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Interesting Time Question

  1. Jun 2, 2007 #1
    1. The A mathematically inclined friend emails you the following instructions: "Meet me in the cafeteria the first time after 2:00 p.m. today that the hands of a clock point in the same direction." When should you meet your friend?


    2: : (hr : min : sec)

    I have attached my attempt at a solution

    << after 6 days with no response from the OP about the attachment, berkeman deleted it out of the Pending Attachments queue >>
     
    Last edited by a moderator: Jun 8, 2007
  2. jcsd
  3. Jun 4, 2007 #2

    berkeman

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    Staff: Mentor

    I'm not able to view your PDF attachment to try to approve it. Even though my Adobe 8 Reader is up to date, your file keeps trying to take me to (supposedly) the Adobe website for a required Flash Player update.

    Can you try again with something more vanilla as the attachment?
     
  4. Jun 4, 2007 #3

    Moonbear

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    Staff Emeritus
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    It seems you've used some method to encrypt the file that requires a special plug in. I'm not inclined to add it since I can't get enough information about what it will do and the system requirements sound very outdated, so might not even be compatible with current versions of Adobe Acrobat or Reader.

    As berkeman suggested, could you upload the attachment or its contents either in a different file format, or without the encryption? Otherwise, we won't be able to approve the attachment.

    Also, to double check, is this really a physics homework question?
     
  5. Jun 5, 2007 #4
    Since any rational number can be expressed as the ratio of two integers, why not just write the fraction that shows exactly how many seconds (or, if you prefer, minutes) after 2:00 that you calculated for the coincidence.

    [BTW - This is the kind of problem that might be fun for your non-physics oriented, puzzle enthusiast friends, since it really doesn't require much more than simple algebra and a some logical thought concerning simple rotational quantities and time. I'll try it on a couple of my non-physics-but-still-genius-type pals this afternoon.]
     
    Last edited: Jun 5, 2007
  6. Jun 5, 2007 #5

    andrevdh

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    This looks very much like the problem that the ancient Greeks discussed at great lengths: The race between Achilles and the tortoise. They had great difficulty in understanding how Achilles could ever pass the tortoise.
     
  7. Jun 5, 2007 #6
    I didn't run into any infinitesimal type scenario. Just a simple solving for elapsed time given the angular separation and angular velocities of the two clock hands. There's a precise elapsed time when both will occupy the same angular position relative to some chosen fixed reference (say, the vertical 12 position). So, its more like the old train A leaves blahville traveling west at 60 kph while train B leaves blah-blahtown, 30 km to the east, travelling in the same direction as A, at 100 kph - what time will B slam into the rear of A :)
     
  8. Jun 6, 2007 #7

    andrevdh

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    The problem with the attachment might be that its name starts with a number - try renaming it.

    Yes, with our present day knowledge it is quite easy to solve it.
     
  9. Jun 7, 2007 #8
    I think simple geometry is enough for this question:

    6x=(x/2)+60
    x=10,909 min
     
  10. Jun 7, 2007 #9
    Just for fun, I chose complete revolutions (vs radians or degrees) as the angular unit and minutes as the unit for time. I also chose the 12 o'clock vertical angular position as the zero reference. In that case, at exactly 2:00 pm, the minute hand is at initial position 0 rev and the hour hand is at 1/6 rev. The minute hand sweeps around with an angular rate of 1 rev/60 min. The hour hand travels around at the rate of 1 rev/7200 min. So, for any elapsed time t (elapsed after 2:00 pm), each hand will be at some angle from the reference (12 o'clock or zero rev). That angle is equal to the sum of its initial angular position plus the product of its angular rate of travel times the elapsed time t. Each hand will have its own equation to calculate this angle of the form:

    angle from reference = intial angular position + (angular rate x elapsed time)

    We need to solve for t when both hands are at the same angle. This is just a matter of equating the two equations and solving for t.

    My equations then looked like this:

    angle minute hand = 0 rev +(1/60 rev/min x t)
    angle hour hand = 1/6 rev + (1/7200 rev/min x t)

    Therefor, when they are at the same angle

    0+t/60 = 1/6 + t/7200

    This will yield a value for t in minutes from which seconds and fractions of a second are easily calculated. Then this can be added to 2:00:00 pm to get the actual time of the occurance.
     
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