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The lifeguard problem: find the direction of sprint.

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data

    This i the usual lifegard problem: we have a lifeguar that has a soil and water velocity, and wants to rescue a person at position (x,y). The lifeguard is ate the origin and th beach ends at position l, he wants to reach destination aa fast as possible.

    We all know that this is accomplished by respecting the snell-descart law but what matters to the lifeguard is the angle he is going to run or the point in the waterline he is aiming, call it c.

    2013-03-25 15.47.03.jpg


    2. Relevant equations

    Snell-Descarte law

    3. The attempt at a solution

    I made the calculations but the expressian became really ugly, can this expression simplify?

    2013-03-25 15.46.46.jpg
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 1, 2013 #2

    BruceW

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    Homework Helper

    yeah, to be honest, I think a direct expression of C in terms of the other distances is going to be ugly, I don't think it will simplify into a nice expression.

    I am trying to check your answer. So is vp the velocity of the man on ground? And vm the velocity of the man in water? I tried to work back from your answer, and I get to
    [tex]\frac{x-c}{y-l} = \frac{y-l}{x-l} \frac{l}{c}[/tex]
    This doesn't make sense, so maybe it is worth checking through your working again?
     
  4. Apr 1, 2013 #3

    rude man

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    Gold Member

    Just a lot of yucky differentiation, right?

    T1 = √[l2 + c2]1/2/v1
    T2 = √[(y-l)2 + (x-c)2]/v2
    d(T1+T2)/dc = 0 etc.
    Yuck!
     
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