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Distance and displacement confusion

  1. Mar 31, 2016 #1
    1. The problem statement, all variables and given/known

    Two objects move with initial velocity -8.00 m/s xfinal velocity 16.0 m/s xand constant accelerations. (a) The first object has displacement 20.0 m. Find its accelera- tion. (b) The second object travels a total distance of 22.0 m. Find its acceleration. x

    2. Relevant equations

    Vf^2=vi^2 + 2ax


    3. The attempt at a solution
    (a) (-8)^2 = (16)^2 + 2a(22)
    a = 4.36m/s

    (b) 0 = (-8)^2+ 2a(11)
    a= -2.91m/s2

    (16)^2= (0)^2 + 2a(11)
    a= 0.73m/s2

    a = -2.91 +0.73
    a= -2.18m/s2
     
  2. jcsd
  3. Mar 31, 2016 #2
    Is this how it is done???
     
  4. Apr 1, 2016 #3

    PhanthomJay

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    You've got your Vf and Vi mixed up...and displacement is 20 m , not 22 m...what's the direction of the acceleration?
    the acceleration is constant, so you can't have 2 values. You incorrectly assumed it goes 11 m each way. The distance it travels from rest varies as the square of the final speed.
     
  5. Apr 2, 2016 #4
    Can you please explain fully how part b is done I dont understand
     
  6. Apr 2, 2016 #5

    jbriggs444

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    For part b, consider splitting the trajectory into two parts. One as the object slows from -8 m/s to zero. And one as the object accelerates from zero to +16m/s. How long does the second part last compared to the first?
     
  7. Apr 3, 2016 #6
    Thats what I done. I got two accelerations
     
  8. Apr 3, 2016 #7

    jbriggs444

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    See post #3. You split it in half. Why half? You are told that the acceleration is constant. That information allows you to determine what fraction of the total time elapses in each part.
     
  9. Apr 3, 2016 #8
    The question asks for acceleration and not time
     
  10. Apr 3, 2016 #9
    Can you please explain by an example. I dont understand what you saying
     
  11. Apr 3, 2016 #10

    jbriggs444

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    The problem says that the acceleration, whatever it is, is constant. So take it as a constant and call it "a".

    How much time elapses while the object slows down from -8 m/s to 0 m/s?
    How much time elapses while the object speeds up from 0 m/s to 16 m/s?
    Express both answers in terms of a.

    Edit: let me expound a bit on the strategy that I have in mind...

    In general terms, we are looking at an object that is moving first to the left and then to the right under a constant rightward acceleration.

    We want to know the acceleration, a. We know the starting velocity, ending velocity, total distance covered and the fact that the acceleration is constant. Ideally we could find a single equation involving starting velocity, ending velocity, distance moved along the path and acceleration and solve for acceleration.

    We have an equation that relates starting velocity, ending velocity, displacement from start to end and acceleration. But that equation does not help much because the path folds back along itself. The distance moved along the path does not match the displacement from start to end.

    If we can split the path into two parts, each of which involves a straight line path that never reverses itself then we can use that displacement equation for each part separately. There are a number of ways to use this general approach. The one I have in mind is...

    Working backward, the hope is that we can figure out what fraction of the total distance is covered in the first (or second -- it does not matter which) part of the event. If we knew either one, we'd be in a position to solve for the required acceleration.

    How do we determine the fraction of the total distance covered? If we knew the fraction of the total time elapsed for each part, that information might come in handy.

    So, what fraction of the total time is required for each part? That motivates the questions I've asked above.
     
    Last edited: Apr 3, 2016
  12. Apr 3, 2016 #11
    Vf = vi + at
    0 = -8 + at
    a = 8/t

    16= 0 +at
    a = 16/t
     
  13. Apr 3, 2016 #12

    jbriggs444

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    That does not answer the questions that I posed.
     
  14. Apr 3, 2016 #13
    Xf = xi + 1/2(Vf + Vi)t
    -11 = 0.5(0 -8)t
    t = 2.75s


    11 = 0.5 ( 16 + 0)t
    t = 1.375s
     
  15. Apr 3, 2016 #14

    jbriggs444

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    Those answers are incorrect and are not expressed in terms of "a".
     
  16. Apr 4, 2016 #15
    I dont see any other way to relate accleration and time
     
  17. Apr 4, 2016 #16

    jbriggs444

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    v = at
     
  18. Apr 5, 2016 #17
    Thats what I used In post #11
     
  19. Apr 5, 2016 #18

    jbriggs444

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    Post 11 claims that 8 = 16.
     
  20. Apr 6, 2016 #19
    I will get the same thing if I use V = at
     
  21. Apr 6, 2016 #20

    jbriggs444

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    Let us review...

    The answer for the first question is "the elapsed time is 8/a"
    The answer for the second question is "the elapsed time is 16/a"

    If the acceleration is the same for both phases, what does that say about the elapsed time for both phases?
     
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