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Particle Problem Inst. Velocity=Avg. Velocity?

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle moves along the x-axis so that at time t≥0 its position is given by:

    x(t)= t^3-2t^2-4t+6.

    For what value(s) of t where 0≤t≤4 is the particle's instantaneous velocity the same as its average velocity on [0,4]?


    3. The attempt at a solution
    So I got the average velocity:

    v(4)-v(0)/ (4-0) = 22-6/(4-0) = 4 = avg velocity for v(t)

    So then would I solve the derivative of the position function to find what value of t in the derivative would make x'(t)= 4?

    Thanks in advance.
     
  2. jcsd
  3. Oct 15, 2009 #2

    Dick

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    That's exactly what you should do.
     
  4. Oct 15, 2009 #3
    so x'(t)= 3t^2-4t-4

    so 3t^2-4t-4 = 4

    = 3t^2-4t-8=0

    Solve for t using the quadratic formula, where I get a positive and negative number. the positive number would be the only number that could fit on this interval. Would that be all?

    Is there anything else I have to do with this problem?

    Thanks
     
  5. Oct 15, 2009 #4

    HallsofIvy

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    But why did you change from "x(t)" to "v(t)" for the postion function?
     
  6. Oct 15, 2009 #5
    Oops. the average velocity should be x(t) not v(t). Just got caught up with the "v"'s.
     
  7. Oct 15, 2009 #6

    Dick

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    That's all.
     
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