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Non constant acceleration

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A car accelerates with the acceleration a(s)=k*s^n m/s^2
    s= distance in m
    k = 5,8
    n=0,7351
    The car starts the acceleration from 0 m/s

    How far has the car travelled when it reaches the speed 4,61m/s?

    3. The attempt at a solution
    When i anti derivate one time i will get dv/dt. Do i put in the distance after that?
     
  2. jcsd
  3. Apr 25, 2012 #2
    adx = adv

    Where x is the distance "s"
    so I get this:

    (1/2)*(x^2) = v2^2 - V1^2

    Where v2 is the final speed and v1 is the starting speed=0..

    Am i right or wrong?
     
  4. Apr 25, 2012 #3

    ehild

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    What do you mean? dv/dt =a, the acceleration.

    You have a differential equation dv(s(t))/dt=k*s^n. Write up the left-hand side with applying the chain rule:

    dv/dt=dv/ds ds/dt. ds/dt=v, so v dv/ds =ks^n. This is a differential equation for v(s), easy to solve.


    ehild
     
  5. Apr 25, 2012 #4
    So I just plug in my values here, or do i have to antiderivate first?

    v dv/ds =ks^n
     
  6. Apr 25, 2012 #5
    So I just plug in my values here and solve for s, or do i have to antiderivate first?

    v dv/ds =ks^n
     
  7. Apr 25, 2012 #6

    ehild

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    How do you want to solve for s? Show your attempt.

    ehild
     
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