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Frames of Reference and Relative Velocity

  1. Sep 21, 2011 #1
    A swimmer who achieves a speed of 0.75 m/s in still water swims directly across a river 72 m wide. The swimmer lands on the far shore at a position 54 m downstream from the starting point.
    (a) Determine the speed of the river current.
    (b) Determine the swimmer?s velocity relative to the shore.
    (c) Determine the direction the swimmer would have to aim to land directly across from the starting position.

    I am having trouble understanding the actual physics of this question. I mean I get the math: a) 72/0.75 = 96s --> 54/96 = 0.56m/s but I don't understand why. If the swimmer is swimming with the current, should it the speed be faster? And are my calculations even right because it says he swims across the river, but is then 54m downstream. So there are two different directions ...

    Can anybody explain the physics of this question to me please?
     
  2. jcsd
  3. Sep 21, 2011 #2

    cepheid

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    Let's say that the river is oriented with the vertical axis in our coordinate system (i.e. vertical = parallel to the river and horizontal = perpendicular to the river). So the swimmer is swimming at 0.75 m/s horizontally. Meanwhile, the current is carrying him at some unknown speed downstream (i.e. vertically). So, his/her velocity relative to the bank is the vector sum of these two (meaning that it points somewhere in between entirely horizontal and entirely vertical). As a result, at the end of the trip, the swimmer will have moved BOTH horizontally (across to the opposite bank) and vertically (downstream). Indeed, that is the case. The swimmer ends up on the opposite shoreline (72 m horizontally from where he was before) but he is 54 m farther along that shoreline compared to when he started. So, relative to the shore, he actually moved in a diagonal line that is the hypotenuse of the right triangle formed by his horizontal and vertical motions (i.e. 54 m is one side of the triangle, 72 m is the perpendicular side, and the third side, the hypotenuse, is his actual net displacement). Does that help?
     
  4. Sep 21, 2011 #3

    PeterO

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    Perhaps a similar situation will make it clearer:

    A conveyor belt runs from a mine to a factory.
    An ant walks across a conveyor belt, 30 cm wide, at a speed of 3mm per second. When it reaches the other side the ant is 5 m closer to the factory than when the ant started.
    a) how fast is the belt running etc etc

    ( in this case the ant is simply not fast enough to get straight across the conveyor mechanism)

    Now the swimmer
    For the first part.
    As the swimmer is about to begin, he aims at a strip of water, say 2m wide, stretching across the river. he swims along that strip. While he is swimming, that strip, and all the rest of the river, moves downstream, so that when he has got to the other side of the strip of water, the strip of water is 54m further down-stream than when he started.

    For the second part,
    As the swimmer is about to start, he aims at a strip of water angled up-stream [given your figures, at about 45 degrees??]. He again swims along that strip, but the strip will have moved downstream just enough that when he reaches the end of the strip, it is positioned exactly opposite the point where he began to swim.
     
  5. Sep 21, 2011 #4

    NascentOxygen

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    I used to have a lot of trouble with these at school, too. I suggest you commit to memory this single formula and it will serve you well for the rest of your life: :smile:

    Va = Va rel b + Vb

    where Va rel b is read as "velocity of a relative to b".

    If this means nothing much to you at the moment, perhaps convert it to money, or weight, or something that is meaningful to you, so that you can work it out and write it down every time you need it. Example, if Fred weighs 25kg more than you, and you weigh 70kg, how much does Fred weigh? Here's the equation:

    Wf = Wf rel u + Wu

    For scalar quantities, these add as scalars. For vector quantities, these add as vectors and you need to draw a vector triangle to perform that vector addition.

    Now, the swimmer makes progress through the water at his swimming speed, this is Vs rel w. To find his progress as far as the fixed reference point of someone on the bank is concerned, we need:
    Vs = Vs rel w + Vw

    Progress (i.e., distance) is directly proportional to speed, so as vector quantities, both displacement and velocity diagrams form congruent (or similar) triangles.
     
    Last edited: Sep 21, 2011
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