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Acceleration vs. time graph

  1. Mar 14, 2012 #1
    I want to draw a graph for a ball thrown directly up in the air.

    I know that when I throw the ball up in the air the acceleration will be constant. So if I draw a graph with the accelation as function of time, I would get a horizontal line. I have read that the line should be drawn below zero at the y-axis, but why? And it cuts the y-axis at -g, what is the explanation to that? Why do I need to look at the acceleration as g?

    Thank you.
     
  2. jcsd
  3. Mar 14, 2012 #2
    Hi Kork!
    If upwards in real life corresponds to upwards on the y-axis, then it would make sense to draw the line at -g. +g would mean the ball was accelerated (attracted) upwards, which it clearly isn't.
    g is the standard gravity, the acceleration (downwards) of any object near the surface of the Earth.
     
  4. Mar 14, 2012 #3
    By the way. I have also made a graph with velocity as a function of time for the ball thrown up in the air. The grapfh starts at max velocity and decreases and crosses the x-axis (time). What can I say about the conservation of energy here? The point that the velocity is negative when the ball falls down confuses me.
     
  5. Mar 14, 2012 #4
    Thank you DennisN,

    So if I start at -g, then it makes sense because I start from low to hight, or how?
     
  6. Mar 14, 2012 #5
    Do you know about kinetic energy and potential energy?
     
  7. Mar 14, 2012 #6
    I'm not sure what you mean. The acceleration will always be -g. The velocity will change; try to think about how it will change. What happens when you throw it up (initial velocity), what happens at the max height, what happens when the ball falls downwards, and what happens when the ball hits the ground? (I know the answers, I just want you to think about it, and I will help you :smile:). And btw, do you have any formulas, or is it just the basic view you're after?
     
  8. Mar 14, 2012 #7
    Yes, I know what potential and kinetic energy is.
     
  9. Mar 14, 2012 #8

    When I throw it up the velocity will be at it's maximum and it will be positiv. In max height the velocity is 0. Downwards the ball will have the same value of velocity as when it came upwards, just negative. When it hits the ground, then all the kinetic energy becomes potential?

    There is so many formulas and i dont know which ones i should use to get the "picture".
     
  10. Mar 14, 2012 #9
    Im still not sure why it starts at -g, when I draw the a-t graph.
     
  11. Mar 14, 2012 #10
    Du er svensk :)
     
  12. Mar 14, 2012 #11
    Ja, jag är svensk! :smile: Vad är du? Norsk, dansk, eller? We have to use English, though, that's the forum rules :frown:.
    Great!
    Let's focus on velocity first, so we don't mix it up and confuse ourselves. When we have gotten the velocity right, the rest will follow more easy. I have to check what you've written, please hang on, will you?
     
  13. Mar 14, 2012 #12
    Im danish :)

    Sure I will hang on, I have confused myself completely.
     
  14. Mar 14, 2012 #13

    sophiecentaur

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    Up or down, + or -, it doesn't matter which you choose as long as you stick to it throughout the calculations.

    If you were considering dropping a ball off a tower then you might possibly choose down to be positive, in which case the acceleration would be g and the distances 'fallen' would be positive. You get the same numerical answer either way and you just need to interpret that answer to fit the physical reality.
     
  15. Mar 14, 2012 #14
    Very good, almost correct. (let's skip what happens when the ball hits the ground, it's not important at this stage). There is a formula for velocity and acceleration (I don't remember the name in English, final velocity, maybe);

    v = v0+a*t

    where

    v is the velocity
    v0 is the initial velocity
    a is the acceleration (that is, a=-g)
    t is the time

    Do you recognize this? This will yield that v = v0-g*t.
     
  16. Mar 14, 2012 #15
    If you are confused, don't hesitate to ask questions. Hint: The solution to the energy conservation question lies in the formulas for kinetic energy (depends on velocity) and potential energy (depends on height). According to the energy conservation principle the total energy during the throw/fall will be conserved, i.e. Etotal = Ekinetic + Epotential = constant.

    What will happen is that

    0. Before the throw, the kinetic energy and potential energy is 0 (if we define the height as 0 at your hand).
    1. When you throw the ball, you give it max kinetic energy (initial velocity).
    2. This kinetic energy will gradually be transformed to potential energy.
    3. At max height, the potential energy will be max and the kinetic energy 0.
    4. Then, gradually the potential energy will be transformed back to kinetic energy.
    (5). (When the ball hits the ground, we can say that the kinetic energy will "dissipate" during the ground collision, but this doesn't really matter in this example).

    Note: All this is actually only true from the time the ball leaves your hand until it falls back into your hand again. If it falls to the ground instead, the potential energy will be a little less than before.

    Was this any help?
     
  17. Mar 14, 2012 #16
    Hi again, Kork. I reread what you wrote before:
    Try not to be confused about that, it is correct; the velocity changes direction. Going up means positive velocity, going down means negative velocity. And when you put a negative velocity into the formula for kinetic energy, it won't matter, the velocity will be squared (v*v), which will result in a positive kinetic energy.

    And what you said before about the velocity:
    is correct.
     
  18. Mar 14, 2012 #17

    sophiecentaur

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    Negative velocity just means you are subtracting a distance every second (i.e. the height is getting less)
     
  19. Mar 14, 2012 #18
    Thank you guys very much. This has helped a lot!
     
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