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Speed of a partical as it approaches stagnation point

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    s = 0.6ft upstream @ t=0

    s = 0.6 e^(-0.5 t)

    where t is in seconds and s is in ft

    a) Determine the speed of the particle as a function of time Vpart(t)
    b) Determind the speed of the of the fluid as a function of the position along the streamline V=V(s)
    c) Determine the fluid acceleration along the streamline as a function of position as = as(s)

    (solutions provided by instructor)

    3. The attempt at a solution

    a) Take the derivative of s with respect to t (simple enough)

    V(t) = -0.3 e^(-0.5t) ft/s <-correct solution

    b) solution = -0.5s ft/s

    So I thought that I would solve the first given equation for t and differentiate, however I was told that all I needed to do was put in t = t(s) into (a).

    My question is how does this work and how do you come up with -0.5s. I am sure it is simple and right in front of me, but I can't get my head around it.

    c) solution = 0.25s

    Material derivative of (b) (using the correct answer)

    DV/DT = dV/dt + u dv/ds
    = 0 + (-.5)*(-.5s)
    = 0.25s <- Correct solution

    Can anyone help be clear up part b?
     
  2. jcsd
  3. Sep 14, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    v = -0.5*0.6e^(-0.5t)
    = -0.5*S
    Now find dv/ds.
     
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