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Calculating Displacement

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm just confused about this. I know that displacement=df-di, but using that formula didn't give me the correct displacment in this case.

    In a 100m sprint, a runner starts from rest and accellerates to 9.6m/s[W] in 4.2 seconds. The runner runs at a constant velocity for the rest of the race, what is the total time?

    I am trying to solve for the time that it takes for the runner to go from their first displacement, when they are accelerating, to the end of the race when they're running at a constant velocity. In a previous calculation, the first displacement is 20m[W]. So, the final displacement- from when the runner is travelling at a constant velocity until they finish the race, should be 80m[W] because the whole sprint it 100m long. But, if you do displacement=d2-d1 and try to solve for d2 that way, you get a displacement of 120m[W]. For my calculations, I plugged in all values in the west directions with negative signs.
    Conventions-- West= negative, East= positive

    How can d2 be 120 m[W] if the sprint itself is only 100m long. Can someone please explain this to me?

    2. Relevant equations

    3. The attempt at a solution
    I showed that above, under the problem statement. Can someone tell me where I went wrong? I'm really confused.
  2. jcsd
  3. Feb 11, 2016 #2
    Since he's already run -20 shouldn't
    d2 = -100 - (-20) ?
    or D = d2 - d1 where D is the distance remaining.
  4. Feb 11, 2016 #3
    Shouldn't you be able to use d=df-di and solve for df by isolating for it. If you do that, you don't get the correct answer.
  5. Feb 11, 2016 #4


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    I think what J Hann is trying to point out is that you are told the runner is running west after he starts the race. The race has a total distance of 100 m, so saying that the runner runs for 120 m is a little ... odd.

    If you draw a diagram of the race, making the starting point the origin, this should help you visualize what happens while the runner accelerates and when he crosses the finish line.
  6. Feb 12, 2016 #5


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    I think your equation for the displacement in the second phase should be...

    Displacement = Dfinal - Dinitial
    = -100 - (-20)
    = -80
  7. Feb 12, 2016 #6


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    The problem statement does not appear to specify which direction he is running so why make it harder by choosing a negative direction?
  8. Feb 12, 2016 #7


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    Actually is see it says he has a positive Velocity and positive displacement at the end of the acceleration phase.
  9. Feb 12, 2016 #8
    Using your sign conventions
    -100 = d1 + d2 where d1 and d2 are the distances traveled.
    Using d2 - d1 you would be referencing coordinates not the distance traveled.
  10. Feb 14, 2016 #9
    Isn't the formula displacement=d2-d1, I subbed in the coordinates for the given points
  11. Feb 15, 2016 #10


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    That depends on what d1 and d2 stand for. If d1 is an initial displacement and d2 is the final displacement then the extra displacement is indeed d2-d1. To apply your formula, you need to set d2=100, as that is the final displacement. With d1=20, the extra displacement will be the length at constant velocity.
  12. Feb 15, 2016 #11


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    You need the clear definition of displacement. It can not be defined by itself.
    Displacement is the change of position. If an object moves along the x axis from position xi to the final position xf, then the displacement is d = xf-xi.
    If the object moves from xi to x1 and then to x1 to x2, then the first displacement is d1=x1-xi the second displacement is d2=xf-x1, and the total displacement is d=xf-xi=d1+d2.
    In this case, you took xi =0 and xf=-100m.The total displacement is d=-100m. You calculated the first displacement, it was -20m. What is the second displacement?
    Note that the velocity is also negative if you take westward direction negative.
  13. Feb 15, 2016 #12


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    That's why in an earlier post I used "final" and "initial". I guess I should have spelled it out a bit more..

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