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Motion in one direction

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A truck on a straight road starts from rest, accelerating at 2.0 m/s^2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0s.
    What is the average velocity of the truck for the motion described?

    2. Relevant equations
    Average velocity = (Velocity 1 + velocity 2 + velocity 3) / 3?
    [s(t2) - s(t1)] / (t2- t1)

    3. The attempt at a solution
    I tried using the first equation stated above, but received an incorrect answer. I know that for specifically average velocity, I have previously used s(t2) - s(t1) / t2- t1, with the t's representing a specific time. However, I cannot seem to apply that equation to this question.
  2. jcsd
  3. Sep 19, 2007 #2


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    It's not as simple as that - you need to integrate the speed while the truck is accelarating or braking.
    Try plotting the speed against time on a graph for each second and you will see.
  4. Sep 19, 2007 #3
    I plotted, however, I do not see how I can derive the average velocity from this chart, unless it is simply 20m/s, which I highly believe to be incorrect.
  5. Sep 19, 2007 #4


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    The average time is the area under the curve divided by total time.
    So just work out the area of the two triangles and the sqaure section and divide by the total time.

    It's sort of intrgration - but you don't actually need to understand that.
    I misread the question, with constant accelaration it's easier because the curve is just straight lines.
  6. Sep 19, 2007 #5
    Brilliant! Thank you so much!
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