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Relative Velocity problem

  1. Jun 1, 2014 #1
    A flat rectangular barge,48m long and 20m wide, is headed directly across a stream at 4.5 km/h. The stream flows at 3.8 km/h. What is the Velocity, relative to the riverbed, of a person walking diagonally across the barge at 5km/h while facing the opposite upstream bank?

    Hi Guys, having some trouble with this one would really appreciate some help.

    Attempted solution-

    Split the figures into three vectors being-

    a (3.8,0)
    b (0,4.5)
    c 5v/|v| relative to the barge, where v=(48,20)
    = sqrt 48^2 + 20^2= sqrt 2704= 52
    =52/(48,20)
    = (48/52,20/52)
    = (12/13,5/13)
    = 5 (12/13,5/13)
    =(60/13,25/13)

    (3.8+60/13,4.5 + 25/13)
    =(8.4,6.4)

    ^ and thats as far as i have got
     
  2. jcsd
  3. Jun 1, 2014 #2

    Nathanael

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    Is it correct that the "60/13" part is in the same direction as the "3.8" part?

    Also it says "while facing the opposite upstream bank" so that means he's walking (partially) in the opposite direction of the flow of water, so there should be a minus sign somewhere.
     
  4. Jun 1, 2014 #3

    tms

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    Disregard.
     
    Last edited: Jun 1, 2014
  5. Jun 1, 2014 #4

    Nathanael

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    That's what he used it for, to break up the 5 km/hr into it's components.
    [itex]\sqrt{(\frac{60}{13})^{2}+(\frac{25}{13})^{2}}=5[/itex]

    He did it correctly.

    Edit:
    His work was misleading (because he put several equal signs when it in fact was not equal) but his answer was correct.
     
    Last edited: Jun 1, 2014
  6. Jun 1, 2014 #5

    tms

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    So he did. I stopped reading when he divided a scalar by a vector.

    He appears to be inconsistent with his directions. He has set up the positive ##x## direction as downstream, so the long axis of the barge should be in the ##y## direction. And, as you said, there is a sign issue.
     
  7. Jun 1, 2014 #6
    Thanks Nathaniel.

    So the velocity in km/h would be?
     
  8. Jun 1, 2014 #7

    Nathanael

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    Well you just made two mistakes (which I pointed out in my first reply).

    Mistake #1:
    When you added the velocities together, the components of the 5 km/hr were added to the wrong part.
    (In other words, you added the "x direction" to the "y direction" and vice versa.)

    It would actually be:
    (3.8+25/13, 4.5 + 60/13)


    Which brings me to mistake #2: (because it wouldn't actually be what I just said it would actually be, lol)
    The other mistake is that the component of the 5 km/hr in the direction of the flow of water would actually be negative (because he's walking upstream)

    So it would ACTUALLY be:
    (3.8-25/13, 4.5 + 60/13) [itex]\approx[/itex] (1.8769, 9.1154)


    If you want the velocity not in "component-form" then simply use the pythagorean theorem.
     
  9. Jun 1, 2014 #8
    Ahhhh, thanks Nathaniel!!

    so converting from component form into a km/h value.

    (1.8769,9.1154) would be the base and adjacent side? with the hypotenuse being the value i'm after?

    so that would mean 1.8769^2 + 9.1154^2 = sqrt 84.613
     
  10. Jun 1, 2014 #9

    Nathanael

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    You're welcome.
    There you go again, using equal signs when it's not equal! Heheh, but yes, the answer would be sqrt(84.613)
     
  11. Jun 1, 2014 #10
    haha! love math boffins! hopefully i'll become one, one day.
    Once again thanks for your patience and help bud
     
  12. Jun 3, 2014 #11
    Nathaniel,
    So far i have the velocity at 9.19 km/h
    i think that i need to provide the direction as well in my answer.
    How would i be able to calculate this?

    Thanks
     
  13. Jun 3, 2014 #12

    Nathanael

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    Remember the (x,y) components of the velocity were (1.8769, 9.1154)

    That's why you could use the pythagorean theorem to find the total speed (because x-direction and y-direction are perpendicular, so it makes a right triangle)

    See if you can draw the triangle (the legs would be x-speed/y-speed/total-speed) and figure out the angle using trig
     
  14. Jun 3, 2014 #13
    ok, i think i have it, 78deg?
     
  15. Jun 3, 2014 #14

    Nathanael

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    but you have to specifcy, 78deg from what?
     
  16. Jun 3, 2014 #15
    from the rear end of the barge?
     
  17. Jun 3, 2014 #16

    Nathanael

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    EDIT:
    Ok I said it all wrong in my first post...

    It would be 78 degrees from the upstream direction (towards the other side of the river, not backwards)
     
    Last edited: Jun 3, 2014
  18. Jun 3, 2014 #17
    wouldn't it be to or facing the upstream direction?
     
  19. Jun 3, 2014 #18
  20. Jun 3, 2014 #19

    Nathanael

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    Yes exactly. That's what I was trying to say. (I meant "78 degrees from the opposite direction of the water flow")

    At any rate, you've got it correct.
     
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