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Complex Vectors

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A boy and a girl both swim at 3.0m/s. They jump into a river 1.0km wide, with a current of 2.0m/s[E].

    a.) The girl swims so that she ends up directly across from her starting point. What is her velocity relative to the ground?

    2. Relevant equations
    Note: V = velocity with arrow above it.

    Vgw = 3.0m/s (Girl relative to water)
    Vwg = 2.0m/s[E] (water relative to ground)
    Vgg = ? [?] (Girl relative to ground)
    Vgw + Vwg = V[gg]

    3. The attempt at a solution
    I believe I've set it up correctly, although I don't see how to solve it. I know you have to the Pythagorean theorem so it's Vgg2 + 22 = 32. Which turns out to be Vgg + 4 = 9, subtract 4 on each side. Vgg = √5. So Vgg = 2.2m/s. But when looking at this equation: Vgw + Vwg = V[gg], the inner 2 w's must match, then the outer 2 letters must match, so the equation is right. But as you can see in my picture at the bottom half of the page, I get a velocity of -2.2m/s. Which doesn't exist but you can't just get rid of the negative sign, can you? How does it just go away like that? Or did I do something wrong?

    Last edited: Feb 6, 2016
  2. jcsd
  3. Feb 6, 2016 #2


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    You seem to have solved this using the triangle for the velocities. Or, at least you got the magnitude of the velocity. I don't undersand how you then went wrong. Using ##g## as a subscript for both "girl" and "ground" adds to the confusion.

    I think it's much better in this case to think in terms of a triangle of vectors, as you did. I would have left it at that. Problem solved!
  4. Feb 6, 2016 #3


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    Correctly: Vgg2 + 22 = 32
    You go on using Vgg instead of its square. You did a lot of subtractions in the picture, and made a sign error ( you wrote 9 instead of -9). You calculated Vgg2 and it can not be negative.
  5. Feb 6, 2016 #4


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    Hi Taco,

    This is a vector addition, so you can't jump from ##\vec V_{gw} + \vec V_{wg} = \vec V_{gg}## to ## V_{gw}^2 + V_{wg}^2 = V_{gg}^2 ## as you can see in the drawing you made.

    Basically you should work out the addition in components, e.g. N and E :
    ##\vec V_{gw, \; N} + \vec V_{wg, \; N} = \vec V_{gg \; N}##
    ##\vec V_{gw, \; E} + \vec V_{wg, \; E} = \vec V_{gg \; E}\quad ## and apply the given info, namely
    ## \vec V_{wg, \; N} = 0 ## and ##\vec V_{gg \; E} = 0 ##.

    The Pythagoras thingy comes in to calculate ##|\vec V_{gg} |##:
    $$ |\vec V_{gg} |^2 = \vec V_{gg \; E}^2 + \vec V_{gg \; N}^2 $$
    Last edited: Feb 6, 2016
  6. Feb 6, 2016 #5
    PeroK: Yeah, I know it was confusing for both g's for ground and girl but I did originally state it in parentheses so there wasn't too much trying to figure out what is what.
    ehild: When the equation is in it's original form, you bring it over to the left side and the hypotenuse (Vgw) over the right. Why are they positive instead of negative? That's what I don't get. It may be a simple question with a simple answer, but that's what get's me on the test, and I'm trying to really understand instead of thinking I understand but make mistakes which will cost me. In the picture, the left side is correct but the right side is what I did my first try.

    IMG_0124 (2).JPG

    Edit: I posted this after BvU commented.

    BvU: so basically, it's also positive? If it's the absolute value and I guess if you're finding velocity, you can solve it and take the absolute value of the answer to give you positive.
  7. Feb 6, 2016 #6
    Hi Frito:

    I see some notation issues with your work, and I suggest fixing these first. I apologize if some of these suggestions come across as nits.

    Your use of "V" is the speed, ignoring the direction. This is OK. I don't think you need an arrow for this, but your teacher may possibly expect an arrow.

    I see that the "[E]" also appears in the problem statement, but it makes no sense to me. Does i t mean "East"? If so, why isn't something designated "[N]" for North?

    The "[?]" seems to be wrong. You know the units. Shouldn't this be "[m/s]"?

    Three issues.
    1. Why V[gg] rather than Vgg?
    2. If this is a vector equation, then you need to distinguish vectors from speeds. If it is a speed equation, then the speeds need top be squared.
    3. Your figure looks OK. You want the hypotenuse vector to be the sum of the side vectors. Or the hypotenuse speed squared is the sum of the squares of the side speeds. Vgg is not the hypotenuse.

    I think your confusion is related not identifying the hypotenuse correctly. You have it right in your figure, but not in your equations.

    Hope this helps.

  8. Feb 6, 2016 #7
    I just had v with an arrow for the sake of having an arrow. :) About the [E], yes it means East, and why not [N] if I have an arrow pointing north in my diagram? Because I'm not exactly sure what bearing of North the girl is traveling, if she wants to go across the river, but the river is pushing her East, she needs to head in the direction of North (as my hypotenuse shows) and Vgg is going up but it never states in the problem the Vgg (Girl relative to ground) is going North. Very confusing but I think that's why. Now for Vgg, shouldn't [] be [m/s]. No, because that's not for units but for direction, like how I have Vwg = 2.0m/s[E]. Now for the not identifying the hypotenuse in my equations. I understand Vgg is not hypotenuse, but when first setting up the equation, it must be set up like this; Vgw + Vwg = Vgg. Notice how the 2 w's are on the inside and the 2 g's on the outside of the equation on the left side. Take those 2 g's and place them on the right side, so now it's equal to Vgg. Vgg isn't the hypotenuse so that's why you see in my previous pictures rearranging it so the hypotenuse does stand alone. :)
  9. Feb 6, 2016 #8
    Hi Frito:

    OK. I am glad you understand that Vgg is not the hypotenuse, but the quote above makes it look like it is. I don't get why you don't have the hypotenuse Vgw on the RHS of the equation.

    OK, [] should not [m/s], but why not [N]? the girl's path goes in a straight line North, right?

  10. Feb 6, 2016 #9
    You don't at first because you have to first set up the equation in a certain order as I showed you above. I don't really know the reason for that because I recently learned this so I'm not sure.
    The girl wants to head North, but the current from the water is going East so she has to be heading the direction Vgw is facing, which is the hypotenuse.
  11. Feb 6, 2016 #10


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    What is the sense to move over those terms, instead of isolating Vgg2? And -Vgg2 is not ( -Vgg)2 .
  12. Feb 6, 2016 #11
    Well since Vgw is the hypotenuse, We can't have it stay in its original form of Vgw + Vwg = Vgg. And you can't set up the equation firstly as Vgg + Vwg = Vgw because my teacher said it must firstly be in this form; Vgw + Vwg = Vgg. You need both letters on the inner side of the left equation to match, so; Vgw + Vwg = Vgg. Then the outer two letters on the left side don't need to match. Vgw + Vwg = Vgg. (Girl is the first one then ground is the second g). Then you bring the outer two letter together from the left equation to the right side of the equation, so; Vgw + Vwg = Vgg. (Girl relative to ground). Then you have to rearrange the equation to find your answer. I hope this is making sense, please reply if it isn't.
  13. Feb 6, 2016 #12
    Hi Frito:

    She may be trying to swim in a roughly NE direction, but her velocity vector "relative to the ground" is in the E direction.
    Thus Vgg = 2.2 m/s [E]. If you have a reason that makes this a wrong answer, then I apologize for not understanding the problem.

  14. Feb 6, 2016 #13
    Actually she is trying to swim in the NW direction because she is trying to swim across the river which is flowing East but she'll wash down the river going NE, so she swims NW. No need to apologize, learning takes effort :). Anyways, Vgg in my previous diagram is pointing North, we got an answer for Vgg which equals 2.2m/s. She is going straight up north and she is oriented NE to flow directly across where she wants to go. I think I understand what you are saying, I drew a picture and tell me if this is your thoughts. I didn't label anything because Vgg I wasn't sure where you might of thought it would've been on the triangle.

    IMG_0125 (2).JPG
  15. Feb 6, 2016 #14


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    Your picture is correct, only your derivation was wrong. ##\vec V_{gg} ##, the girl's velocity with respect to the ground is the "north" leg of the right triangle. Applying Pythagoras's Vgg2+22=32. You can subtract 4 from both sides, and you get Vgg2=5. So the speed is √5 and the direction follows from the text of the problem: across the river, perpendicular to the river flow. I do not understand what your teacher said.
  16. Feb 6, 2016 #15
    Alright, thanks. I'm still new to this but I think I understand it more than I did before. :)
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