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Using quadratic interpolation in matlab how ?

  1. Apr 17, 2008 #1
    [​IMG]

    Can anyone explain to me please how this problem is solved ?
     
  2. jcsd
  3. Apr 17, 2008 #2

    Integral

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    Two ways, take your pick.

    Set v(t) = 35, then use the quadratic formula. OR

    set t = each of the possible answers, which is closest to 35?
     
  4. Apr 17, 2008 #3
    Well this is basically what i have done But im not sure why im getting to positive values for t ?
    I would like to :
    Please feel free to tell me if the quadratic equation i used is wrong:-

    a=8.66; b=-349.67;c=3523.18;
    discr=qrt(b^2-4*a*c)
    qf1=(-b+discr)/(2*a);
    qf2=(-b-discr)/(2*a);
    qf=[qf1 qf2]

    I got qf=

    21.0571 19.3025


    Is the function in matlab defined as:
    Ax^2 +bx + c

    or

    Ax^2 -bx +c
     
    Last edited: Apr 17, 2008
  5. Apr 17, 2008 #4
    Anyone please ? I have to submit this in 2 days please help...
     
  6. Apr 17, 2008 #5
    Found it finally :


    [​IMG]

    This line basically i dont understand:-

    p(3)=p(3)-35

    Why minus ? I ued trial and error so i dont really know how i got it. May anyone explain please ?
     
  7. Apr 17, 2008 #6
    In addition to what i posted:

    Ok now i can see i figured out the answer But what does not makes sense to me is that the velocity 35m/s is obtained twice between 18 & 22 seconds , 22 & 24 seconds. ! So theoretically i should be having 2 answers. Is there a Physical meaning of why im having one answer ?
     
  8. Apr 19, 2008 #7
    Anyone please help me submission is tommrow
     
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