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blackstreet23

  • one year ago

In this exploration, acceleration was assumed to be constant. Consider the case where the acceleration is changing with time. This is known as "jerk." Jerk is often used in engineering, especially when constructing roller coasters. Humans can be safely subjected not only to a maximum acceleration, but also to a maximum jerk. Which of Shakina's statements regarding jerk is not correct? "If the acceleration is decreasing over time, the jerk is negative." "Jerk is the area under the acceleration vs. time graph." "The units of jerk are m/s3."

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  1. blackstreet23
    • one year ago
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    In this exploration, acceleration was assumed to be constant. Consider the case where the acceleration is changing with time. This is known as "jerk." Jerk is often used in engineering, especially when constructing roller coasters. Humans can be safely subjected not only to a maximum acceleration, but also to a maximum jerk. Which of Shakina's statements regarding jerk is not correct? 1- "If the acceleration is decreasing over time, the jerk is negative." 2- "Jerk is the area under the acceleration vs. time graph." 3- "The units of jerk are m/s3." 4- "Jerk is the slope of the acceleration vs. time graph."

  2. blackstreet23
    • one year ago
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    @PlasmaFuzer @Shalante @freckles

  3. anonymous
    • one year ago
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    Ok so I take it you are able to select multiple answers that are correct yes?

  4. anonymous
    • one year ago
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    Oh darn nm you are supposed to select the INCORRECT statements.... ok then no problem

  5. anonymous
    • one year ago
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    So let me ask you a question... If velocity changes with respect to time does this mean there is an acceleration?

  6. anonymous
    • one year ago
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    Assuming of course the mass of the object isn't changing that is :D

  7. blackstreet23
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    ok then what are the units of velocity and what are the units of acceleration.... and how are the two related?

  9. blackstreet23
    • one year ago
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    units of velocity are m/s and of acceleration m/s^2

  10. anonymous
    • one year ago
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    Correct... how are the two related to one another?

  11. blackstreet23
    • one year ago
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    they are related because the derivative of one of them gives us the other one

  12. blackstreet23
    • one year ago
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    right?

  13. anonymous
    • one year ago
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    Exactley the derivative of the velocity function gives us the acceleration function.... tell me what does it mean to find the derivative of a curve at a point?

  14. anonymous
    • one year ago
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    And before you start sweating this is actually a lot easier to imagine and explain than the integral before :D

  15. blackstreet23
    • one year ago
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    umm. the derivative of a curve point?

  16. anonymous
    • one year ago
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    Yes.... so we said that: \[a=\frac{ dv }{ dt }\] But to be more precise it is really: \[ a(t)=\frac{ d }{ dt } v(t) \] For an object who has a constant mass.... Therefore at the point t=T what is the meaning of : \[ a(T)=\frac{ d }{ dt } v(T) \]

  17. anonymous
    • one year ago
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    Geometrically, what is its meaning... Just like the integral is the area under the curve... what does this derivative at the point T mean?

  18. anonymous
    • one year ago
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    If you dont know you can just say btw

  19. anonymous
    • one year ago
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    No judgements here :D we all have to learn it at some point I don't hold it against people for not knowing

  20. anonymous
    • one year ago
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    Or getting it wrong or making a mistake for that matter

  21. blackstreet23
    • one year ago
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    the derivative is the slope of the tangent

  22. anonymous
    • one year ago
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    Bingo nailed it.... So you should have all the pieces to get this one... Please go through the choices for me and explain to me why one first or why one doesnt please? :D

  23. anonymous
    • one year ago
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    fits*

  24. blackstreet23
    • one year ago
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    if the acceleration is decreasing, the jerk is negative? is that so?

  25. anonymous
    • one year ago
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    I dont know is it? Please try and reason it out.... Start with this... We already agreed on the relationship between velocity and acceleration... If the velocity is decreasing what can you say about acceleration?

  26. blackstreet23
    • one year ago
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    ohh I think i got it. Since the derivative means the change of something over time. That means that the jerk has to be the change of acceleration over time. Right?

  27. anonymous
    • one year ago
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    Exactley :D :D

  28. blackstreet23
    • one year ago
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    if the velocity is decreasing is accelaration has a different sign

  29. anonymous
    • one year ago
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    Different?

  30. blackstreet23
    • one year ago
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    like if the velocity is negative and acceleration positive, it starts backwards, slow down and then goes to the front and vice-versa

  31. blackstreet23
    • one year ago
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    ohh so the same principle applies to acceleration?

  32. anonymous
    • one year ago
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    Exactley correct, but note the use of the word DECREASING.... if the velocity is negative and is DECREASING it is getting more negative no?

  33. anonymous
    • one year ago
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    And in this case the acceleration is also negative so it happens to be the SAME sign as that of velocity

  34. blackstreet23
    • one year ago
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    i guess

  35. anonymous
    • one year ago
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    Your situation is when velocity is negative and begins to INCREASE.... then it has a positive acceleration

  36. blackstreet23
    • one year ago
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    but if the acceleration is positive and the jerk negative, the acceleration decreases?

  37. anonymous
    • one year ago
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    Exactley :D:D

  38. anonymous
    • one year ago
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    but also if the acceleration is NEGATIVE and DECREASING the jerk is also negative Does this make sense?

  39. blackstreet23
    • one year ago
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    and if the acceleration is negative and the jerk is negative as well it becomes more negative, just as the same as (velocity and acceleration)

  40. anonymous
    • one year ago
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    change in acceleration over time is what is important not the absolute sign

  41. blackstreet23
    • one year ago
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    yes it does!

  42. blackstreet23
    • one year ago
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    i mean yes it does make sense!

  43. anonymous
    • one year ago
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    Oh yes sorry didnt see that post... yes that is correct

  44. blackstreet23
    • one year ago
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    I get it now :D

  45. anonymous
    • one year ago
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    Ok so is option 1 true or false?

  46. blackstreet23
    • one year ago
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    true

  47. blackstreet23
    • one year ago
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    1-T

  48. anonymous
    • one year ago
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    bingo.... next

  49. anonymous
    • one year ago
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    because we are of course looking for what is INCORRECT

  50. blackstreet23
    • one year ago
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    2 is the false one right?

  51. anonymous
    • one year ago
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    Correct yes that is the answer

  52. blackstreet23
    • one year ago
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    jerk is the derivative of the graph (acceleration vs time) and not the integral of it (area under the curve)

  53. anonymous
    • one year ago
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    What would that represent?

  54. blackstreet23
    • one year ago
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    that would be velocity

  55. anonymous
    • one year ago
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    Exactley on both posts :D :D

  56. anonymous
    • one year ago
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    So I take it you got it but just to be thorough do the units make sense of jerk?

  57. blackstreet23
    • one year ago
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    You are a physics genius lol. I am going to look for you in any question i ask haha

  58. anonymous
    • one year ago
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    Hahahaha why thank you, but unfortunately I am in school myself so I only come here when I have some spare time to help out :(

  59. blackstreet23
    • one year ago
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    ohh no why the units are m/s^3?

  60. anonymous
    • one year ago
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    But I do check my emails so if I have a moment and I see your post I will try to help, but sometimes I wont have the same amount of free time so keep that in mind

  61. anonymous
    • one year ago
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    Ahh well let me ask you a question... why are the units of acceleration m/s^2 ?

  62. blackstreet23
    • one year ago
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    \[\frac{ m }{ s } * \frac{ 1 }{ s }\]

  63. blackstreet23
    • one year ago
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    velocity / time

  64. blackstreet23
    • one year ago
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    so it would be (acceleration / time)

  65. anonymous
    • one year ago
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    so you really mean: \[\frac{ \frac{m}{s}}{s} \] ... Yes exactley

  66. anonymous
    • one year ago
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    Yes exactley....

  67. anonymous
    • one year ago
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    \[\frac{ \frac{m}{s^2}}{s}\]

  68. blackstreet23
    • one year ago
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    ohh ok. Thanks a lot! Btw can i ask what is your major and if you are getting your Associates, Bachelors or Masters? lol

  69. anonymous
    • one year ago
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    Im actually in a Ph.D program for plasma physics

  70. blackstreet23
    • one year ago
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    sorry just got curious

  71. anonymous
    • one year ago
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    Its alright :D

  72. blackstreet23
    • one year ago
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    ohhh ok. I am talking to a Doctor! that is so cool!!!

  73. anonymous
    • one year ago
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    hahaha not yet no but maybe in a few years

  74. blackstreet23
    • one year ago
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    But what is plasma physics about?

  75. anonymous
    • one year ago
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    Well I have always been interested in fusion energy... So I decided to pursue it and study man made plasmas specifically tokamak plasmas

  76. anonymous
    • one year ago
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    Its hard to give a simple explanation, but if you think it is interesting I invite you to check out wikipedia.... It is an invaluable resource to get a once over viewpoint of a topic, though I should not you always need to go deeper if you are asked to actually write a paper or do research since it is only really a summary of things.

  77. anonymous
    • one year ago
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    This should be a good place to start: http://fusionforenergy.europa.eu/understandingfusion/

  78. anonymous
    • one year ago
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    And this one also: https://en.wikipedia.org/wiki/Tokamak if you want to get a little more

  79. anonymous
    • one year ago
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    But I would suggest saving it till you have a little free time.... make sure to finish your hw first so you dont have it hanging over your head... that is a terrible feeling

  80. blackstreet23
    • one year ago
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    haha yeah i know ...

  81. blackstreet23
    • one year ago
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    and btw what motivated you to go for your masters?

  82. blackstreet23
    • one year ago
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    sorry P.h.D

  83. anonymous
    • one year ago
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    Hmmm well that's a good question and it doesn't have a simple answer; but I guess it ultimately came down to the fact that I decided to pursue physics so I could help out in this field. If I were to stop at a bachelors level then I felt I couldn't really contribute the way I wanted to (or at least doing the things I liked to do). So yea, suffice it to say I always found this stuff interesting and it always drove me to want to know more.

  84. anonymous
    • one year ago
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    A real turning point was when I stopped looking at hw problems and study problems as just a means to an end (a good grade) and started looking at them as valuable opportunities to practice my reasoning skills on a specific topic... If I couldn't do it then obviously I didn't understand it.

  85. anonymous
    • one year ago
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    But at the end of the day I'm with Feynman on this one.... I just got interested in this stuff and decided to want to pursue it. I really feel anyone can do it, it just comes down to finding an interest and passion for it that can carry you through the dull bits in order to be able to enjoy the more satisfying bits (which also happen to be the more difficult bits).

  86. blackstreet23
    • one year ago
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    ohh i loved your answer is pretty enlightening!

  87. blackstreet23
    • one year ago
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    btw do you have to go. Because it would be helpful if you could explain to me how to do the previous problem. The one I told you to skip. Without the simulator haha

  88. anonymous
    • one year ago
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    Sure no problem, but give me a min I have been going back and forth from the kitchen I'm making something to eat.

  89. blackstreet23
    • one year ago
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    ohh no problem

  90. anonymous
    • one year ago
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    are u here?

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