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Swimming Across a River

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A river is 400 feet wide and flows at 1 ft/s. A swimmer swims at 2 ft/second, straight across the river and back to where he started on the original shore. Find the time required to complete his trip.

    To clarify, the turn around point is straight across the river.

    (I have already been given the answer: 461 seconds)

    2. Relevant equations

    t=(2d/vs)*(1/sqrt (1-(vr/vs)^2)), where vs is the velocity of the swimmer (√3 from Pythagorean theorem) and vr is the velocity of the river (1), and d is distance (800ft). Sorry I couldn't figure out how to attach or embed an image (novice here).

    3. The attempt at a solution
    I used this equation and got 566 seconds, which is decidedly not 461 seconds. Where did I go wrong?

    Thanks in advance guys!
  2. jcsd
  3. Dec 2, 2014 #2


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    I'm not getting anything from your attached image, it just shows up as a broken image link. Can you embed it instead of linking to it?
  4. Dec 2, 2014 #3


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    It's not completely clear, but it seems the 2f/s is relative to the water, not to the land.
    In that equation, vs is the 2f/s speed.
  5. Dec 3, 2014 #4
    Using vs= 2f/s, I got 327s. (??)

    It would seem that, using the Pythagorean theorem, I would calculate that he's swimming √3 ft/s relative to the land, for a distance of 800ft total, so 800/√3= 461 seconds- which is what I'm looking for. However, our professor told us to use that relativity equation that I wrote out, and I can't seem to get the right answer with that. What am I doing wrong?
  6. Dec 3, 2014 #5


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    So I suspect that your professor's point is that the relativity equation is essentially the same as this "crossing the river against a current".
    The formula you give is [itex]\frac{2d}{v_s}\frac{1}{\sqrt{1- \left(\frac{v_r}{v_s}\right)^2}}[/itex]

    With d= 400, [itex]v_s= 2[/itex], and [itex]v_r= 1[/itex], that is [itex]\frac{800}{2}\frac{1}{\sqrt{1- \frac{1}{4}}}[/itex].

    Work that out and you should see that you get the same thing.
    Last edited by a moderator: Dec 3, 2014
  7. Dec 3, 2014 #6
    Okay, I got the right answer with those numbers, but he swam 800ft total, not 400ft. Is the x2 in the equation to account for there-and-back?
  8. Dec 3, 2014 #7


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    Please post your working. I get 800/√3=461.
    That equation uses Pythagoras' formula, so either use the equation or apply Pythagoras directly, not both.
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