The RHP Ferry

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Posers and Puzzles 14 Apr '21 23:18
  1. R
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    14 Apr '21 23:18
     On a normal day a Ferry takes 10 minutes to makes its journey across the great RHP river. The speed of the ferry in still water is 12 m/s. The current of the river is 5 m/s. The ferry's path is straight line directly across the river ( shortest possible path length ).

     Today the river is flowing much faster that normal - double its normal current. The ferries' path across the river is the same, how long will it take to make the journey today?
  2. Standard memberBigDogg
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    15 Apr '21 04:161 edit
    @joe-shmo
    Reveal Hidden Content
    5 * sqrt(119) / sqrt(11) ~= 16.45 minutes
  3. R
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    15 Apr '21 11:23
  4. Standard memberBigDogg
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    15 Apr '21 17:15
    I can show my work, if anyone would like to see it.
  5. R
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    15 Apr '21 18:33
    @bigdoggproblem said
    I can show my work, if anyone would like to see it.
    This problem doesn't appear to have generated much interest. It's up to you.
  6. Subscribervenda
    Dave
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    15 Apr '21 19:01
    @bigdoggproblem said
    I can show my work, if anyone would like to see it.
    It all helps the learning process Bigdogg.
    As has become obvious, I'm out of my depth with Joe's problems but I still try from time to time.
    I might even fluke a right answer once in a while!!
    One way to learn is to look at the answer and then try and work out how it is derived!!
  7. Standard memberBigDogg
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    16 Apr '21 02:59
    One way to sum up force vectors is to create a triangle, using the two vectors, as shown on this site:

    https://www.school-for-champions.com/science/force_vectors.htm

    So we get a right triangle with a hypotenuse of 12 [boat speed in still water] and one side equal to 5 [river current]. Use the Pythagorean theorem to solve for the remaining side [the actual rate of the boat as it goes straight across, fighting the current somewhat].

    5² + R² = 12²
    R = sqrt(119) m/s

    We were given the time (10 min, which equals 600 s) for crossing the river, so Rate * Time gives a distance of:

    600 * sqrt(119) m

    Now, we bump the current up to 10 m/s. We have a right triangle, similar to above, but with a longer side. Friend Pythagoras tells us that:

    10² + RP² = 12² [where RP is 'rate prime']

    RP = sqrt(44) = 2 * sqrt(11) m/s

    Since we found the distance across the river above, we can solve for the new crossing time.

    2 * sqrt(11) m/s * TP = 600 * sqrt(119) m
    TP = 300 * sqrt(119) / sqrt(11) s
      = 5 * sqrt(119) / sqrt(11) min
  8. Subscribervenda
    Dave
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    16 Apr '21 11:41
    @bigdoggproblem said
    One way to sum up force vectors is to create a triangle, using the two vectors, as shown on this site:

    https://www.school-for-champions.com/science/force_vectors.htm

    So we get a right triangle with a hypotenuse of 12 [boat speed in still water] and one side equal to 5 [river current]. Use the Pythagorean theorem to solve for the remaining side [the actual rate of ...[text shortened]... 1) m/s * TP = 600 * sqrt(119) m
    TP = 300 * sqrt(119) / sqrt(11) s
      = 5 * sqrt(119) / sqrt(11) min
    Thanks.
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