The jogger's dilemma - A fun little problem.

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The jogger's dilemma - A fun little problem.

Postby fryman » Thu Oct 04, 2007 8:19 pm UTC

Adam leaves his house, jogs on a flat, level road, up a hill, then back the same way. When he comes back, he has found his jog lasted exactly one hour. Adam can jog at 8 mph on flat surfaces, 6 mph uphill, and 12 mph downhill. How far did Adam jog?

Hint: It may seem like not enough information has been given, but it can still be solved.
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Re: The jogger's dilemma - A fun little problem.

Postby quintopia » Thu Oct 04, 2007 8:26 pm UTC

Spoiler:
8 mi

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Re: The jogger's dilemma - A fun little problem.

Postby SimonM » Thu Oct 04, 2007 8:30 pm UTC

Spoiler:
8 miles? not concentrating too well at the moment
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Re: The jogger's dilemma - A fun little problem.

Postby Hamorad » Thu Oct 04, 2007 8:35 pm UTC

Spoiler:
8 miles. He spends the same amount of time going uphill as downhill because he takes the same route back.

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Re: The jogger's dilemma - A fun little problem.

Postby skeptical scientist » Thu Oct 04, 2007 8:57 pm UTC

Solution 1: cheat
Spoiler:
Assume the problem is solvable. Then it must not matter how much of the run is flat, so we can assume the whole run is flat, and the problem is trivial.


Solution 2: do it properly
Spoiler:
Show the problem is solvable. By inverting the speeds, we find that Adam spends 10 minutes per uphill mile, 5 minutes per downhill mile, and 7.5 minutes per flat mile. Therefore, one mile of hill takes a total of 15 minutes to go up and down during the two halves of the run, and a mile of flat takes 15 minutes to go both ways during the two halves of the run, so any mile of run length accounts for two miles traveled and 15 minutes of running. At an hour of running, the distance of the run must be four miles each way, for a total of 8 miles.
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Re: The jogger's dilemma - A fun little problem.

Postby Govalant » Thu Oct 04, 2007 8:59 pm UTC

Spoiler:
He spent half the time going down (up 6mph, down 12mph) than going up. And he did it all in one hour.

So-> he went down in 20minutes, and 12mph * (1/3) = 4 miles.

if he did 4 miles down, he also did 4 down.

so 8.
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Re: The jogger's dilemma - A fun little problem.

Postby McHell » Thu Oct 04, 2007 10:37 pm UTC

The question is ambiguous: he got N miles far, but he ran 2N miles because he returned --- that's the strict meaning of `far'. But `how far along the trajectory' is the whole run... all posters before went for the 2N version, I stick to N.

Spoiler:
Extreme Case 1:
All flat with a 0m long hill ---> 4 miles far in half an hour, plus home stretch in an hour.

Extreme Case 2:
A 0m flat followed by a hill ---> 4 miles far in 40min (@6mph), plus back home in 20min (@12mph).

Conclusion:
Any linear combination (a, 1-a) of the above is a*4miles + (1-a)*4miles = 4miles far.

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Re: The jogger's dilemma - A fun little problem.

Postby ATCG » Thu Oct 04, 2007 11:59 pm UTC

A fellow Car Talk fan, perhaps?

Let Tom and Ray explain it all (a two-hour jog instead of a one-hour jog accounts for the factor of two in the answer):

The Puzzler for July 2, 2007

(Spoiler) Answer (Spoiler)
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Re: The jogger's dilemma - A fun little problem.

Postby Yakk » Fri Oct 05, 2007 12:07 am UTC

Spoiler:
0 miles -- he ended up at the same place he started.

Neglecting, of course, orbital motion, gravity, etc. We should pay attention to this.

Using a constantly replaced inertial local frame of reference, we find that he is averaging about 9.8 m/s^2 of acceleration strait up over the 1 hour.
d = 1/2 a t^2 = 4.9 * 3600^2 = 6.3504 * 10^7 m = roughly 63504 km
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Re: The jogger's dilemma - A fun little problem.

Postby Token » Fri Oct 05, 2007 12:15 am UTC

McHell wrote:The question is ambiguous: he got N miles far, but he ran 2N miles because he returned --- that's the strict meaning of `far'. But `how far along the trajectory' is the whole run...

I would contest that. When talking about journeys with no important destination - the obvious and relevant example being running or cycling for exercise - how far will refer to the total distance traveled. But that's off-topic.
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Re: The jogger's dilemma - A fun little problem.

Postby t0ksik » Fri Oct 05, 2007 12:32 am UTC

Spoiler:
Let Tf = total time traveled on a flat road and let Tu time traveled uphill
where h is hours and m is miles

Since Adam was jogging for one hour then we have 1h = Tf + Tu + (Tu)/2
=> Tf = 1 - Tu - (Tu)/2
=> Distance = 8m/h*(1h - Tuh - (Tu)/2)h + 6(Tuh) + 12((Tu)/2)
= (8m/h)*(1h) - (8m/h)*Tuh - (4*8m/h)*(Tuh)/2) + 6m/h(Tuh) + 6*12m/h((Tuh)/2)
= 8m - 8m*Tu - 4m*Tu + 6m*Tu + 6m*Tu
= 8m


Sorry it is so jacked up, hard to denote math in plaintext

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Re: The jogger's dilemma - A fun little problem.

Postby Xavier » Fri Oct 05, 2007 4:35 am UTC

Spoiler:
Let the distance along the flat surface be d1 - let the distance up the hill be d2.

time = distance/speed

(d1/8) + (d2/6) + (d2/12) + (d1/8) = 1

Add the pairs of fractions.

(d1/4) + (d2/4) = 1

d1 + d2 = 4

Therefore the distance travelled on the journey there is 4 miles, and the total distance travelled is 8 miles.

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Re: The jogger's dilemma - A fun little problem.

Postby fryman » Fri Oct 05, 2007 4:56 am UTC

ATCG wrote:A fellow Car Talk fan, perhaps?

Let Tom and Ray explain it all (a two-hour jog instead of a one-hour jog accounts for the factor of two in the answer):

The Puzzler for July 2, 2007

(Spoiler) Answer (Spoiler)


Nope, I don't watch Car Talk. This problem was posed to my Calc I class as a bonus question, and the TA just laughed at us. I thought it was fun to do, so I shared.
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Re: The jogger's dilemma - A fun little problem.

Postby Pesto » Fri Oct 05, 2007 2:12 pm UTC

I don't watch Car Talk either, but I do listen to it on the radio all the time.

*snerk*

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Re: The jogger's dilemma - A fun little problem.

Postby fryman » Fri Oct 05, 2007 6:00 pm UTC

Pesto wrote:I don't watch Car Talk either, but I do listen to it on the radio all the time.

*snerk*


Neither do I watch the radio while not listening to Car Talk.
[/offtopic]

Here's the answer I got for this...
Spoiler:
Let's not assume the hill and flat are equal in length. We can use distance/rate=time.
For x=flat distance and y=hill distance, we get x/8+y/6+y/12+x/8=1 hour.
Simplify, 2x/8+3y/12=1, and simplify even further to get x/4+y/4=1. By multiplying both sides by 4, we see that any distances will do, as long as x+y=4. Since the total distance traveled is defined as 2x+2y, it is obvious that the total distance will always be equal to 8.
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Re: The jogger's dilemma - A fun little problem.

Postby Lord Bob » Sat Oct 06, 2007 3:57 am UTC

Yakk wrote:
Spoiler:
0 miles -- he ended up at the same place he started.

Neglecting, of course, orbital motion, gravity, etc. We should pay attention to this.

Using a constantly replaced inertial local frame of reference, we find that he is averaging about 9.8 m/s^2 of acceleration strait up over the 1 hour.
d = 1/2 a t^2 = 4.9 * 3600^2 = 6.3504 * 10^7 m = roughly 63504 km


That's the perfect answer XD
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Re: The jogger's dilemma - A fun little problem.

Postby Pathway » Sat Oct 06, 2007 4:31 am UTC

fryman wrote:Adam leaves his house, jogs on a flat, level road, up a hill, then back the same way. When he comes back, he has found his jog lasted exactly one hour. Adam can jog at 8 mph on flat surfaces, 6 mph uphill, and 12 mph downhill. How far did Adam jog?

Hint: It may seem like not enough information has been given, but it can still be solved.


The real question is: who runs 5:00/mile on a downhill, 7:30/mile on the flats, and only 10:00/mile uphill? Well, Adam, obviously, but who is Adam and why is he so irrational? You'd have to have incredibly steep hills for that to work.
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Re: The jogger's dilemma - A fun little problem.

Postby no-genius » Sat Oct 06, 2007 1:54 pm UTC

Spoiler:
8 miles. His average speed is 8 miles per hour, and the time is one hour.
distance = rate times time
distance = 8 * 1 = 8 miles


the real question is why you didn't use SI
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Re: The jogger's dilemma - A fun little problem.

Postby fryman » Sat Oct 06, 2007 2:17 pm UTC

no-genius wrote:*snip*
the real question is why you didn't use SI

Probably because I didn't feel like converting from mph to km/h, along with the fact that the numbers would not have been so pretty and whole.
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Re: The jogger's dilemma - A fun little problem.

Postby McHell » Sat Oct 06, 2007 2:33 pm UTC

Pathway wrote:The real question is: who runs 5:00/mile on a downhill, 7:30/mile on the flats, and only 10:00/mile uphill?


That was indeed also my thought but I accidentally deleted it in editing. It doesn't make sense to average the same speed over a hilly circle than over flats. And having steeper hills makes it worse.

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Re: The jogger's dilemma - A fun little problem.

Postby Tchebu » Sat Oct 06, 2007 9:01 pm UTC

Isn't this... grade school stuff?

Spoiler:
8x+6y = 8x+12z (where x - time spent on flat (just one way), y - time spent on uphill, z - time spent on downhill)
2x+y+z = 1
y = 2z
2x+3z = 1
16x+6y+12z = 16x+24z = 8(2x+3z) = 8
Answer: 8 miles
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Re: The jogger's dilemma - A fun little problem.

Postby no-genius » Sat Oct 06, 2007 9:07 pm UTC

fryman wrote:
no-genius wrote:*snip*
the real question is why you didn't use SI

Probably because I didn't feel like converting from mph to km/h, along with the fact that the numbers would not have been so pretty and whole.

But you could have just substituted 'kmh-1 for mh-1 - without converting. It's only a puzzle!

edit: Tchebu - its easier to use average speed (the way I did it) - your method is needlessly complicated.
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Re: The jogger's dilemma - A fun little problem.

Postby Tchebu » Sat Oct 06, 2007 9:36 pm UTC

If you write out the calculation for the average speed, you'll use up pretty much the same amount of space, and do what will essencially be the same exact steps as I did. But yes I guess when you put it as
Spoiler:
"well the average speed is 8, cuz it's obvious... therefore 8*1 = 8, answer: 8 miles"


then yes, it is simpler. Also it's easier to use the average speed when doing things mentally. But properly written out, we end up doing pretty much the same thing.
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Re: The jogger's dilemma - A fun little problem.

Postby fryman » Sun Oct 07, 2007 1:56 am UTC

no-genius wrote:But you could have just substituted 'kmh-1 for mh-1 - without converting. It's only a puzzle!

8km/h is roughly 5mph. Adam can jog faster than that, but for the purposes of the problem, does it really matter?
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Re: The jogger's dilemma - A fun little problem.

Postby genewitch » Sun Oct 07, 2007 4:26 am UTC

yah it pretty much makes sense that the 12 and 6 cancel out, making it irrelevant. good one, OP :-)
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Re: The jogger's dilemma - A fun little problem.

Postby no-genius » Sun Oct 07, 2007 12:50 pm UTC

fryman wrote:
no-genius wrote:But you could have just substituted 'kmh-1 for mh-1 - without converting. It's only a puzzle!

8km/h is roughly 5mph. Adam can jog faster than that, but for the purposes of the problem, does it really matter?

no... :cry:
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Re: The jogger's dilemma - A fun little problem.

Postby pkuky » Sun Oct 07, 2007 3:06 pm UTC

0 miles. he ended up where he started, and the qustion is how far did he go.
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Re: The jogger's dilemma - A fun little problem.

Postby SimonM » Sun Oct 07, 2007 3:15 pm UTC

pkuky wrote:0 miles. he ended up where he started, and the qustion is how far did he go.


The question is "how far did he jog"

and you need to post in the introduction thread
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Re: The jogger's dilemma - A fun little problem.

Postby no-genius » Sun Oct 07, 2007 7:16 pm UTC

pkuky wrote:0 miles. he ended up where he started, and the question is how far did he go.

You think you're being clever - but you're not. and go post in the introduction thread.
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Re: The jogger's dilemma - A fun little problem.

Postby Geekthras » Mon Oct 08, 2007 12:41 am UTC

That post reminds me of a physics test problem I got once:
Sarah needs to go to the video store to return a video, 6km away. It takes her 10 minutes to drive to the store, spends 3 minutes there, and takes 12 minutes to get back. What is her average velocity for the entire trip?
Wait. With a SPOON?!

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Re: The jogger's dilemma - A fun little problem.

Postby genewitch » Mon Oct 08, 2007 2:08 am UTC

Geekthras wrote:That post reminds me of a physics test problem I got once:
Sarah needs to go to the video store to return a video, 6km away. It takes her 10 minutes to drive to the store, spends 3 minutes there, and takes 12 minutes to get back. What is her average velocity for the entire trip?

is that a trick question - in all seriousness? why wouldn't the answer be 6km in 25 minutes in kmh?
ergo 3.6 kmh...
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Re: The jogger's dilemma - A fun little problem.

Postby Yakk » Mon Oct 08, 2007 2:24 am UTC

genewitch wrote:
Geekthras wrote:That post reminds me of a physics test problem I got once:
Sarah needs to go to the video store to return a video, 6km away. It takes her 10 minutes to drive to the store, spends 3 minutes there, and takes 12 minutes to get back. What is her average velocity for the entire trip?

is that a trick question - in all seriousness? why wouldn't the answer be 6km in 25 minutes in kmh?
ergo 3.6 kmh...


Yes, it is a trick question.
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Re: The jogger's dilemma - A fun little problem.

Postby Geekthras » Mon Oct 08, 2007 5:02 am UTC

genewitch wrote:
Geekthras wrote:That post reminds me of a physics test problem I got once:
Sarah needs to go to the video store to return a video, 6km away. It takes her 10 minutes to drive to the store, spends 3 minutes there, and takes 12 minutes to get back. What is her average velocity for the entire trip?

is that a trick question - in all seriousness? why wouldn't the answer be 6km in 25 minutes in kmh?
ergo 3.6 kmh...



*sigh*

Velocity =/= speed
Velocity=displacement/time
Speed=distance/time

So, since she ended up where she started, she went nowhere i.e 0/25 minutes.


Ignoring that, it would be 12 km/25 minutes because she went there and back.
Wait. With a SPOON?!

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Re: The jogger's dilemma - A fun little problem.

Postby MFHodge » Mon Oct 08, 2007 4:41 pm UTC

SimonM wrote:. . . and you need to post in the introduction thread
no-genius wrote:. . . and go post in the introduction thread.

While there is nothing wrong with giving a new member a gentle nudge towards the Intro Thread, it is completely unnecessary and inappropriate for non-moderators to demand that people do so. There is no need for any member to take an unfriendly and condescending tone with a new member.
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Re: The jogger's dilemma - A fun little problem.

Postby SimonM » Mon Oct 08, 2007 4:53 pm UTC

Sorry, I was just a bit annoyed with the content of their post
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Re: The jogger's dilemma - A fun little problem.

Postby Nimz » Mon Oct 08, 2007 7:39 pm UTC

Yakk wrote:
Spoiler:
0 miles -- he ended up at the same place he started.

Neglecting, of course, orbital motion, gravity, etc. We should pay attention to this.

Using a constantly replaced inertial local frame of reference, we find that he is averaging about 9.8 m/s^2 of acceleration strait up over the 1 hour.
d = 1/2 a t^2 = 4.9 * 3600^2 = 6.3504 * 10^7 m = roughly 63504 km
Spoiler:
If you assume the acceleration is EXACTLY 9.8 m s-2 and you take the initial velocity to be opposed to the direction of the acceleration and with a magnitude of about 17640 +/- 36.756 m s-1 (in the standard sense of +/-, not the error bar sense), you would still arrive at a displacement of 8 miles. ;) Of course, that's assuming I did the conversions in my head correctly. If the problem had been 4 orders of magnitude larger, relativistic effects would have to be taken into consideration for more than just the acceleration, as the CAE's would indicate faster-than-light speeds for small initial speeds. Of course, relativistic effects would appear well before that...
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