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Mechanics ( velocity - time graph )

  1. Aug 3, 2015 #1
    So I have problem with understanding graphs in kinematics. Right now I am solving one task and it goes like this :

    Capture 111.JPG

    Solution : a) positive , because of positive slope of tangent
    b) negative
    c) at 3 and 5
    d) a2=a6 > a3=a5 > a1=a4 Magnitudes of the acceleration are the greatest at points 2 and 6. Magnitudes of the acceleration at points 3 and 5 are eaqual and lesser than magnitudes at points 2 and 6. At points 1 and 4 magnitudes of the acceleration is 0.

    This is how it is in solution and this is how I understood it ( If it is not ok please correct me or tell me how it should be ) :
    By some logic this curve is starting at positive part of graph and it is growing until point 1 ( so particle is moving forward and it is accelerating) and suddenly at point 1 acceleration is equal zero ( This is in solution, a=0 and it is confusing me ). So it was moving ,at point 1, with constant velocity and now as curve goes, the particle started slowing down until point 3 ( so it was moving forward but slowing down because it has negative acceleration ). At point 3 direction of moving is changing and now the particle is moving backwards ( I guess because it was slowing down now velocity is negative and it is moving backwards and accelerating ) until point 4. At point 4 as it is said in solution a=0 so now particle is moving with constant velocity. From 4 to 5 it is has acceleration and the particle is moving forward but in negative area ( this is sooo much confusing me. Does this mean that it has negative velocity but positive acceleration, and if it has, how come it is changing direction at point 5 ? I just cant imagine this ) And as it is said in solution at point 5 it changes direction, and now it is accelerating foward with positive acceleration.

    So what is confusing me is : 1)This with changing direction of moving, How particle was moving from point 4 to point 5 ?
    2) Are the magnitudes of the negative acceleration at point 2 and positive acceleration at point 6 greatest?
    3) Should particle stop moving when it has acceleration and it suddenly wants to start moving slowly ?
    4) If it is slowing down and then changes direction of moving it doesnt have to stop ?
     
  2. jcsd
  3. Aug 3, 2015 #2
    Here is my explanation of your problem.
    1. At point 3 to point 4 your particle has negative velocity and negative acceleration, meaning your particle is speeding up in the negative direction. At point 4, acceleration becomes zero (the particle is still moving at the negative direction, however with constant velocity). Between point 4 and point 5, the particle has negative velocity and positive acceleration, that is, the particle is slowing down (but still moving in the negative direction).
    2. Most likely, yes. I have assumed here that particle's velocity is sinusoidal in behavior, so if points 2 and 6 are equal to your initial velocity, then yes.
    3. No, there is still velocity and acceleration present.
    4. Instantaneously, it stops. However, since acceleration is continuously acting on the particle, it starts to gain velocity.
     
  4. Aug 3, 2015 #3
    The problem didn't mention the initial condition.
    I think what you want to say is that 2 and 6 are at the center line of this sinusoidal-like curve.
     
  5. Aug 3, 2015 #4
    This is wrong . Slope in this graph would represent acceleration . The direction of travel is +ve because , simply , velocity at t = 0 is some +ve value .
    You did (a) wrong but you got this correct ? How ?
    Here again ?
     
    Last edited: Aug 3, 2015
  6. Aug 3, 2015 #5
    What do these mean ?
    Yes .
    Yes it does - but only instantaneously .
     
  7. Aug 3, 2015 #6

    I dont know if points 2 and 6 are equal to initial velocity. Can I read it from graph? Or maybe because they are in the same level ?
     
  8. Aug 3, 2015 #7
    They are at the same level - therefore they have the same velocity .

    But having same velocity as initial doesn't mean you will necessarily have maximum acceleration .
     
  9. Aug 3, 2015 #8

    These are solutions from Chegg.com so I was trying to understand this by their directions.

    a) ok, you are right
    b) and c) Those were their solutions but for me it was logical from graph


    1) It means, how moving from point 4 to point 5
    3) Particle is accelerating (+) and now it wants to slow down.What I am asking is should it stop and start moving with negative acceleration or it will move with constant velocity.
     
  10. Aug 3, 2015 #9
    From 4 to 5 , object has velocity in -ve direction , and acceleration in +ve direction . So it is moving from say x1 to x2 , where x2 < x1 .

    I still didn't get (3) . But still -
    If object has +ve velocity along with +ve acceleration , it will not slow down .
    If object has -ve velocity along with +ve acceleration , it will slow down , stop for an instant , and then sstart moving with +ve velocity .
     
  11. Aug 3, 2015 #10


    It doesnt matter I think ecastro has explained it :) Thanks for everything :)
     
  12. Aug 3, 2015 #11
    I have one more problem to solve can someone help me? It goes like this :
    At t=0, a particle moving along an x axis is at position x0 = - 20 m. The signs of the particle's initial velocity v0 (at time t0) and constant acceleration a are, respectively, for four situations: (1) +, +; (2) +, - ; (3) -, +; (4) -, -. In which situations will the particle (a) stop momentarily, (b) pass through the origin, and (c) never pass through the origin?

    I started like this :
    WP_20150803_026[1].jpg

    As I have understood this, the particle is starting to move from negative area. Now in the first situation ( +, +) beacuse it is negative area it is starting foward ( because initial velocity is + ) but slowing down because acceleration is + so it should stop momentarily or it is moving foward and slowing down but moving away from the origin so it wont pass through the origin ( I am not sure if any of these is correct )
    In the second situation ( +, -) the particle is driving foward (away from the origin) but accelerating ( because acceleration is - ) so it wont pass through origin.
    In the third situation ( -, + ) the particle is starting backward but slowing down (acceleration is + ) so it should stop momentairly.
    In the fourth situation ( -, -) the particle is moving backward but accelerating so it should pass through the origin .

    So the answers are : (a) is (3) and maybe (1)
    (b) is (4)
    (c) is (2) and maybe (1)
    Am I right ?
     
  13. Aug 3, 2015 #12
    I'm sorry for the misunderstanding. What I meant here is that point 2 and point 6 should be halfway from points where acceleration is zero, as tommyxu3 has said.

    As for your new question, I think you misunderstood some points.

    A positive velocity and a positive acceleration denotes an increase in velocity, and traveling to the positive axis.
    A positive velocity and a negative acceleration denotes a decrease in velocity, and traveling to the positive axis.
    A negative velocity and a positive acceleration denotes a decrease in velocity, and traveling to the negative axis.
    A negative velocity and a negative acceleration denotes an increase in velocity, and traveling to the negative axis.

    In summary, if both the velocity and acceleration have the same sign, it is speeding up. The velocity's sign will tell you where the particle is traveling to (opposing where the particle is currently is).
     
  14. Aug 4, 2015 #13
    So If a particle is in a positive area of graph and it has negative velocity and positive acceleration it is traveling to negative axis but slowing down and if it has negative velocity and negative acceleration it is traveling to negative axis but getting more speed because acceleration is positive. In the other two cases if velocity is positive the particle is traveling away from negative axis ( it will move deep in positive axis ).
    And if the particle is in negative area of graph and it has positive velocity and positive acceleration it is traveling to positive axis and accelerating and if the particle has positive velocity and negative acceleration it is traveling to positive axis but slowing down. In other two cases if velocirry is negative the particle is traveling away from positive axis ( it will move deep in negative axis)

    Did I get it right ?

    Can you please suggest me a book or some tutorial on internet so I can learn these graphs in all situations ? I was searching for it but all the time there was missing something or it was not good explained.
     
  15. Aug 4, 2015 #14

    A.T.

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  16. Aug 4, 2015 #15
    I think you meant here is that acceleration is negative.

    A. T.'s suggestion is a good reference, and here's another one: http://www.physicsclassroom.com/mmedia/kinema.

    As a side note, I think you're getting confused on the particle's position and the signs of velocity and acceleration. The particle's position has no relation whether the particle is slowing down or speeding up. The sign of both velocity and acceleration matters.
     
  17. Aug 4, 2015 #16
    Yes, I meant that acceleration is negative. My mistake

    Thank you, you have helped me a lot :)
     
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