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Velocity with air resistance help

  1. Sep 29, 2007 #1

    PHK

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    I need to find the velocity with the following information. V=Vo+AT and A=G-KV

    so how would i find what V equals from this: V=Vo+(G-KV)T
     
  2. jcsd
  3. Sep 29, 2007 #2
    Thats right. Is there something else to the question too? What velocity do you have to find?
     
  4. Sep 29, 2007 #3

    PHK

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    theres nothing else to the problem. i plugged in G-KV in for A in the first equation but now i have V=Vo+(G-KV)T. and i want to find velocity, but cant because there is velocity in the equation. so i want to know how to simplify the equation so i get V only on one side.
     
  5. Sep 29, 2007 #4

    PHK

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    did anyone figure it out yet?
     
  6. Sep 29, 2007 #5

    learningphysics

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    V=Vo+(G-KV)T

    V = Vo + GT - KVT

    V + KVT = Vo + GT

    etc...
     
  7. Sep 29, 2007 #6
    I think this is right:

    V=Vo+(G-KV)T = Vo+TG-TKV
    V-Vo=TG-TKV
    V-Vo+TKV=TG
    V+TKV=TG+Vo
    V(1+TK)=TG+Vo

    V=(TG+Vo)/(1+TK)


    [EDIT]oops, looks like learningphysics beat me to it.
     
  8. Sep 30, 2007 #7

    lightgrav

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    "Replacement by an equal expression" is a procedure that is used a lot.
    ( like you replaced the "A" by "G-kV" )
    "Multiplication, Distributive" ( Phoenix line 1 ) and
    its inverse , "Factoring" (Phoenix line 5) are the basis of proportion.
    When you add the negative of some term to both sides
    (so that term cancels the original term on that side)
    and/or divide both sides by the same factor
    (so as to "move the factor to the other side")
    ... there's always more than one path that you can take
    notice how Phoenix "undid" in line 4 , what he had done in line 2 ...
    it is okay to take more steps, so long as you keep the goal in mind.
     
  9. Sep 30, 2007 #8

    PHK

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    thanks the equation works. i got one question tho. how did you get from this V+TKV=TG+Vo
    to this V(1+TK)=TG+Vo?
     
  10. Sep 30, 2007 #9

    PhanthomJay

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    Where did you get the equation V = Vo + AT? This equation is derived from calculus for the special case of constant acceleration. In your case, the acceleration is changing with speed, and is therefore not constant.
     
  11. Sep 30, 2007 #10

    learningphysics

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    he factored out the V... try multiplying out: V(1+TK)... what do you get?
     
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