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The average velocity of the point - numerical value of t

  1. Feb 7, 2015 #1
    1. The problem statement, all variables and given/known data

    During the first t seconds of motion, the average velocity of the point is zero. What is the numerical value of t?

    2. Relevant equations


    3. The attempt at a solution

    I don't understand this question. I understand that the average velocity is zero if the displacement in the denominator is zero. E.g. if a body is moving in a circle. Or if a body moves to the right and then to the left and distances cancel out. But how could you give the numerical value of t?
    I thought t tended to zero (as a limit): 0.0000000001 but it is wrong.
    Am I overlooking something obvious?
     
  2. jcsd
  3. Feb 7, 2015 #2

    gneill

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    Staff: Mentor

    I think we're missing some context here. Is this part of a larger question? (The problem mentions "the point", not "a point", so it seems to be referring to something previously defined).
     
  4. Feb 7, 2015 #3
    Oh yes, silly me. This is an equation from the previous question. I will do it now. That's English for you. Many thanks.

    x(t)=2t^2−3t−4
     
  5. Feb 7, 2015 #4

    gneill

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    Staff: Mentor

    So at time t=0, what's the initial position according to your equation?

    When is the position again that value?
     
  6. Feb 7, 2015 #5
    It is simple now.
    I need the displacement = 0 so

    at t = 0 initial position x(i) is -4.

    The displacement:
    (2t^2−3t−4)-(-4)=0
    2t^2−3t=0
    t=3/2

    I can't assume that t=0 because you can't divide by 0. (the average velocity = displacement/time elapsed)

    2*(9/4)-3*(3/2)=9/2-9/2=0 :)

    Many thanks. :) :) :)
     
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