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Non constant acceleration

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A car accelerates with a(s) = k·sn m/s2, from 0 m/s. What is the velocity of the car when s=1,3m?

    2. Relevant equations

    k = 7,96 och n = 0,454

    3. The attempt at a solution

    I know I have to intergrate in some way, but I really don't know how to start? I don't have the time?
     
  2. jcsd
  3. Apr 24, 2012 #2
    What are you trying to find? Distance or velocity?

    for velocity: [tex]v = v_{0} + \int_{t_{0}}^{t}a(t)dt[/tex]

    for distance: [tex]x = x_{0} + \int_{t_{0}}^{t}v(t)dt[/tex] - using the velocity found above.
     
  4. Apr 24, 2012 #3
     
  5. Apr 24, 2012 #4
    Yes you should be able to use the integral, although I'm having a bit of trouble understanding the functional relationship of your acceleration. What is the dependent variable? I'm not sure what "s" is.
     
  6. Apr 24, 2012 #5
    s is the distance travelled
     
  7. Apr 24, 2012 #6
    Oh ok, well in that case you have to use a little trick:
    [tex]a = \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/tex]

    then the integral becomes:

    [tex]\int vdv=\int adx[/tex]

    Hopefully you can figure it out from there.
     
  8. Apr 24, 2012 #7
    I'm sorry but I don't understand :(
     
  9. Apr 24, 2012 #8
    Hmm...if a(s)=k*s*n and the car starts at s=0 with v=0, then the car will have an acceleration of 0 and will just stay stationary. Am I missing anything?
     
  10. Apr 24, 2012 #9
    Im sorry!!

    I wrote the expression wrong, it should be:

    A(s)=k*s^n m/s^2
     
  11. Apr 24, 2012 #10
    No, this only means that v_0 = x_0 = 0. As long as the acceleration is nonzero, the particle will have nonconstant motion.

    But let s = x and integrate a over x and you'll have your answer.
     
  12. Apr 25, 2012 #11
    But the acceleration is a function of the displacement such that if the displacement is zero, then the acceleration is also zero. It also happens that the velocity is zero initially, so the displacement would never change from zero and hence the acceleration would also never change from zero. Although this is only if the initial displacement is zero, which the question never actually states (it never tells us initial displacement).
     
  13. Apr 25, 2012 #12
    The answer should be 4,0m/s. I got 8m/s when I integrated 1 time. What did i do wrong?
     
  14. Apr 25, 2012 #13
    What do you mean by 'integrated 1 time'? Integrating a(s) from s=0 to s=1.3 is not going to give a velocity, the units will be [itex]\frac{m}{s^2}*m = \frac{m^2}{s^2}[/itex], while velocity has units of [itex]\frac{m}{s}[/itex].

    I'm not actually sure what grindfreak means when he says [itex]\int vdv=\int adx[/itex] then "integrate a over x and you'll have your answer", which seems to be implying that [itex]\int vdv=\int vdt[/itex], which is not correct as far as I know.

    But his application of the chain rule does give something useful:
    [itex]a =\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}[/itex]
    If you'll recall, the question tells us what a is in terms of s:
    [itex]a = k*s^n[/itex]
    So:
    [itex]v\frac{dv}{ds}=k*s^n[/itex]
    Which is a first order non-linear ordinary differential equation for v(s) from which we can obtain our desired answer v(s=1.3), which leads me to believe that there must be some other way to do this question because I only started learning that kind of maths in a second year university maths unit. I know I'm not suppose to give you the answers, but you could have just put this straight into Wolfram Alpha anyway, like this, so I might as well just show what it spat out:
    [tex]v(s)=\sqrt{\frac{2*(c*n+c+k*s^{n+1})}{n+1}}[/tex]
    To find the constant c, remember that v(0)=0 from the information in the question.
     
  15. Apr 25, 2012 #14
    Here's how it should integrate. Since the initial values for both velocity and displacement are 0, the right integral looks like:
    [tex]\int_{0}^{s}ads^{'}=k\int_{0}^{s}(s^{'})^{n}ds^{'}=\frac{ks^{n+1}}{n+1}[/tex]
    The right integral is:
    [tex]\int_{0}^{v}v^{'}dv' = \frac{1}{2}v^{2}[/tex]
    Now don't be confused about the primes, it's just used to discern between the limits and the variables you're integrating over (you'll see this quite a bit in more upper level courses).

    Now, equating these two answers together you get:
    [tex]\frac{ks^{n+1}}{n+1} = \frac{1}{2}v^{2}[/tex]
    Now solve for v and plug in your numbers and you're done :D

    *Just realized EmittingLight came up with the same conclusion.
     
    Last edited: Apr 25, 2012
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