average speed math

[mostsmooth]

[ QUOTE ]
maybe im stupid, but wouldnt the correct answer be "its impossible" rather than "infinitely fast"?

[/ QUOTE ]
well?

[andyfox]

"Haven't you ever watched Rocky 5?"

I tried, but I found it impossible.

Even when running it infinitely fast.

[Josh W]

hypothetically, if you go the speed of light, time stands still. so by going around the track at the speed of light, it takes no time. as such, you've gone around the track twice in 1 minute, averaging 120 mph.

josh

[usmhot]

No.
Firstly, time flows normally for you regardless of what speed you're travelling at. Time in other frames of reference that are travelling at different speed relative to you appear to flow at a different rates.

If you travel at the speed of light (impossible, but never mind) relative to the track then it will take exactly 1/c seconds (for c - the speed of light measured in miles per second) to cover the distance and 1/c is a finite number.

Aside: other limitations also exist, such as the assumption that you can accelerate from 60 mph to c in no time is false.

[Nathan183]

[ QUOTE ]
maybe im stupid, but wouldnt the correct answer be "its impossible" rather than "infinitely fast"?

[/ QUOTE ]

It depends on who's giving you the answer. If you consult a mathematician, he will tell you you must drive infinitely fast.

If you ask an engineer, he will tell you that it is impossible.

Both answers are correct, one is theoretical, the other is practical.

[NutzyClutz]

It depends how the arcade owner set up the machine. If completing the lap 1 at 60 miles per hour gave you bonus time..

[robokop]

[ QUOTE ]
If you travel at the speed of light (impossible, but never mind) relative to the track then it will take exactly 1/c seconds (for c - the speed of light measured in miles per second) to cover the distance and 1/c is a finite number.

[/ QUOTE ]
Err, but wouldn't the length of the track contract for you once you go c, so that you travel the distance at faster than 1/c seconds?

[usmhot]

The length of the track would appear to contract, but then time in the same frame as the track would also appear to move more slowly.

Part of the confusion here is the fact that the frames of reference aren't properly established. In fact we've got three frames of reference - with three different clocks - the track, the motion of the first time around, and the theoretical motion of the second time around.

Anyway, the point is, time would have to be measured in only one of the frames of reference for the entire attempt - so assuming its measured by an observer standing beside the track then the second circuit of the track (assuming it was at the speed of light) would take 1/c seconds.

Anyway, we can only go so far with this question before it really begins to break down - for example, we're also ignoring the fundamental problems such as getting from a small velocity to the speed of light in zero seconds.

[Nytecaster]

<<1 mile oval
you average 60mph for the first of 2 laps
how fast must you go for the second lap to average 120mph for the two laps combined?
i apologize if this is too easy >>

180mph for the second lap. Simple algebra to find the answer.

60 + x = 120 * 2

[usmhot]