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Motion in Two Dimensions

  1. May 8, 2015 #1
    Question:
    In an anniversary celebration of Marilyn Bell's 1954 feat- she was the first person to swim across 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 2 m/s (E 25 degrees S) - 25 degrees South of East. Calculate her velocity with respect to the crowd observing from the beach (with respect to the ground).

    My Attempt at a Solution:

    First I drew a triangle. 4m/s (N) is the left side of the triangle. 2m/s (E 25 degrees S) is the top of the triangle. And the right side is the side that I think that I am solving for.

    Next I divided the triangle into two right angled triangles.

    Solving for the top triangle:

    sin=Opposite/Hypotenuse
    =2sin(65)
    =1.81 m/s

    cos=Adjacent/Hypotenuse
    =2cos(65)
    =0.845 m/s
    =0.85m/s

    To solve for the bottom triangle:

    4m/s - 0.85m/s = 3.15m/s

    Use Pythagorean theorem to solve for the unknown side:

    3.15^2 + 1.81^2
    =3.63m/s

    Calculate the direction of the velocity:

    tan=ay/ax
    tan=1.81/3.15
    =29.88
    =30 degrees

    Therefore her velocity, with respect to the crowd observing from the beach is 3.63m/s (N 30 degrees E)

    Could someone please let me know if my calculations are correct? I would appreciate the help. :)
     
  2. jcsd
  3. May 9, 2015 #2
    Looks right to me.
     
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