 300 tonnes hits a fly on the exact opposite direction the fly squashes on the train window and continues in the direction of the train, so for however short a period the fly has changed direction eg. for east to west into west to east therefore at some point it must have been stationary to reverse direction? As it was in contact with the train at this point the train must have been stationary?
I dont think I have explained this well but do you get the gist of the question?
If a train travelling at 100mph and weighing 300 tonnes hits a fly on the exact opposite direction the fly squashes on the train window and continues in the direction of the train, so for however short a period the fly has changed direction eg. for east to west into west to east therefore at some point it must have been stationary to reverse direction? As it was in contact with the train at this point the train must have been stationary?
I dont think I have explained this well but do you get the gist of the question?
14 November 2006, 05:26 PM
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Originally Posted by WHEELSHOP0_0
A question for all you maths/physics wizards here.
If a train travelling at 100mph and weighing 300 tonnes hits a fly on the exact opposite direction the fly squashes on the train window and continues in the direction of the train, so for however short a period the fly has changed direction eg. for east to west into west to east therefore at some point it must have been stationary to reverse direction? As it was in contact with the train at this point the train must have been stationary?
I dont think I have explained this well but do you get the gist of the question?
If a train travelling at 100mph and weighing 300 tonnes hits a fly on the exact opposite direction the fly squashes on the train window and continues in the direction of the train, so for however short a period the fly has changed direction eg. for east to west into west to east therefore at some point it must have been stationary to reverse direction? As it was in contact with the train at this point the train must have been stationary?
I dont think I have explained this well but do you get the gist of the question?
It would be inappropriate to model the fly as a rigid body under these circumstances, you'd probably need to use something like CFD (computational fluid dynamics) to work out what happens. Sounds like overkill to me, but what would I know? :-D
15 November 2006, 10:50 PM
#21
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Originally Posted by tim hardisty
No - it's not directly proportional.
1s difference: 7.6m
2s difference: 13.4m
3s difference: 17.9m
4s difference: 21.5m
I might regret asking this, but why do you want to know?
1s difference: 7.6m
2s difference: 13.4m
3s difference: 17.9m
4s difference: 21.5m
I might regret asking this, but why do you want to know?
15 November 2006, 11:10 PM
#22
Fair enough! More useless info then...
If your Scooby does 0-60 in 5 seconds, it will have travelled about 67m in this time. If you indulge in a traffic-light grand prix - on a private road of course - then this is the lead you'll have on cars with lesser 0-60 times than you:
0-60=6s; lead=11m
0-60=7s; lead=19m
0-60=8s; lead=25m
0-60 greater than 8s...why did you bother in the first place
If your Scooby does 0-60 in 5 seconds, it will have travelled about 67m in this time. If you indulge in a traffic-light grand prix - on a private road of course
- then this is the lead you'll have on cars with lesser 0-60 times than you:0-60=6s; lead=11m
0-60=7s; lead=19m
0-60=8s; lead=25m
0-60 greater than 8s...why did you bother in the first place
16 November 2006, 01:54 PM
#25
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Very good..!
It's just that in real world conditions people are often surprised with the difference between cars that, on paper anyhow, are in different performance leagues.
You hear it all the time on here about a hot hatch that 'stayed in touch' with a 300bhp scoob and conversely a 300bhp scoob that stayed with a 'supercar'.
Given the figures you so kindly supplied, showing the relatively small distances involved, it's no wonder these things happen. Especially when you factor in driver response/hesitation/ability..
As I have said in other threads, it takes a car with a big performance advantage to 'trounce/leave for dead' another car.
Interesting IMHO..
It's just that in real world conditions people are often surprised with the difference between cars that, on paper anyhow, are in different performance leagues.
You hear it all the time on here about a hot hatch that 'stayed in touch' with a 300bhp scoob and conversely a 300bhp scoob that stayed with a 'supercar'.
Given the figures you so kindly supplied, showing the relatively small distances involved, it's no wonder these things happen. Especially when you factor in driver response/hesitation/ability..
As I have said in other threads, it takes a car with a big performance advantage to 'trounce/leave for dead' another car.
Interesting IMHO..
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