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Homework Help: Finding the distance from a velocity-time graph

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data
    The velocity graph of a particle moving along the x-axis is shown. The particle has zero velocity at t=0.00 s and reaches a maximum velocity, vmax, after a total elapsed time, ttotal. If the initial position of the particle is d0=8.29 m, the maximum velocity of the particle is vmax=47.9 m/s, and the total elapsed time is total=43.2 s, what is the particle's position at t=28.8 s? The graph that is provided with the question shows a constant slope on the

    find the final distance when t=28.2s

    2. Relevant equations
    a(avg)=delta v/t
    v(avg)=delta d/t

    3. The attempt at a solution
    So since the slope of the vt graph is constant, I found v(avg) by dividing the final+initial by 2.

    V(avg) = d2 - d1 / t
    23.95m/s = d2 - 8.29m / 28.8s
    d2 = 698m

    I still got the answer wrong... I don't really care for the correct answer, I'm just worried that I'm really misunderstanding some basic concepts.
  2. jcsd
  3. May 26, 2009 #2


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    Gold Member

    Well, without really thinking it over, based on what you've given us we have uniform acceleration, a known final velocity, a known initial velocity, known initial postion, time elapsed etc.

    I think an ideal canidate for this sort of problem looks like the equation: s = at2/2 + v0t + s0.

    The reason your attempt did not work is that your average velocity value applies over the entire elapsed time and does not represent the average velocity in the time interval [0,28.8].
  4. May 26, 2009 #3
    This is a very important point that regularly catches people out.

    If the slope is constant, then you know that the acceleration is constant, so you can use the SUVAT equations.

    jgens pointed out the relevant one to use - and don't forget to add the extra displacement due to the initial position (which is included in jgens' equation).

    The acceleration can be found from the equations that you have shown.
    Remember that if it is constant, then the constant value is equal to the average value.
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