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Fluid dynamics - find velocity of a particle in a vector fie

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the speed of a fluid particle at the origin and at a point (1,-2,0) for each of the following velocity fields when t = 2s. All distances are in meters and t is in seconds.

    V = (x+2)i + xtj - zk m/s

    2. Relevant equations
    V x dr = 0
    dV/dt = u(dV/dx) + v(dV/dy) + w(dV/dz) + dV/dt

    3. The attempt at a solution
    Started with trying to:

    dx/dt = x+2 ==> dx/(x+2) = dt

    dy/dt = xt ==> dy/x = tdt

    This is as far as I have gotten, my textbook doesn't give a single example when dealing with 3-dimensional vector fields and my professor doesn't give any clarification on this program in his notes nor has he given any examples based on 3-dimensional problems or a vector field problem in general.
     
  2. jcsd
  3. Feb 8, 2015 #2

    SteamKing

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    V is already a velocity vector (Hint: Look at the units!). The velocity at a point depends on the coordinates of the point (x,y,z) and the time t. You are given both in the problem statement.

    Velocity is a vector quantity, but the OP asks for the 'speed' of the particle at those two points. How is 'speed' related to 'velocity'?

    Computing dV/dt is not giving you a velocity but something else.
     
  4. Feb 8, 2015 #3
    So then just plugging in the points gives me the velocity, taking the absolute value would give me the speed I am guessing?

    My memory of year to year physics classes goes away quickly so I don't remember much from my first year courses.
     
  5. Feb 8, 2015 #4

    SteamKing

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    Sounds like a plan.

    Remember, the hardest part of solving a problem is sometimes understanding what you are given to work with.

    Except for knowing the difference between velocity and speed, you don't need to know any physics to work this problem.
     
  6. Feb 8, 2015 #5
    Well just did that and checked one of my answers, and its wrong so I am still missing something.
     
  7. Feb 8, 2015 #6

    SteamKing

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    Well, post your calculations (and the answer, if you have it.)
     
  8. Feb 8, 2015 #7
    Well there was a part b which gave a different vector field

    b) V = xy i - 2y^(2)j + tyzk m/s

    answer is: 0, 8.246 m/s

    My answers are 0 and 10m/s
     
  9. Feb 8, 2015 #8
    Show us your work please.

    Chet
     
  10. Feb 8, 2015 #9
    All I did was what was talked about earlier.

    Using the points (0,0,0) and (1,-2,0) I plugged them into the vector field b then took the magnitude for the speed:

    V(1,-2,0) = {(1)(-2) - 2(-2)^2 - 0} = -10

    speed = |-10 m/s| = 10m/s

    But I am guessing I am missing something to do with a sqroot
     
  11. Feb 8, 2015 #10

    SteamKing

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    Remember, V = xy i - 2y^(2)j + tyzk is a vector. That's what those i, j, k mean. Absolute value has no meaning here. You want to calculate the magnitude.
     
  12. Feb 8, 2015 #11
    Sigh, right. I knew there was supposed to be a sqroot in there.. Haven't done magnitude in a while.
     
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