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Acceleration when velocity is a function of distance

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A car has a variable veloctiy given as a funtion of distance s by 1 +2s. What its accleration when it has travelled 10m and how far has it travelled in 4s.

    2. Relevant equations

    [tex]a(s)=v\frac{dv}{ds}[/tex](i think?)


    3. The attempt at a solution

    accleration at 10m

    [tex]a(s)=v\frac{dv}{ds}=(2s+1)(2) = (4s+2)[/tex]

    [tex] a(10) = 42ms^-2[/tex]

    however i am unsure how to do the second part as velocity is already a function of distance, is it solved using the chain rule?
    Last edited: Apr 29, 2008
  2. jcsd
  3. Apr 29, 2008 #2


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    Welcome to PF!

    Hi Fyei! Welcome to PF! :smile:

    Nice solution for the first part! :biggrin:

    In the second part, the 4s means 4 seconds, doesn't it?

    You'll have to solve a differential equation in t. :smile:
  4. Apr 29, 2008 #3
    yes s means seconds, sorry about the confusion

    and differential in t, you mean something along the lines

  5. Apr 29, 2008 #4


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    Hi Fyei! :smile:

    No … more like ds/dt = 1 + 2s. :smile:
  6. Apr 29, 2008 #5
    I tried to do that but I must be solving it wrong

    ds/dt = 1 + 2s


    ds/dt-2s = 1

    now in the form dy/dx+Py=Q

    therefore can be solved with intergrating factors

    thefore If = [tex]e^{\int-2ds} = e^{-2s}[/tex]





    which is obviously wrong
    Last edited: Apr 29, 2008
  7. Apr 29, 2008 #6
    Separate and integrate if you want s(t)--

    [tex]\int ds/(1+2s)=\int dt[/tex]

  8. Apr 29, 2008 #7
    Separation of variables is easier. your method should work but in this equation:


    - you should have had t instead of s as the dependent variable
    - you forgot about the integration constant
    - the integral of [itex] e^{-2t} [/itex] is [itex] -\frac{1}{2} e^{-2t} [/itex]
  9. Apr 29, 2008 #8
    thank you, I'm feeling a bit silly at all my school boy errors
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