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Deriving displacement, velocity and acceleration projecticle motion

  1. Aug 20, 2009 #1
    1. An object of mass, M, is projected into the air with an initial vertical component of velocity, Vo. If the air resistance is proportional to the instantaneous velocity, with the constant of proportionality being, K, derive the equations for the displacement, velocity and acceleration as functions of time.

    appologies for the way this is wrote if anyone has a better way of notating this it would be much appreciated.

    my work
    m dv/dt=(-mg-kv)

    so first I got my V's on one side and my T's on the other

    mdv/(mg+kv)=-dt then added the integral sign to this equation and integrated it too get

    m/k ln(mg+kv)=-t+c

    next I tidied it up

    mg+kv=e^-((k/m)*t)+c that goop in the middle is meant to be exp to the power of whats in the brackets

    mg+kv=ce^-((k/m)*t) movin the c for more tidiness

    now I tried to get V on its own


    sub in t=0 and V=Vo

    Vo=C-mg/k therfore c is

    C=Vo + mg/k

    subbing back into my previous I got

    v=(Vo + mg/k)e^-((k/m)*t)-mg/k that is my velocity part right

    thats pretty much it for me any help would be appreciated for displacement and acceleration[/b]
    Last edited: Aug 20, 2009
  2. jcsd
  3. Aug 20, 2009 #2


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    Staff: Mentor

    I'm not tracking this part:

    Integrated with respect to what? You have a differential equation... what method are you trying to use to solve this DiffEq?
  4. Aug 20, 2009 #3
    Sorry I just took that part from my notes I just have first


    then adding the intergration symbol

    integrate mdv/(mg+kv)= integrate -dt

    So I am quessing it's with respect to time as its to be as a function of time I really have only a vague idea whats going on here.

    I believe its a separable equation yes no
    Last edited: Aug 20, 2009
  5. Aug 20, 2009 #4


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    Staff: Mentor

    Ah, confusing lecture notes, eh? Okay, writing your original post's first equation in the traditional form for inhomogeneous differential equations:

    [tex]m\frac{dv(t)}{dt} + kv(t) + mg = 0[/tex]

    You can then solve it using the traditional techniques discussed here:


    Hope that helps.
  6. Aug 21, 2009 #5




    ah I cant for the life od me figure this fancy way of displaying the functions like you would on paper so I will gracefully withdraw.
  7. Aug 22, 2009 #6


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    Staff Emeritus
    Science Advisor
    Homework Helper

    Use a forward slash in the [noparse][/tex][/noparse] tag:




  8. Aug 22, 2009 #7
    Thanks for the tip man
  9. Aug 22, 2009 #8







    [tex]mg+kv=\exp^\frac{-k t}{m}+c[/tex]

    [tex]v=C\exp^\frac{-k t}{m}-\frac{mg}{k}[/tex]

    [tex]v=v0[/tex] [tex]t=0[/tex]




    [tex]V=(v0-\frac{mg}{k})\exp^\frac{-k (0)}{m}-\frac{mg}{k}[/tex]
    Last edited: Aug 22, 2009
  10. Aug 22, 2009 #9













    [tex]V=(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}[/tex] Velocity

    displacement is [tex]s=\int vdt[/tex]

    [tex]s=\int(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}dt[/tex]

    [tex]s=\frac{-m}{k}(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}t-c[/tex]


    [tex]0=\frac{-m}{k}(v0+\frac{mg}{k})\exp^\frac{-k (0)}{m}-\frac{mg}{k}(0)+c[/tex]




    [tex]s=\frac{-m}{k}(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}t+\frac{m}{k}(v0-\frac{mg}{k})[/tex]

    [tex]s=\frac{-m}{k}(v0+\frac{mg}{k})(1-\exp^\frac{-k t}{m}-\frac{mg}{k}t[/tex] Displacement
    Last edited: Aug 22, 2009
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