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Rising acceleration

  1. Dec 27, 2006 #1

    disregardthat

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    I am just wondering what this concept is called.
    (I have no education is this sort of math...)

    Acceleration is: (v2-v1)\(t2-t1)=average accalearation

    Like: (10m\s-5m\s)\(3s-1s) = 5m\s\2s = 2.5m\s^2

    Right?

    Ok, I was wondering what the rising acceleration is called, and if this is the correct way of calculating it:

    Average rising acceleration: (20m\s^2-5m\s^2)\(7s-2s) = 15m\s^2/5s = 15m\s^3

    Where 15m\s^3 is the rising acceleration. Is this way correct and what is it called?
     
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  3. Dec 27, 2006 #2

    cristo

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    The time derivative, or rate of change, of acceleration is technically called "jerk." Average jerk will be calculated in an analagous way to average acceleration, and so, yes, your method of calculation is correct.

    (Although, you have a typo in your last line: j=15/5=3ms-3).
     
  4. Dec 27, 2006 #3

    disregardthat

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    Thanks!

    So, the jerk is 3m\s^3. I believe that it is "jerk" that is used when calculating gravity when you find out how much the acceleration is at different length's from the earth.

    Hmmm, when there is no mention of time, is there a method of calculating jerk, or acceleration with speed and 'length instead of time'?

    And I guess this way of multiplying it works in higher dimensions too? Like if the jerk is accelerating. Is the a n' definition of this?
     
    Last edited: Dec 27, 2006
  5. Dec 27, 2006 #4

    cristo

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    I'm not sure what you mean here. If you mean is it possible to derive an analog of the kinematic equations for constant acceleration, and get a set of equations for constant jerk then, yes, this should be possible.
    Do you mean taking the rate of change of jerk? If so, yes, this is done in a similar manner.
     
  6. Dec 28, 2006 #5

    disregardthat

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    I meant for example, if you are to measure the jerk the earth's gravity. 0 km over the sea level, the acceleration is 9.8m\s^2. 1000 km up in the air the acceleration is (i don't know but let's assume) 3m\s^2. So with no mention of the time an object uses from this point up in the air, to the ground, could we measure the jerk?
     
  7. Dec 28, 2006 #6

    cristo

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    If I understand the question correctly, then no, the change in time must be known.
     
  8. Dec 28, 2006 #7

    disregardthat

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    Uhm, are you sure? Since if in the very unit of Xm\s^3 both second and meter is mentioned. And if you knew the number in front (x) you should be able to use it as an equation to find out the s (second)... Right?

    Time after velocity:

    You travel 50 meter at a speed of 5m\s
    5=m\s 5=50\s 5=50\s |*s 5s=50 5s=50 |/5 s=10
    You use 10 seconds after 50 meter if you travel at 5 m\s Correct

    Time after acceleration:
    You travel 50 meter at an acceleration of 5m\s^2
    5=m\s^2 5=50\s^2 5=50\s^2 |*s^2 5s^2=50 5s^2=50 |/5 s^2=10 s^2=10 |quadr s=3.16
    You use 3.16 seconds to travel 50 meter if you accelerate at 5m\s^2 Correct

    Time after Jerk(or Jolt as I saw on wikipedia after searching on jerk)
    You travel 50 meter at a Jerk\Jolt of 5m\s^3

    5=m\s^3 5=50\s^3 5=50\s^3 |*s^3 5s^3=50 5s^3=50 |/5 s^3=10 s^3=10 |cubicroot(revers ^3) s=2.15-2.16 (around that)
    You use 2.15-2.16 seconds to travel 50 meter at a Jerk\Jolt of 5m\s^3


    Wouldn't that be correct?
     
    Last edited: Dec 28, 2006
  9. Dec 28, 2006 #8

    cristo

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    I really don't understand what you're getting at here! IF you knew the average jerk, then sure, you could calculate the time between two points of different acceleration. Consider the case of (constant) acceleration; if you know the acceleration, then you can calculate the time it takes for the velocity to change from v_0 to v_1.

    Therefore, if you have know the jerk, then you can calculate the time taken for an object's acceleration to increase from a_0 to a_1. However, I'm not sure whether this is your question.
     
  10. Dec 28, 2006 #9

    Doc Al

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    I assume by measuring the "jerk of earth's gravity" you mean measuring the jerk of something that is falling due to gravity? I have no idea why you'd want to do that, but whatever. If you wanted to measure the rate of change of acceleration, then you would certainly need the time. But you can easily calculate how the acceleration due to gravity depends on distance from the earth--making some simplifying assumptions, of course. (Note that since the acceleration due to gravity just depends on position, the time rate of change of that acceleration will depend on how fast you are falling.)
     
  11. Dec 28, 2006 #10

    disregardthat

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    I don't want to measure it, I have little interest in knowing it, I'm just trying to find out how gravity works. What makes it true that it is only jerk that works in gravity? Maybe the the jerk is rising too?

    Well, the assumptions is that the earth is a perfect sphere, and that it is just as dense all over. And excluding air resistance. And excluding other matter that effect the falling object.
     
  12. Dec 28, 2006 #11

    Doc Al

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    I assure you, the concept of "jerk" will not help you understand gravity.

    What is the point of your questions?

    Given those assumptions, the gravitational field above the earth's surface is proportional to the inverse square of the distance from the earth's center.

    Rather than waste time with "jerk", you'd be better off learning a bit about inverse square laws. Try this: Inverse Square Law, Gravity. (You'll find plenty of other information about gravity on that site as well.)

    (Since this thread is about physics, not math, I will move it.)
     
  13. Dec 28, 2006 #12

    disregardthat

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    First of, it was not originally about physics, but anyway...

    It doesn't seem that you are even trying to understand what I mean. I just wanted to know if Jerk or Jounce (m/s^4) had anything to do with gravity. And in a way, it does, if you take away some factors that's insignificant on small scales... An object's acceleration does accelerate as it get's closer to the earth. But since my questions 'doesn't matter' or has no point, I will stop asking them...

    I do understand gravity or at least the basics of it, I just wanted to know if there are a simplicified way of doing it more easy...

    I will read your link, thanks.

    Edit: It was very informative :) I searched wikipedia for radius and weight of the earth, and got around 9.8 N\kg
    I have a question thought. What equation must you use if the gravitational object is not squared?
     
    Last edited: Dec 28, 2006
  14. Dec 28, 2006 #13

    Doc Al

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    Realize that "jerk" is the time rate of change of acceleration. Much better would be to simply say that the acceleration due to gravity gets weaker as distance from the earth increases. (Refer to that inverse square law equation.) No need to mention time or "jerk".
    I think you are complicating things, not making them simpler.
    Yes, that's the gravitational field strength at the earth's surface; note that it's also the acceleration due to gravity at the earth's surface: g = 9.8 m/s^2.
    I don't understand the question. The "square" is a property of the law of gravity, not the object. (Of course, it's the simple geometry of the sphere that allows us to write a simple formula.)
     
  15. Dec 28, 2006 #14

    disregardthat

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    Yeah, never mind my Jerk question then...

    N/kg=m/s^2 then?

    I did not mean squared, sorry, I mean what if the gravitational object is not formed as a sphere? Is there any equations if the object is a cube for example? I guess that would make things very much more difficult.
     
  16. Dec 28, 2006 #15
    N= 1 kg*(m/s^2) from F=ma so yes N/kg=m/s^2
     
  17. Dec 29, 2006 #16

    Doc Al

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    Just as you suspect, if the earth were a cube instead of a sphere the gravitation field near the surface would be much more complicated to describe--no simple formula. But, as you get further from the "cube-shaped" earth you will find that the gravitational field begins to take on the familiar "inverse square" form. If you are far enough away from a given mass, it doesn't matter what shape it has--its gravitational field at that distance will look just like it was a point mass.
     
  18. Dec 30, 2006 #17

    disregardthat

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    Yeah, if course, at a large scale it would not matter. But there has to be some factors that minimalisticly matters at large distances, and matters a lot at short distances. Are there any equation?
     
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