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)

j_sum1 - 11-3-2016 at 16:49

teleport

blogfast25 - 11-3-2016 at 16:56

Well, I can tell you one way

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]

annaandherdad - 11-3-2016 at 21:17

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.

IrC - 11-3-2016 at 21:24

Does the ghost of Johnny Cash enter into the equation? I could swear he used to sing about a boy named Sue.

Metacelsus - 11-3-2016 at 21:30

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]

blogfast25 - 12-3-2016 at 06:01

Quote: Originally posted by annaandherdad |

So my absurdity:

$$\frac1v=0$$

... was in fact the truth!

Tsjerk - 12-3-2016 at 06:27

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?

Tsjerk - 12-3-2016 at 06:41

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]

annaandherdad - 12-3-2016 at 07:32

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.

Metacelsus - 12-3-2016 at 07:40

I think, though, that if Sue himself actually traveled at the speed of light, it would take zero time in

[Edited on 3-12-2016 by Metacelsus]

blogfast25 - 12-3-2016 at 07:54

Quote: Originally posted by Metacelsus |

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.

aga - 12-3-2016 at 09:15

20uS at 300mph, cos the route was a big circle.

[Edited on 12-3-2016 by aga]

Tsjerk - 12-3-2016 at 09:43

uS? united Siblings? You're getting your capitals wrong again Aga...

blogfast25 - 12-3-2016 at 09:48

Quote: Originally posted by Tsjerk |

I think he means:

$$\mu s$$

Darkstar - 12-3-2016 at 09:55

Quote: Originally posted by blogfast25 |

We will give partial credit.

Tsjerk - 12-3-2016 at 10:18

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]

aga - 12-3-2016 at 10:50

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]

mayko - 12-3-2016 at 11:42

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!

IrC - 12-3-2016 at 11:46

Quote: Originally posted 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.

blogfast25 - 12-3-2016 at 12:21

Quote: Originally posted by IrC |

Huh? I'd say the meaning is that it can't be done, physically.

Thanks, Mayko!

aga - 12-3-2016 at 12:36

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]

aga - 12-3-2016 at 13:34

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]

crystal grower - 12-3-2016 at 14:01

Quote: Originally posted by aga |

Aga, how could he go 90mph for 1hour if he was only 30miles from point A?

aga - 12-3-2016 at 14:04

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]

mayko - 12-3-2016 at 16:34

Quote: Originally posted by aga |

Ok, I might be losing my thouch. I implied but never explicitly defined a property called the

The intended topology was not necessarily a straight line segment, but something that is "line-sement like" (topologists would say that the space is

There are actually lots of other topologies we could give him (for example, a grid-like network of roads, like in New York City). You give the example of a circular race track; a circle (or a thing homeomorphic to a circle) is a topologically different space than a line segment; for example, you can take a point out of a circle-like track while preserving a property called "path connectedness" (if we put up a roadblock on the circular track, Sue can still drive to any other point on it - not so with a line segment!). First, let's look at the most general description of the problem, then look at two alternate topologies.

For any particular part of the trip x, the speed $$S_x$$ is related to the distance traveled $$D_x$$ and the time taken $$T_x$$:

$$S_x = \frac{D_x}{T_x}$$

Thus, for the forward, return, and total trips, respectively:

$$S_f = \frac{D_f}{T_f}$$

$$S_r = \frac{D_r}{T_r}$$

$$S_t = \frac{D_t}{T_t}$$

Moreover, the total distance and time are the sums of the component distances and times:

$$D_t = D_f + D_r $$

$$T_t = T_f + T_r $$

Merging these identities and churning the algebra, I get the identity:

$$S_r = \frac{D_r}{\frac{D_r+D_f}{S_t}-\frac{D_f}{S_f}}$$

The intended interpretation fixes the forward and return distances as equal: $$D_r = D_f = 1mi$$ Combined with the constraint that $$S_t = 2S_f$$ the denominator implodes.

Let's suppose he has an ATV, though, and can offroad. This means the topology available is "flat-plane-like". In particular, there is no particular limit on how far back Sue can drive in order to complete the circuit. With $$D_r$$ no longer constrained, the above identity works out:

$$S_r = \frac{D_r S_t}{D_r-D_f}$$

Which has a meaningful solution so long as Sue takes a longer drive back than forward.

Now let's consider the closed-loop track. This constrains the return distance to some whole-number multiple of loops around the track : $$D_r=w D_f$$ (if he doesn't do this, he hasn't returned to the starting point, and hasn't "driven back" ).

Topologists often call w a winding number

Combing this constraint with the above identity, I get the relation:

$$S_r = \frac{w}{w-1}S_t$$

Note that the intended interpretation is more or less equivalent to w=1.

mayko - 12-3-2016 at 16:43

(PS - I'm sure you meant no ill will, but you should be aware that tr*nny is generally considered a pretty bad slur, and I would caution against using it casually, or at all.)

Often in economics, problems are by default written around women because it tends to shake the reader a bit; I can think of a couple where the riddle actually hinges upon unconcious sex prejudices. I sometimes use the boy named Sue for the same disorienting effect. I'm surprised IRC and I are the only Cash fans about- good to have something in common at last

aga - 13-3-2016 at 12:19

So the route is circular then ?

P.S. We all like Cash. Lots and lots of Cash every day is great.

Not so sure about the Johnny, whoever she is.

IrC - 13-3-2016 at 13:54

Quote: Originally posted by aga |

https://www.youtube.com/watch?v=-OVHnWyDmdw

https://www.youtube.com/watch?v=ZE9KWU-ntBk

https://www.youtube.com/watch?v=KHF9itPLUo4

A few years ago I downloaded the movie "I walk the line" which was free but now all I see are scams and pay links. The movie would tell you about him and understanding it would give you an idea of this country as it was for generations. Alas, no longer.

mayko - 13-3-2016 at 15:47

Quote: Originally posted by aga |

Like I said, the structure of space was implied but not specified. Under the implied structure, there are no finite solutions. A circle-like course will work, though your proposed solution of "90mph for 1 hour" is off. (90 mph for 1 hr gives 90 mi, for a total distance of 91 miles and a total time of 62 minutes, ie a trip average of 87.6 mph ) However, 90 mph for 2 minutes (ie w=3) should do the trick.

Here's another example of a puzzle where (possibly unstated) assumptions about the structure of space can determine if an answer even exists:

Quote: |

As it happens, on a sphere like planet Earth, this is not possible, for reasons related to the Four Color Map Theorem. However, it's totally feasible on Toroidia, the legendary doughnut shaped planet!

aga - 13-3-2016 at 23:48

Doh !

> drives for one mile at thirty miles per hour.

I wrote that down as 1 hour at 30mph.

There's a basic assumption gone wrong already : that i could read straight.