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mayko
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A puzzler: "Fun" with Averages
Sue gets into a car and drives for one mile at thirty miles per hour. Then, he drives back. How fast does Sue have to drive on the return trip to
average sixty miles per hour over the whole trip? (Ignore turnaround time.) Show your work!
(I have a symbolic solution and a conceptual solution)
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j_sum1
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teleport
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blogfast25
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Well, I can tell you one way not to do it, because I tried it!  
Time needed for outbound journey:
$$\frac{1}{30}$$
Time needed for return journey at unknown speed v:
$$\frac{1}{v}$$
Total time:
$$\Delta t=\frac{1}{30}+\frac{1}{v}$$
Average speed over outbound + return journey:
$$\frac{2}{\frac{1}{30}+\frac{1}{v}}=60$$
... which leads to an absurdity.
[Edited on 12-3-2016 by blogfast25]
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annaandherdad
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Infinity.
The problem only deals with ratios, so the answer doesn't depend on the distance travelled. So to simplify it, suppose he drives 30 miles at 30
miles/hr, which takes 1 hour. Now he wants to drive 30 miles back, for a total of 60 miles, to make an average speed of 60 mph. That means he has
to get back instantaneously.
Any other SF Bay chemists?
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IrC
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Does the ghost of Johnny Cash enter into the equation? I could swear he used to sing about a boy named Sue.
"Science is the belief in the ignorance of the experts" Richard Feynman
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Metacelsus
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I'm not entirely sure about this, but I think he would "only" have to travel at the speed of light, due to time dilation. Of course, this would be
equally impossible as traveling infinitely quickly.
Edit: You could also look at this as length contraction.
[Edited on 3-12-2016 by Metacelsus]
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blogfast25
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So my absurdity:
$$\frac1v=0$$
... was in fact the truth!
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Tsjerk
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Wouldn't travelling at the speed of light be the same as infinitely fast? As the speed of light is the absolute limit and therefore comparable with
infinitely fast?
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Tsjerk
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Wouldn't travelling at the speed of light be the same as infinitely fast? As the speed of light is the absolute limit and therefore comparable with
infinitely fast?
Edit: A bit like infinitely cold would be the same as 0 Kelvin, as that is also an absolute limit.
[Edited on 12-3-2016 by Tsjerk]
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annaandherdad
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No, if Sue ("his" name) shined a light beam back toward his starting point, instead of driving back in his car, it would still take some time for the
light beam to arrive (distance divided by the speed of light). The speed of light is just the speed of light---miles/second or whatever units you
want to use.
Any other SF Bay chemists?
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Metacelsus
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I think, though, that if Sue himself actually traveled at the speed of light, it would take zero time in his reference frame (it would be
different to a stationary observer). However, I'm a bit rusty on my special relativity, and so I could be wrong about this.
[Edited on 3-12-2016 by Metacelsus]
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blogfast25
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Quote: Originally posted by Metacelsus  | I think, though, that if Sue himself actually traveled at the speed of light, it would take zero time in his reference frame (it would be
different to a stationary observer). However, I'm a bit rusty on my special relativity, and so I could be wrong about this.
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No, I think you're right. Velocity time dilation means that 'proper' time for the fast moving observer is lower than for the stationary one.
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aga
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20uS at 300mph, cos the route was a big circle.
[Edited on 12-3-2016 by aga]
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Tsjerk
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uS? united Siblings? You're getting your capitals wrong again Aga...
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blogfast25
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I think he means:
$$\mu s$$
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Darkstar
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We will give partial credit.
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Tsjerk
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uS? united Siblings? You're getting your capitals wrong again Aga...
Edit: Sorry for all these double posts, it seems to be happening when I'm over a VPN connection. Does anyone know what could cause this?
[Edited on 12-3-2016 by Tsjerk]
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aga
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uS is underpant Siemens, defined as the root mean resistivity of my undercrackers over 1cm of cotton
Pint<sup>-1</sup>minute<sup>-1</sup> on a Saturday evening.
[Edited on 12-3-2016 by aga]
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mayko
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Nice team effort, folks! My symbolic solution was only a little more general than blogfast's, with the known quantities substituted at the end (my
physics teachers thrust this habit upon me.)
The more conceptual solution:
Driving 2 miles at an average speed of 60 mph requires 2 minutes. So does driving one mile at 30 mph; thus by the time the first leg of the journey is
complete, there is no time left to return on schedule!
I like this problem because it illustrates how partial uncertainties don't necessarily preclude knowledge about a global system: even with the speed
of the return trip completely unconstrained upon the positive real numbers, we can still assert strong bounds on the trip average.
I admit that I had not considered relativistic effects! 
al-khemie is not a terrorist organization
"Chemicals, chemicals... I need chemicals!" - George Hayduke
"Wubbalubba dub-dub!" - Rick Sanchez
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IrC
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| Quote: Originally posted by Metacelsus | I think, though, that if Sue himself actually traveled at the speed of light, it would take zero time in his reference frame (it would be
different to a stationary observer). However, I'm a bit rusty on my special relativity, and so I could be wrong about this.
[Edited on 3-12-2016 by Metacelsus] |
Sue was not the one solving the problem, we are supposed to. It is us who were asked not Sue. Relative to us time would pass even if Sue was frozen in
time. Since all allowed time had already passed during the first half of the trip, this entire exercise is totally without meaning.
"Science is the belief in the ignorance of the experts" Richard Feynman
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blogfast25
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| Quote: Originally posted by IrC | Since all allowed time had already passed during the first half of the trip, this entire exercise is totally without meaning.
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Huh? I'd say the meaning is that it can't be done, physically.
Thanks, Mayko!
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aga
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You learnt about underpant-Sieverts at least, so not all is lost.
Edit :
Eeek ! those are the glow-in-the-dark ones.
meant Siemens.
[Edited on 12-3-2016 by aga]
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aga
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Spending a few (tens of) minutes with a pen and paper, i get 90mph for 1 hour to work.
... provided that the route is not a straight line from A to B, but a circle where start and finish are both point A.
No wonder the turnaround time can be ignored - tranny Steve just accellerates.
The algebra works out, so you're saved mayko.
(just say this is what you meant all along and you'll get away with it)
Edit:
Can't post the workings-out as i did them on my underpants, which became unreadable when they passed the 400uS mark.
[Edited on 12-3-2016 by aga]
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crystal grower
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| Quote: Originally posted by aga | Spending a few (tens of) minutes with a pen and paper, i get 90mph for 1 hour to work.
... provided that the route is not a straight line from A to B, but a circle where start and finish are both point A.
No wonder the turnaround time can be ignored - tranny Steve just accellerates.
The algebra works out, so you're saved mayko.
(just say this is what you meant all along and you'll get away with it)
Edit:
Can't post the workings-out as i did them on my underpants, which became unreadable when they passed the 400uS mark.
[Edited on 12-3-2016 by aga] |
Aga, how could he go 90mph for 1hour if he was only 30miles from point A?
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aga
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He/She did not drive in a straight line for an hour then go back on the same straight line.
That's the only way it can work (that i can see).
In total, the route from Start back to Start is 120 miles.
30 miles first, then 90 miles after that.
A bit like driving round a circle, mountain, racetrack : something like that.
Simply cannot work if it was a straight line.
Edit:
When the maths do not add up, something is wrong.
Sometimes (often) it is with the assumptions.
Nowhere in the question did mayko say it was a straight line.
Specifically 'Ignore turnaround time' was said.
I think he just fucked up, but i'm really happy to have found a solution !
[Edited on 12-3-2016 by aga]
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