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Mechanics Problem.

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Swimmer A swims due east at a constant speed of 3m/s along a straight stretch of river with neglible current, keeping a constant distance of 10m from the southern bank. A second swimmer, B, starts swimming from this bank when A is a disnace L down the river from her (i.e. L is the distance measured along the bank.) Swimmer B swims with a constant speed of 2m/s and at an angle of N60E in order to intercept A.

    i) what is the velocity of b as observed from a?
    ii) what is the distance L, and how long does B swim for?
    iii) If a strong current is flowing how, if at all, would the above results be alterd?

    3. The attempt at a solution

    I have calculated A and B's vector velocities.
    Va = 3i m/s
    Vb = sqrt(3)i + 1j
    Then surely the velocity of b as observed from is just the difference between the two vectors.
    Vdiff. = (3-sqrt(3))i + 1j

    Then i run into trouble for the next bit, surely if swimmer b's vertical component of velocity is 1m/s then no matter what distance A swims before B starts. B will always have to swim through 10m of water vertically. So if B's vertical component of velocity is 1, then he has to swim 10s before he contacts A.
    Its the distance L which confuses me. B's horizontal component is sqrt(3) which is less than 3, so if B starts after A has passed B will never reach A. The distance will get progressively larger. So surely B has to start before A passes him on the river?
     
    Last edited: Nov 6, 2007
  2. jcsd
  3. Nov 6, 2007 #2

    Doc Al

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    Staff: Mentor

    Recheck the velocity of B. [Edit: Your velocity is fine!]
    Right. A is a distance L down river from B. When B starts moving, A is to the west of B. (As you realize, that's the only interpretation that makes sense.)

    [Edit: Ignore my comment about velocity.]
     
    Last edited: Nov 6, 2007
  4. Nov 6, 2007 #3
    The velocity of B is right (im sure), as the triangle is a 30,60,90 degree triangle, since 2 is the hypotneuse then root3 and 1 are the other sides. The angle used is 30, as N60E is 60 degrees FROM the north so its 30 degrees measured from the bank counterclockwise.

    Correct me if im wrong. But im preety sure that's right. as the triangle is of the form:

    [​IMG]
     
  5. Nov 6, 2007 #4
    Ahhh, going mad. I can't see where im going wrong as the vertical component from the triangle is 1m/s and the horizontal is sqrt(3)m/s. Unless ive got my i's and j's mixed up. Please help me :(
     
    Last edited: Nov 6, 2007
  6. Nov 6, 2007 #5

    Doc Al

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    Staff: Mentor

    D'oh! I just realized that I'm full of it. Your Vb components are correct. :redface: (Sorry about that!)
     
    Last edited: Nov 6, 2007
  7. Nov 6, 2007 #6
    Hehe np, would the time taken be 10s and the distance L be 15.58m then?
    (as |10*sqrt(3) - 3*10| = 15.58 ?)
    sorry to badger, but im quite unconfident in my physics atm as im having to learn this all by myself as ive missed alot of lectures.
     
  8. Nov 6, 2007 #7

    Doc Al

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    Staff: Mentor

    The time is 10s, but check your arithmetic on that subtraction.
     
  9. Nov 6, 2007 #8
    Sorry 12.68m :S
     
  10. Nov 6, 2007 #9

    Doc Al

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    Staff: Mentor

    Looks good.
     
  11. Nov 6, 2007 #10
    Cheers for all your help mate :)
     
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