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Homework Help: Relative Motion (Swimmer Crossing River Question)

  1. Feb 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A swimmer who can swim at a speed of 0.80m/s in still water heads directly across a river 86m wide. The swimmer lands at a position on the far bank 54 m downstream from the starting point. Determine:

    (C) The direction of departure that would have taken the swimmer directly across the river.


    (s - swimmer
    g - ground
    w - water

    Vsw = 0.8 m/s
    d across stream = 86m)

    2. Relevant equations

    (n/a - see bellow)

    3. The attempt at a solution

    From the previous two parts of the question I determined that it took the swimmer 107.5s to cross the river and thus the speed of the current is 0.5 m/s [E]. And that the velocity of the swimmer relative to the shore was 0.94 m/s [58 N of E]. (Which according to my textbook is correct.)

    (t=107.5s
    Vwg = 0.5 m/s [E]
    Vsg=0.94 m/s [58 N of E])

    Now, I figured that in order to end up straight across where you start from, you would have to swim [58 N of W] since the current resulted in the swimmer following a path of [58 N of E]. (So this would negate the effect of the current?) However, according to the textbook the answer is [W 51 N] and I have no clue how else to approach this question. Any help/tips please?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 22, 2007 #2
    the situation isn't symmetric as one might intuit at first glance. In the first case, all the velocity goes to crossing the river, in the second some is lost; ie

    t=86/.8=107.5 as you posted. The velocity of the current downstream then just 54/107.5=0.5 m/s

    so now you need to aim upwards to compensate for the drift--in other words total velocity^2=0.64=Vx^2+0.5^2 solve for Vx, then angle should fall out.
     
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