The jogger's dilemma - A fun little problem.

For the discussion of math. Duh.

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fryman
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The jogger's dilemma - A fun little problem.

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.

Spoiler:
8 mi

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

Spoiler:
8 miles? not concentrating too well at the moment
Last edited by SimonM on Thu Oct 04, 2007 9:10 pm UTC, edited 1 time in total.
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Re: The jogger's dilemma - A fun little problem.

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.

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.

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.

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.

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):

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

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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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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

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.

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

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

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

(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.

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

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.

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.

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.

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

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

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.

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.

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.

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.

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

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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The Mighty Thesaurus wrote:Why? It does nothing to address dance music's core problem: the fact that it sucks.

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

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.

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.

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.

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...
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The Mighty Thesaurus wrote:Why? It does nothing to address dance music's core problem: the fact that it sucks.

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

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.

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.

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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The Mighty Thesaurus wrote:Why? It does nothing to address dance music's core problem: the fact that it sucks.

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

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.

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.

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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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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

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.

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.

Sorry, I was just a bit annoyed with the content of their post
mosc wrote:How did you LEARN, exactly, to suck?

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

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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