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Homework Help: Swimmer in River

  1. Apr 8, 2005 #1
    Q. A river 590 ft wide flows with a speed of 6 ft/s with respect to the earth. A woman swims with a speed of 3 ft/s with respect to the water.
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    a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
    d1 = ft *
    1179.96 OK


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    b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. If she heads 29° upstream, how far downstream is she swept before reaching the opposite bank?
    d2 = ft

    Any ideas on part b

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    c) For the conditions of part (b), how long does it take for her to reach the opposite bank?
    Any ideas on part c

    Please help me !!!
     
  2. jcsd
  3. Apr 8, 2005 #2
    Aha I remember this from vectors in grade 10.

    Remember relative velocity law:

    V woman,earth = V woman,water + V water,earth

    Draw your diagrams, look at the components, add them. find scalar multiple k so that the vertical component is 590 (the width of river) and the horizontal component is how far she is swept downstream. k is the time it takes to get the the other side.

    Part b: exactly the same as part a, except change your vectors so she heads 29 degrees instead of being perpendicular to the current of the river.

    Part c: same as part a; find k a scalar multiple such that the vertical component is the width of the river. k is the time it takes.

    By the way it's so funny seeing things in feet and feet per second. I kind of gained an impression about how annoying it must be for poor americans learning science and having been brought up to know the old-fashioned units and not metric!!
     
  4. Apr 9, 2005 #3
    Here is what I did for part b,

    d = v0 cos theta * t

    width of the river, d = 590 m

    velocity of the river = 6ft /s

    velocity of the woman = 3ft/s

    so, t = d / vwoman * cos (29)

    t= 590 /3 * cos(29)

    After finding t , we can multiply this t with 6 ,which is the velocity of the river.

    I doubt if this right, but please help!
     
  5. Apr 11, 2005 #4
    No luck with this problem for the last 2 parts, can anybody tell me if something is wrong here.
     
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