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How to make sense of this velocity-time graph question?

  1. Feb 4, 2016 #1
    1. The problem statement, all variables and given/known data

    http://m.imgur.com/tAbHL6S



    2. Relevant equations

    Velocity=displacement/time

    3. The attempt at a solution

    I thought B at first because they start at the same position, so their relative velocities would determine their relative positions, but I guess not since the answer is D. I know that the area below the lines is also equal to the displacement, but how can you even be sure that they pass each other if you don't know what their velocities are?
     
  2. jcsd
  3. Feb 4, 2016 #2

    RUber

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    Homework Helper

    You can see that the velocity from 0 to 4 is a straight line, and T. Swift has a constant velocity. So, since the velocity plots cross at 2 seconds, you know that the average velocity from 0-4 is the same for both of you...which means displacement at t=4 will be the same...indicating you will be at the same location at t=4.

    So, do you pass her? Yes...since at t=4, you are moving faster.
    Use the same logic for the last 4 seconds...seeing that you start the interval at the same location.
     
  4. Feb 4, 2016 #3

    Merlin3189

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    Gold Member

    You are right to consider the area under the graph: that tells you the distance run.
    Can you give a commentary on these areas at 2 second intervals, 2,4,6,8?

    Edit: and do you know the area of a trapezium?
     
  5. Feb 4, 2016 #4
    You started slower than swift, so where the two lines intersect at T=2, you're moving at the same speed, but you haven't caught up yet.. you are still accelerating.. you need to double that time to actually catch up lost ground and pass her.
    Distance covered is the area under the curve, not the height of the curve.

    The question is a little ambiguous.. it never explicitly says if you started at the same point in space... if you didn't, it's impossible to know when anyone passes the other...
    D is still the best answer, B is definitely not.
     
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