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Frames of reference

  1. Aug 19, 2007 #1
    In an anniversary celebration of Marilyn Bell's 1954 crossing of Lake Ontario a swimmer set out from the shores of New York and maintained a velocity of 4m/s [N]. As the swimmer approached the Ontario shore, she encountered a cross current of 2m/s [E 25deg S]. Find her velocity with respect to the crowd observing from the beach.

    Let North and East be positive and let S represent the swimmer, W for water and G for the observers on the ground.

    To be honest I don't have a clue where to start. I'm not even sure that my vector diagram is correct. I'm having trouble grasping this concept. My text is very discreet on explaining this concept.
  2. jcsd
  3. Aug 19, 2007 #2


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    Mmm, not too clear where you are stuck. Call NS the y axis and EW the x axis. Then before entering the current the velocity vector is (0,4m/sec). After entering the current her velocity vector will be added to the velocity vector of the current. Can you work out the velocity vector of the current?
  4. Aug 19, 2007 #3
    I got 2.4m/s [E 69deg N] using trig and pyth. theor. If anyone has a moment to work this out I would really appreciate it!!!!
  5. Aug 19, 2007 #4


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    The magnitude of your velocity is too small and the direction is only roughly right. If you show more intermediate steps there is a much better chance someone can help.
  6. Aug 19, 2007 #5
    Let North and East be positive and let S represent the swimmer, W for water and G for the observers on the ground.
    So wVg= 2m/s [E 25 deg S] and sVg = 4 m/s [N]
    I drew a diagram of both the swimmers vector and the cross current vector. The resulting diagram is a triangle with 2m/s [E 25 deg S]on the top side 4m/s [N] on the left side and the third side is what I thought I was solving for.
    I divided the triangle into to right angled triangles. To find the lengths of the sides of the top RA tri. I used sin and cos. sin 65deg=opp/2m/s....opp=1.81m/s and then cos65deg=adj/2m/s.....adj=0.845m/s or 0.85. Then I subtracted 4m/s-1.81m/s=2.19m/s. Now using the pyth. theor. to find the length of sVg. 2.19m/s^2+0.85m/s^2=sVg^2..........sVg=2.349m/s. Then tan^-1=2.19/2.35 to get 68.735.
    Am I completely of base or what????
  7. Aug 19, 2007 #6
    Depending on what the angle is, you got it right. I interpreted the angle as it being 25º south of east, but you looked at as east of south. Do you know which one it is?

    Be careful about rounding errors though, I solved for the equations algebraically and had mathematica find the numeric results, of which give v = 2.345 m/s and ø = 68.87.
  8. Aug 19, 2007 #7
    By the way, I might as well give you the other answers since you obviously figured out how to do the problem and it might save you time if it ends up being my interpretation (though check it real quick just for practice). The angle of 25 degrees south of east gives

    v = 3.638
    ø = 60.12º
  9. Jan 23, 2008 #8
    I also got V=3.6

    but got theta as 30 degrees, instead of 60.

    using the picture below:
    red = current E 25 S
    blue = swimmer going N
    green = observer on the ground

    The question being worded as it is....does it make sense that we calculate the smaller angle where the blue and green lines meet? rather than the outter angle?

  10. Jan 23, 2008 #9

    it could be said 30degrees east of N

    or 60degrees N of east

    ..lol..dumb question.
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