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Help with Calc/Physics Question - Integration Methods

  1. Jan 29, 2014 #1
    1. The Problem
    The acceleration function for a particle is given as a(t)=60t+18. v(1)=5, find the distance traveled by the particle from t=0 to t=2

    3. The attempt at a solution

    First, I integrated the acceleration function to get the velocity function, I found:

    v(t)=30t2+18t+c
    v(1)=5=30+18+C
    5=48+C
    C=-43
    So v(t)=30t2+18t-43

    Well, now I scratched my head a little bit. I know that the distance traveled is the absolute value of the integral of this function on my interval, but with the -43 hanging out on the end of this equation, I will not be able to factor. I approached my prof. about this and he said there was a formula that I should use, and that the quadratic equation solution that I produced wasn't necessary, and that this problem could be done on a scientific calculator.

    What am I missing?

    (I had hoped that v(1) would be 0 so it would split right down the middle :cry: )

    TIA for the help.

    Chris
     
  2. jcsd
  3. Jan 30, 2014 #2

    SteamKing

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    Well, I'm scratching my head, too. You apparently know that integrating the acceleration w.r.t. time will give velocity. How is velocity related to knowing distance and time? From what I recall of rectilinear motion, the answer doesn't involve the quadratic formula.
     
  4. Jan 30, 2014 #3
    If you find the net area under the velocity curve (absolute value), then you end up with your distance traveled.

    To determine where the velocity curve may change from positive to negative, we need to know where velocity is zero. This would allow one to create two separate integrals to find the absolute value of the area under the curve.
     
  5. Jan 30, 2014 #4

    SteamKing

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    I'm still scratching my head. The OP just wants the distance travelled from t = 0 to t = 2. It doesn't say distance travelled w.r.t. some fixed point, just distance travelled. I think you are overthinking this problem.
     
  6. Jan 30, 2014 #5

    vela

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    I don't think you're missing anything. I'm curious to know what your professor has in mind.
     
  7. Jan 30, 2014 #6
    What makes you feel like you have to factor something?

    Chet
     
  8. Jan 30, 2014 #7

    SammyS

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    Let's see:

    The particle leaves its initial position, which we can assume is the origin, then travels in the negative direction to position x1 where it changes direction passing the origin and arrives at position x2 2 seconds after leaving the origin. The distance traveled is arrived at quite easily from this.

    Using a graphing calculator and the position function, one can easily answer the question being asked here. Perhaps that's what the prof. has in mind.
     
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