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Why Velocity is different in different equations?

  1. Dec 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate velocity. x=position. t=time

    2. Relevant equations
    x=16-12t+2(t^2)

    3. The attempt at a solution
    Derivative of x: v=4t-12
    Ok let's try:
    Equation of x: x=16-12t+2(t^2)
    Equation of v: v=4t-12

    With the equation of x: x(3)=-2 then v(3)=-2/3.
    Ok, but with the equation of v: v(3)=0.
    There are different results! Why?
    What am I doing wrong?
     
    Last edited: Dec 12, 2015
  2. jcsd
  3. Dec 12, 2015 #2

    Samy_A

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    Using your equations: x(3)=16-12*3+2*9=16-36+18=-2 and v(3)=0.
    How do you get v(3)=3/2?
     
  4. Dec 12, 2015 #3
    So x(3)=-2 in t=3.
    v=x/t
    v(3)=-2/3
    Sorry, I was wrong.
    But it keep being different
    v(3)=0: v(3)=-2/3
     
  5. Dec 12, 2015 #4

    Samy_A

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    You are computing two different quantities.
    The formula v=4t-12 gives you the velocity at t=3

    The formula (distance travelled)/time would give you the average speed.
    The formula (change in position)/time would give you the average velocity.
    Try it out, by computing how much the travelling object has moved from t=0 to t=3, and then dividing this by 3. Take into account that x(0)≠0.
     
    Last edited: Dec 12, 2015
  6. Dec 12, 2015 #5
    -13/3 is the average speed from t=0 to t=3 OK thanks a lot.
     
  7. Dec 12, 2015 #6

    Samy_A

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    How did you get -13?

    Not saying it is wrong, but I find something different:
    x(0)=16, x(3)=-2
    Change in position=-2-16=-18
    Average velocity = -18/3 =-6
     
  8. Dec 12, 2015 #7
    v is not equal to x/t. This is only correct if x =0 at t = 0 (which in your problem it is not) and if v is constant (which in your problem it is not).
     
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