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Calculate the average velocity of the car

  1. Nov 14, 2015 #1
    1. The problem statement, all variables and given/known data

    A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=bt2−ct3, where b = 3.00 m/s2 and c = 0.130 m/s3 .

    Calculate the average velocity of the car for the time interval t= 0 to t= 10.0 s

    2. Relevant equations

    vav = V2-V1/T2-T1


    3. The attempt at a solution

    x(t) = 3.00 t^2 - .130 t^3
    differentiation \/
    v(t) = 6.00t - .39 t^2
    a(t) = 6.00 - .78t

    V(0) = 6.00 (0) - .39 (0)
    v(10) = 6.00 (10) - .39 (10)^2
    = 60.0 - 39.
    = 21

    21-0/ 10-0
    21/10
    2.1

    it tells me its wrong, why?
     
  2. jcsd
  3. Nov 14, 2015 #2

    Astronuc

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    Staff: Mentor

    What are the units of (V2-V1)/(T2-T1) or of velocity/time?

    What are the units of distance divided by time?

    What are the units of speed or velocity?
     
  4. Nov 14, 2015 #3

    BvU

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    Gold Member

    The average velocity is not the same as the change in velocity divided by time (because that is the average acceleration).
     
  5. Nov 14, 2015 #4
    oh right, I forgot, so I tried dividing it my 2 instead of the time interval and i get 10.5, not the answer either
     
  6. Nov 15, 2015 #5

    BvU

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    That's because the velocity doesn't depend linearly on time: ##\Delta v\over 2 ## doesn't work then.
    So all you have to go by in this exercise is the definition of average velocity: $$<\vec v> \equiv {\Delta \vec x \over \Delta t}$$ and with the given information it turns out to be a fairly easy exercise :smile:

    --
     
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