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Homework Help: Kinematic question

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is projected vertically upwards, with an initial speedo of u, in a medium which exerts a resistance of magnitude mkv, where k is a positive constant and v is the speed of the particle after time t. Express, in terms of k, g and u, the time taken for the particle to reach its greatest height


    2. Relevant equations
    u - gt = 0


    3. The attempt at a solution
    u - gt - kvt = 0
    I don't know how to express v in terms of u, g and k.
     
  2. jcsd
  3. Apr 21, 2010 #2

    kuruman

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    Can you find an expression for the acceleration of the particle?
     
  4. Apr 22, 2010 #3
    a = g + kv?
     
  5. Apr 22, 2010 #4

    kuruman

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    Basically yes, but I would put a negative sign in front of both terms. Now can you write an expression for the rate of change of velocity with respect to time?
     
  6. Apr 22, 2010 #5
    Isn't dv/dt = a = -g - kv?
     
  7. Apr 22, 2010 #6

    kuruman

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    Correct. Now can you solve this equation to find v(t)?
     
  8. Apr 22, 2010 #7
    dv/dt = -g - kv
    dv = -g - kvdt
    Integrating both side
    v = -gt - kvt
    v = -gt / (1 + kt)

    I just learnt about differential equation, I am not sure if this is correct
     
  9. Apr 22, 2010 #8

    kuruman

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    It is not correct. You need to separate the two variables before you integrate. This means that you need to rearrange the equation algebraically so that the left side has only v (and constants) in it and the right side has only t (and constants) in it.
     
  10. Apr 22, 2010 #9
    Sorry, I am quite new in differential equation. I don't know how to rearrange the equation such that v is on one side and t on the other side. Can you please teach me? Thanks
     
  11. Apr 22, 2010 #10

    kuruman

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    This part of the process is "algebra" not "differential equations".

    [tex]\frac{dv}{dt}=-(g+kv)[/tex]

    Cross multiply to get

    [tex]\frac{dv}{g+kv}=-dt[/tex]


    Voila. The variables have been separated. Now you can integrate. Differential equations are not as scary as they sound. Now you can integrate.
     
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