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Question On Displacement - Velocity-Time Graph

  1. Mar 19, 2008 #1
    [SOLVED] Question On Displacement -- Velocity-Time Graph

    When calculating displacement, if you have a graph that starts with a non zero velocity, do you calculate the area from zero still, or do you calculate only from where it starts?

    For instance, a linear problem with a velocity starting at 20m/s 0s and goes to 30m/s 2s.

    Do I go Area=1/2bh=1/2(30m/s-20m/s x 2s-0s)=10m and then.. this is the part, do I calculate the left over rectangle under it even though the velocity started at a non zero number? 20m/s x 2s=40m

    40m+10m=50m, or does it just stay with where the linear velocity started (10m)?
     
  2. jcsd
  3. Mar 20, 2008 #2
    Displacement is independent of the path taken.
    It depends only on the initial and the final positions.

    Using the equation.:

    v^2=u^+2as

    we get s= 50m

    It does not matter if the particle started with a non zero velocity.You must include that part of the rectangle also for calculating the displacement.The negative part of the graph does not mean you have to reduce that from the final answer.It must be added to get the final Result.
     
  4. Mar 21, 2008 #3
    Thank you very much!
     
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