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Homework Help: Solved it; Don't understand it.

  1. Sep 9, 2004 #1
    I did the problem, got the correct answer but dont' know why I was supposed to do it this way! Please help me understand the reasoning behind all this:

    A swimming is capable of swimming .45m/s in still water. She is swimming across a 75 m wide river whose current is .40 m/s. A) at what upstream angle must the swimmer aim if she is to arrive at a poitn directly across the stream? B) How long would it take her?

    I drew a picture and set the x and y components separately. So V being velocity, WL = water with respect to Land, SL swimmer respect to land, and SW swimmer respect to water .. I got

    Vwlx= 0 Vslx = ? Vswx = ?
    Vwly = .4 Vsly = 0 Vswy = ?

    A) to find the angle I took the inverse sin of .40/.45 to get 62.74 . However doesn't this depend on the picture? The way my profesor explained it, he drew it so that it would be inverse cos .. BEcause this confused me I drew it differently so that inverse sin would make sense. According to the drawing, the answer shouldn't be different but then why does it seem like it should be?

    B) then I took Vswy = VwLy + VsLy to get
    Vswy = .40 + 0 = .4 < this I understand
    .4 = .45sin(angle) << this I do not. Why do we set this up this way?
    Inv sin (.4/.45) =angle
    62.74 degrees ..

    then to get the x component <<again, I don't understand why we set this up this way.
    Vswx = .45cos(62.74)
    Vswx = .206

    distance formula:
    75= .206t
    75/.206 = t
    t= 363.8 s

    Any help would be appreciated.
  2. jcsd
  3. Sep 10, 2004 #2
    Before you read this post, please be forewarned that this is coming from a 15-year-old who has not been taught or self-learned relative velocities. I'm posting to bump your thread, and also to seek help for myself.

    I'm thoroughly confused by your working. I intuitively feel that your professor was right, and that you should take the inverse cos. I would say that the answer to (A) would be cos-1(0.4/0.45) = 27.3 degrees.

    I used the Pythagorean Theorem for (B). With x as the speed of the total effort crossing the river (bank to bank), which is the speed at which the swimmer appears to be crossing the river because to our eyes the swimmer is swimming directly from bank to bank without deviation,

    x2 + 0.42 = 0.452
    x = 0.206155281 m/s

    So the total time taken to cross the river is 75/0.206155281 = 363.8034376 seconds.

    Your final answer was correct because you took a short cut. Unfortunately this led to your first answer being wrong.
  4. Sep 10, 2004 #3


    User Avatar
    Science Advisor

    To quote the problem
    Your angle is complimentary to mine, so maybe you did it the other way.
    Here's the way I see the situation


    It's 75m across, the river flows 0.4m/s to the right, the swimmer is going 0.45 to the left at angle theta based on the shore (did you base your angle off the normal instead?).

    Horizontal speeds (finding the angle):

    [tex]V_x = 0 = 0.4 - 0.45 \cos(\theta)[/tex]

    [tex]\cos(\theta) = \frac{0.4}{0.45}[/tex]

    [tex]\theta = 27.266[/tex]

    Then finding vertical speed:

    [tex]V_y = 0.45 \sin(\theta)[/tex]

    [tex]V_y = 0.20615[/tex]

    Then the time:

    [tex]t = \frac{75}{0.20615}[/tex]

    time = 363.809852 seconds

    Looks like I get the same answer
    My angle is between the shore and the swimmer. Your angle is between the swimmer and an imaginary line perpendicular to the shore. No problem doing it that way; it's just weird. :wink:
    Last edited by a moderator: Apr 21, 2017
  5. Sep 10, 2004 #4
    You may have a look at something I just scanned here:

    Code (Text):
    [/PLAIN] [Broken]

    Copy and paste the address in the address bar for it to work. Thanks ShawnD!

    And sorry for the atrocious handwriting!
    Last edited by a moderator: May 1, 2017
  6. Sep 10, 2004 #5


    User Avatar
    Science Advisor

    recon, I can sort of understand what you're doing, but I'm too tired to figure it out.

    For your link to work, it has to be manually copied and pasted into the address bar. Just clicking that link gives a 403 error.
  7. Sep 10, 2004 #6
    thank you both!
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