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lilg132 Group Title

help with question

  • 2 years ago
  • 2 years ago

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  1. lilg132 Group Title
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    \[v = {g \over k} (1 - e ^{-kt})\] v = velocity g = gravitational constant which is \[9.8 ms ^{-2}\] k = (measured in \[s^{-1}\] t = seconds what is e? in the equation

    • 2 years ago
  2. lilg132 Group Title
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    lanaaa to the rescue :p

    • 2 years ago
  3. lalaly Group Title
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    which course are u taking lilg/?

    • 2 years ago
  4. lilg132 Group Title
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    you know already ee can you help me?

    • 2 years ago
  5. lalaly Group Title
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    differentiation?

    • 2 years ago
  6. lilg132 Group Title
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    i think it involves differentiation

    • 2 years ago
  7. lalaly Group Title
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    this is a differential equation and e^-kt is the homogeneous solution

    • 2 years ago
  8. lilg132 Group Title
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    the value for k (s^-1) is 0.15 the value of t you can say is 1 second

    • 2 years ago
  9. lilg132 Group Title
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    what would it be -kte^-kt?

    • 2 years ago
  10. lilg132 Group Title
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    0? :s

    • 2 years ago
  11. lalaly Group Title
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    I still dont get what ur question is, why did u differentiate that? write down the full question lilg

    • 2 years ago
  12. lilg132 Group Title
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    its really long one minute

    • 2 years ago
  13. lalaly Group Title
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    Take ur time im not leavin:)

    • 2 years ago
  14. lilg132 Group Title
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    It is already known that objects falling under the influence of gravity obey the equations of constant acceleration, with a gravitational constant g (usually taken to be 9.8 ms ^-2) that applies to all objects independent of their size or mass Under more realistic conditions, objects falling through air will quickly reach a state of constant velocity, this velocity, which depends on the size, shape, and mass of the falling object, is called terminal velocity. you are on a team developing parachutes for skydivers the velocity v (in ms^-1) of a typical skydiver after t seconds is modelled by your team using the equation \[v = {g \over k} (1 - e ^{-kt})\] the value of k (measured in s^-1) is used to model the different stages of the skydivers decent. The table shows some values found through the experimentation: TYPE OF MOTION k(s^-1) Human in free fall 0.15 Parachute deployed 2.00 1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist 2) what is the terminal velocity for the skydiver? theres more questions but yes im stuck on 1 :p

    • 2 years ago
  15. lilg132 Group Title
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    anyone?

    • 2 years ago
  16. JopHP Group Title
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    do u k anything abt Euler's formula ?

    • 2 years ago
  17. lilg132 Group Title
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    no

    • 2 years ago
  18. JopHP Group Title
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    e^x=e^z(cosy + i siny) this is the formula

    • 2 years ago
  19. lalaly Group Title
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    Jop euler cant be used here:S

    • 2 years ago
  20. lilg132 Group Title
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    http://en.wikipedia.org/wiki/Euler%27s_formula

    • 2 years ago
  21. JopHP Group Title
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    WHY?

    • 2 years ago
  22. lilg132 Group Title
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    e^2.00(1 second) = cos(1) + 2sin(1)

    • 2 years ago
  23. JopHP Group Title
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    did it work with u?

    • 2 years ago
  24. lilg132 Group Title
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    that would give me -1.223244275 (btw i made a mistake it should -2sin(1))

    • 2 years ago
  25. lilg132 Group Title
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    so \[{9.8 \over 2.00} (-1.223244275)\]

    • 2 years ago
  26. lilg132 Group Title
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    but shoudn't it be \[{9.8^{-2} \over 2.00^{-1}}(-1.223244275)\]

    • 2 years ago
  27. lilg132 Group Title
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    which gives you -0.02547364171

    • 2 years ago
  28. lilg132 Group Title
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    -0.02547364171^-1

    • 2 years ago
  29. lilg132 Group Title
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    is that right?

    • 2 years ago
  30. lilg132 Group Title
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    which is -39.25626384

    • 2 years ago
  31. JopHP Group Title
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    i think it makes sense But i'm nt sure

    • 2 years ago
  32. lilg132 Group Title
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    but i also forgot to take in account that 2.00 should 2.00^s-1 :S that confuses me the most

    • 2 years ago
  33. SmoothMath Group Title
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    Is your question really just what e is?

    • 2 years ago
  34. lilg132 Group Title
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    thats one of my questions but i can see gets taken out using the eulers formula, if thats what im supposed to use in this. but i need help working out question 1 really

    • 2 years ago
  35. lilg132 Group Title
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    if someone can show me how they would write the equation out with the values that would help me understand alot better

    • 2 years ago
  36. experimentX Group Title
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    http://www.wolframalpha.com/input/?i=y+%3D+10%281+-+e^%28-2t%29%29+from+t%3D0+to+5

    • 2 years ago
  37. lilg132 Group Title
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    where did you get 10 from? and what about when it says k is measured in s^-1?

    • 2 years ago
  38. SmoothMath Group Title
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    Okay okay. Well, the question of what e is is not too hard. e is a mathematical constant. Its value is approximately 2.71828

    • 2 years ago
  39. lilg132 Group Title
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    ok thanks for that smoothmath

    • 2 years ago
  40. lilg132 Group Title
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    also g = 9.8 ms^-2

    • 2 years ago
  41. SmoothMath Group Title
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    Yup. If you want to read more about it or understand why it's important, wikipedia is a good source, as always. http://en.wikipedia.org/wiki/E_(mathematical_constant)

    • 2 years ago
  42. experimentX Group Title
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    just putting arbitrary values ... e is Euler's constant ... your velocity should be increasing with constant acceleration ... but since air resistance is directly proportional to velocity .... after certain velocity .. your velocity remains constant ... just another decay equation!!

    • 2 years ago
  43. lilg132 Group Title
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    {lilg scratches his head}

    • 2 years ago
  44. SmoothMath Group Title
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    1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist

    • 2 years ago
  45. SmoothMath Group Title
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    I mean... I'm not fantastic at spreadsheets, but this is a simple problem as long as you are inputting the equation correctly.

    • 2 years ago
  46. lilg132 Group Title
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    what im mainly confused about is when i use these values in the equation FOR EXAMPLE g = 9.8 ms^-2 to put it in the equation as 9.8^-2 or just 9.8

    • 2 years ago
  47. SmoothMath Group Title
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    just 9.8 =)

    • 2 years ago
  48. lilg132 Group Title
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    and the same for k when it says it is measured in s^-1 so i should just leave it as 0.15 or 2.00 instead of 0.15^-1

    • 2 years ago
  49. SmoothMath Group Title
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    When they say "ms^-2" that's the unit only. \[ms^{-2} = \frac{m}{s^2}\] Which is the saem as "meters per second squared."

    • 2 years ago
  50. lilg132 Group Title
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    also the end result v is in ms^-1 so the end answer e.g. is 20 would i write it as 20^-1 or just 20

    • 2 years ago
  51. SmoothMath Group Title
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    Yup. Don't worry too much about the units here =)

    • 2 years ago
  52. lilg132 Group Title
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    ok :)

    • 2 years ago
  53. lilg132 Group Title
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    soooo if you can just check one answer for me if its correct im going to use k = 0.15 and t = 1 second so i would get \[v = {9.8 \over 0.15}(1 + \cos(1) - 0.15\sin(1))\] or is it - cos(1)

    • 2 years ago
  54. SmoothMath Group Title
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    Those are units of measurement. If I ask you how much something costs, and you answer "20," I will get angry and ask you "20 WHAT?" Same thing here. Velocity is measured in meters per second, which can be written as \[\frac{m}{s}\]

    • 2 years ago
  55. SmoothMath Group Title
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    Oh my god no no no. Ignore that fancy pants crap those other people were giving you. That's just making the problem complicated. Just plug in the right values to the original equation and solve.

    • 2 years ago
  56. lilg132 Group Title
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    can you show me how you would plus in the values in the equation please it will help me understand much better

    • 2 years ago
  57. lilg132 Group Title
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    use the value of k = 0.15 and t = 1 second

    • 2 years ago
  58. rs32623 Group Title
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    are u asking the value of e? it is approximately equal to 2.71..

    • 2 years ago
  59. SmoothMath Group Title
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    Sure. k = .15 t=1 \[\frac{g}{k}(1-e^{-kt}) = \frac{9.8}{0.15}(1-e^{-0.15*1})\]

    • 2 years ago
  60. SmoothMath Group Title
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    Easy peasy. Plug in and solve. Make sure to use the correct order of operations and keep track of signs =)

    • 2 years ago
  61. SmoothMath Group Title
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    Still confused?

    • 2 years ago
  62. lilg132 Group Title
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    do i not use eulers formula then?

    • 2 years ago
  63. lilg132 Group Title
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    when i have e^-0.15*1 what do i do with that?

    • 2 years ago
  64. SmoothMath Group Title
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    You don't need to, so I don't see why you should.

    • 2 years ago
  65. SmoothMath Group Title
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    \[a^{-x} = (\frac{1}{a})^x\]

    • 2 years ago
  66. lilg132 Group Title
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    so would it be \[({1 \over e})^{-0.15}\]

    • 2 years ago
  67. SmoothMath Group Title
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    That's the rule for negative exponents. Take the reciprocal and make the exponent positive. Alternatively, you could just put in that expression to a calculator and it'll do the work for you.

    • 2 years ago
  68. SmoothMath Group Title
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    Not quite. \[(\frac{1}{e})^{0.15}\]

    • 2 years ago
  69. lilg132 Group Title
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    oh ok yes my mistake

    • 2 years ago
  70. lilg132 Group Title
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    so it would be \[(1 - ({1 \over e})^{0.15})\]

    • 2 years ago
  71. SmoothMath Group Title
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    Yessir.

    • 2 years ago
  72. lilg132 Group Title
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    thanks alot mate appreciate your great help

    • 2 years ago
  73. SmoothMath Group Title
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    My pleasure =)

    • 2 years ago
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