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Differentiation of displacement and time to get velocity

  1. Jun 4, 2008 #1
    [SOLVED] Differentiation of displacement and time to get velocity

    A moving body is related to a fixed point by the displacement equation: s = 12( 1 - e^-t).
    Assuming the body is moving in a straight line on a flat plain.

    1. obtain an expression for the velocity after time t seconds.

    2. obtain an expression for the acceleration also after t seconds

    3. Calculate the time at which the velocity of the body is 6 ms-1.




    So i know that s = v t. So v = s/ t and ds/dt = v



    My attempt has gone round in circles because I am unsure of how exactly to differentiate it do I ignore the 12 first and diff. it, or times out the bracket then diff. it....? Too confused..
     
  2. jcsd
  3. Jun 4, 2008 #2

    nicksauce

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    Well to simplifiy s = 12 - 12e^(-t)

    What if the derivative of a constant? What is the derivative of an exponential?
     
  4. Jun 4, 2008 #3
    constant = 0 and derivative of an exponential would be - e ^-t so its just 12 e ^-t ?
     
  5. Jun 4, 2008 #4

    nicksauce

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    I would agree.
     
  6. Jun 4, 2008 #5
    but then to differentiate it a second time it becomes what - 12 e^-t ? how do I work out what the time is then for part 3. i would have e^-t = 6/12 ? I am stuck where to go from there how do I do it?
     
    Last edited: Jun 4, 2008
  7. Jun 4, 2008 #6

    nicksauce

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    Do you know how to use logarithms?
     
  8. Jun 4, 2008 #7
    I have in the past but am not overly keen on them. Would it be something along the lines of
    -t log e = 6/12 and then 6 /12 log e = -t well that looks wrong, what else..?
     
  9. Jun 4, 2008 #8

    nicksauce

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    e^-t = 6/12
    -t log e = log(1/2) (taking the log on BOTH sides)
    and since we know log e = 1 (when using base e, ie the 'ln' button on your calculator) and knowing that log(1/2) = -log(2), we get
    t = ln 2

    It might be a good time to remind yourself about all the properties of logarithms, as they are mucho important.
     
  10. Jun 4, 2008 #9
    Yeah you are right, I do - many thanks. Is the second derivative right that i wrote earlier?
     
  11. Jun 4, 2008 #10

    nicksauce

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    Yes it is right.
     
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