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Given acceleration relation, find variables

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    The acceleration of a particle is defined by the relation a=0.4(1-kV), where k is a constant. Knowing t=0 the particle starts from rest at x=4m and that when t=15s, V=4 m/s, determine....
    a)the constant k
    b)the postion when V=6m/s
    c) the maximum velocity of the particle


    2. Relevant equations



    3. The attempt at a solution

    a=0.4(1-kV)
    0.4(1-kV)=dV/dt.........im lost after this, can someone just simply point me in the right direction.
     
  2. jcsd
  3. Jan 7, 2010 #2

    ehild

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    This is a differential equation for v, you can solve it by separation of variables:

    [tex]\int{\frac{dv}{1-kv}=\int{0.4 dt}[/tex]

    ehild
     
  4. Jan 7, 2010 #3
    let me see if I can take it from here if I can't ill post what work I have done.
     
  5. Jan 7, 2010 #4
    I am stuck again, I solved the right side of the equation you created and got .6s. But on the left side i solved by substitution. Where u=1-kV du=k dV so the left side becomes....

    (1/k)ln(1-kV) over the intgral of V0=0 V1=4 m/s .....after everything cancels i have (1/k)(ln4k)=.6 is this right? Sorry if it is hard to follow.
     
  6. Jan 7, 2010 #5
    which is wrong because those sides do not equal each other when I plug in a number for k
     
  7. Jan 7, 2010 #6
    i tried solving with out using u, and i got ln(0) at one point which is impossible to solve so i am once again stuck.
     
  8. Jan 8, 2010 #7

    ehild

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    You need the function v(t) to get x(t). Integrate with respect to v between 0 and v(t), and between 0 and t on the right side.

    [tex]
    \int_{0}^{v(t)}{\frac{dv}{1-kv}=-\frac{1}{k}(\ln(1-kv(t))-\ln(1-0))=0.4t
    [/tex]



    [tex]
    -1/k\ln(1-kv(t))=0.4t\rightarrow 1-kv(t)=\exp(-0.4kt)\rightarrow v(t)=1/k(1-\exp(-0.4kt))
    [/tex]

    At t=15 s, v=4 m/s, so 4=1/k(1- exp(-6k)), k=0.25(1-exp(-6k)) It is a bit tricky, to get k from here. Apply some numerical method.


    ehild
     
  9. Jan 8, 2010 #8
    There are two things you can do from here:
    1) Solve for k numerically via computer box
    or
    2) You can get an approximation by expanding the exponential with a Taylor Series.
    [tex] e^{-0.4 k *15} \approx 1 - 0.4 k*15 + \frac{(-0.4 k*15 )^2}{2} [/tex]
    Throw this in your eq. and you should get a decent approximation for k.
     
  10. Jan 8, 2010 #9

    ehild

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    it is: k=0.1458...

    ehild
     
  11. Jan 9, 2010 #10
    I was right, i just couldnt figure out how to solve. That mkes me feel better! Thanks for everything you guys!
     
  12. Jan 9, 2010 #11
    for the second part I am stuck.....here is my work.... this is problem 11.25
    dynamics047.jpg
     
  13. Jan 9, 2010 #12

    ehild

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    I do not understand your last line. It is all right up to then.

    After integration, you got:

    [tex]0.4x-1.6=1/k^2[1-kv-ln(1-kv)]_0^6[/tex]

    Substitute first v=6, then v=0, and subtract. But how did you get that 48.32?

    1-6k-ln(1-6k)=0.1258+2.07317=2.1989, and divided by k^2: 103.6

    ehild
     
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