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Similar to the swimmer problem:

  1. Apr 4, 2012 #1
    Hello. I have a problem that I can't solve:
    1. The problem statement, all variables and given/known data
    A boat (boat one) is in the position (0,0) and he sees a friend's boat (boat 2) which is in the position (a,b) and sails with velocity ω2 in a straight line (I mean, a is always the same coordinate and b changes with time: b = b0 + ω2*t).
    Boat one has to approach at least at a distance r for being seen by boat 2. If boat 1 starts sailing diagonally with velocity ω1, ¿what angle is the correct for approaching the objective at distance r in the less time posible?

    3. The attempt at a solution
    I think coordinates of boat one change with this equations: x= ω1cos(angle)*t and y = ω1sin(angle)*t; I have calculated the disance: d= √((x-a)^2 + (y-b)^2), then set it to r, put everything in fuction of t, derivate and set to cero but it's a way I think it's not correct because the result looks like a monster. I have to give the result in fuction of the problem's parameters. I have thought maybe we could do it with ODE or with Hamilton... but I don't know.

    Thank you for every suggestion.
     
  2. jcsd
  3. Apr 4, 2012 #2

    tms

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    It sounds as though you're on the right track, though I don't think a derivative is necessary; just set [itex]d = r[/itex]. Don't let long and ugly algebra discourage you; look for simplifications.
     
    Last edited: Apr 4, 2012
  4. Apr 5, 2012 #3
    But I need to derivate because it asks me for the angle which minimizes the time... :S
     
  5. Apr 5, 2012 #4
    The problem will become much easier, if you think of the second boat as a buoy, anchored in a stream with a velocity w.
     
  6. Apr 5, 2012 #5
    I cannot see how it gets simplified and why it could be reduce to that case. Could you explain your idea a little more?
     
  7. Apr 5, 2012 #6

    tms

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    But you already know that the shortest time will be along the shortest path (since the velocity is constant), and that the shortest path will be a straight line. The problem is to find out where to aim the straight line.

    BTW, the proper word is `differentiate'.
     
  8. Apr 8, 2012 #7
    Sorry, I thought that we could say derivate because we're doing it respect one variable (and also english is not my language).
    I understand what you're saying but when I do d=r and we find the value of the angle, that depends on the time so... it is not the way :S
     
  9. Apr 12, 2012 #8

    tms

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    No need to apologize. I assumed English was not your native language. I just thought that, for future reference, I'd tell you the right word. As far as I know, 'derivate' was not a word until you used it, but this is English, so making up new words is perfectly okay.

    Think geometrically. Imagine a circle of radius [itex]r[/itex] centered on the location of the second boat at the moment of sighting. Then think of a circle centered on the origin with a radius equal to the distance the first boat travelled in that time. The point the first boat reached at the time of sighting is on both circles. What can you say, geometrically, about the origin, the sighting point, and the location of the second boat, bearing in mind the least-time requirement?
     
  10. Apr 17, 2012 #9
    Please be more explicit because the only thing that comes to my mind is that they're aligned.:confused:
     
  11. Apr 17, 2012 #10

    tms

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    That's it. That means that at the moment of sighting you have a nice right triangle. One side's length is [itex]a[/itex], one side's length is [itex]b[/itex] plus the distance boat 2 travelled, and the other side is [itex]r[/itex] plus the distance boat 1 travelled.
     
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