1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?)

    [tex]a=\frac{d^2s}{dt^2}[/tex]


    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    [tex]\frac{dv}{dt}=\frac{d^2s}{dt^2}[/tex]?
     
  5. Apr 29, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    therefore

    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]

    [tex]s=\frac{\int1*e^{-2s}ds}{e^{-2s}}[/tex]

    therefore

    [tex]s=\frac{1*-2e^{-2s}}{e^{-2s}}[/tex]

    [tex]s=-2[/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]

    [tex]s=\frac{1}{2}(1-e^{2t})[/tex]
     
  8. Apr 29, 2008 #7
    Separation of variables is easier. your method should work but in this equation:

    [tex]
    s=\frac{1*-2e^{-2s}}{e^{-2s}}
    [/tex]

    - 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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?