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lilg132

  • 2 years ago

help with question

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

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

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

  5. lalaly
    • 2 years ago
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    differentiation?

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

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

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

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

  10. lilg132
    • 2 years ago
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    0? :s

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

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

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

  14. lilg132
    • 2 years ago
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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

  15. lilg132
    • 2 years ago
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    anyone?

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

  17. lilg132
    • 2 years ago
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    no

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

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

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

  21. JopHP
    • 2 years ago
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    WHY?

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

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

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

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

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

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

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

  29. lilg132
    • 2 years ago
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    is that right?

  30. lilg132
    • 2 years ago
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    which is -39.25626384

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

  32. lilg132
    • 2 years ago
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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

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

  34. lilg132
    • 2 years ago
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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

  35. lilg132
    • 2 years ago
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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

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

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

  38. SmoothMath
    • 2 years ago
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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

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

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

  41. SmoothMath
    • 2 years ago
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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)

  42. experimentX
    • 2 years ago
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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!!

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

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

  45. SmoothMath
    • 2 years ago
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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.

  46. lilg132
    • 2 years ago
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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

  47. SmoothMath
    • 2 years ago
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    just 9.8 =)

  48. lilg132
    • 2 years ago
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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

  49. SmoothMath
    • 2 years ago
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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."

  50. lilg132
    • 2 years ago
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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

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

  52. lilg132
    • 2 years ago
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    ok :)

  53. lilg132
    • 2 years ago
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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)

  54. SmoothMath
    • 2 years ago
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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 \[\frac{m}{s}\]

  55. SmoothMath
    • 2 years ago
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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.

  56. lilg132
    • 2 years ago
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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

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

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

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

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

  61. SmoothMath
    • 2 years ago
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    Still confused?

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

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

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

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

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

  67. SmoothMath
    • 2 years ago
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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.

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

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

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

  71. SmoothMath
    • 2 years ago
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    Yessir.

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

  73. SmoothMath
    • 2 years ago
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    My pleasure =)

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