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Crossing a river.

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    A swimmer wants to cross a river, swimming directly from point A to point B, as shown in the figure. The distance d_1 (from A to C) is 179 m, the distance d_2 (from C to B) is 137 m, and the speed v_r of the current in the river is 5.00 km/h. Suppose that the swimmer's velocity relative to the water makes an angle of 34.0 degrees with the line from A to C, as indicated in the figure.

    What is the angle between the velocity vector for the swimmer with respect to a stationary observer (i.e. an observer standing at rest on the shore) and the riverbank? Hint: you should draw a triangle expressing the necessary addition of velocities for this problem.

    http://session.masteringphysics.com/problemAsset/1000050755/6/M2K_rm_2_v1_001.jpg

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I thought it was 56 degrees because 90-34=56 but this is wrong.

    i dont even understand the question i guess. Can anyone better explain this to me? i dont get what angle i am looking for.

    thank you
     
  2. jcsd
  3. Feb 2, 2010 #2

    Delphi51

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    Homework Helper

    I am finding the wording a little confusing. It looks to me like the velocity diagram looks like this:
    river.jpg
    and the angle they are asking for is the one marked A.
    A little odd that they would say the swimmer is going straight from A to B and then ask for the angle of AB . . .
     
  4. Feb 2, 2010 #3
    52.6 was correct. thank you. may i ask how you found the 37.4 degrees?
     
  5. Feb 2, 2010 #4
    mastering physics is satan. tell your prof to drop it.
     
  6. Feb 2, 2010 #5
    yeah i know. it is so stupid and we have a ton of hw from it. like way to much. its ridiculous and he doesnt teach anything so we have to figure it all out ourselves :(
     
  7. Feb 2, 2010 #6
    i just tried using law of cosine to find that angle of 37.4 but that was a fail for me.
     
  8. Feb 2, 2010 #7
    here is the next part of the problem that also i do not understand how to find at all.

    What is the angle between the velocity vector for the swimmer with respect to the water and the velocity vector for the swimmer with respect to a stationary observer?

    this makes absolutely no sense to me. i have no idea what angle they are asking for. This is one of my last problems too and its really frustrating :(
     
  9. Feb 2, 2010 #8
    Anyone know how the 37.4 was found by Delphi51 in the image that was posted?

    Also can anyone help me out with the second part please?

    What is the angle between the velocity vector for the swimmer with respect to the water and the velocity vector for the swimmer with respect to a stationary observer?

    I do not understand it at all.

    Thanks for any help
     
  10. Feb 2, 2010 #9
    Can anyone please help me with this second part?

    What is the angle between the velocity vector for the swimmer with respect to the water and the velocity vector for the swimmer with respect to a stationary observer?

    It is due tomorrow and i really need to get it done!

    Thank you :)
     
  11. Feb 3, 2010 #10
    ok I have no idea how this makes sense, I think something's wrong but I somehow got it all right using these equations for each step
    for Delphi51's Triangle I used similar triangles of water velocity vector and the triangle he actually wanted to swim. so it'll look something like this:
    (c-b)/velocity vector=(a-c)/ (a different vague side) which we will call y
    for you it'll be: (y)=((a-c)*velocity vector)/(c-b)=(179*5)/137
    that will get you the y to plug into inverse tangent formula
    so angle=arctan(y/5)= 52.6

    now for part 2 I added the original angle 34 to 90=124 and subtracted the angle from part 1: 124-52.6= answer for part 2

    Hope this helps it worked for me, maybe someone else can explain it but I hate mastering physics too not learning, just guessing
     
  12. Feb 3, 2010 #11
    yes that helped a lot. thank you :)
     
  13. Feb 3, 2010 #12

    Delphi51

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    Sorry I was away last night. The 37.4 degree angle is angle CAB in the diagram you attached. You can find it easily using an inverse tan.

    The swimmer aims a bit upstream, but when the water flow vector is added, he is actually going a bit downstream in the specified AB direction.
    The swimmer's vector with respect to the water is the 34 degrees upstream.
    The swimmer's vector with respect to the shore is the 37.4 degrees downstream. Total of 71.4 degrees between them. You can see all that on the original diagram you attached.
     
  14. Feb 3, 2010 #13
    Ok i get it all now.....finally haha :) Thank you for the explanation.
     
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