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B Integrating velocity equation problem

  1. Apr 10, 2016 #1
    upload_2016-4-9_23-55-27.png
    I've already completed the first question, but with number two, it's a different case. Here's my attempt:
    [itex]\frac { d{ v }_{ y } }{ dt } \quad =\quad -g\quad -\quad \beta { v }_{ y }\\ \frac { d{ v }_{ y } }{ -g\quad -\quad \beta { v }_{ y } } \quad =\quad dt\\ \int { \frac { d{ v }_{ y } }{ -g\quad -\quad \beta { v }_{ y } } } \quad =\quad \int { dt } \\ \frac { -1 }{ \beta } ln(-g-\beta { v }_{ y })\quad =\quad t\\ [/itex]
    natural log of a negative?
     
  2. jcsd
  3. Apr 10, 2016 #2
    EDIT: I seem to have figured out the answer right after posting this?
    [itex]\frac { -1 }{ \beta } ln(-g-\beta { v }_{ y })\quad =\quad t\\ ln(-g-\beta { v }_{ y })\quad =\quad -t\beta \\ -g-\beta { v }_{ y }\quad =\quad { e }^{ -t\beta }[/itex]
     
  4. Apr 10, 2016 #3

    Ssnow

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    In general ##\int \frac{1}{x} dx= \log{|x|}+c##
     
  5. Apr 10, 2016 #4
    log base e right?
     
  6. Apr 10, 2016 #5

    Ssnow

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    yes, ##\int \frac{1}{x}=\ln{|x|}+c##
     
  7. Apr 10, 2016 #6

    Ssnow

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    if ##-g-\beta v_{y}>0## this is part of your solution, you must use the condition that ##v_{y0}=0##, but attention that you forgot the integration constant in your solution ...
     
  8. Apr 10, 2016 #7
    The integration constant is just the initial condition, which is 0
     
  9. Apr 10, 2016 #8

    Ssnow

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    mmm, I don't think, assuming ##-g-\beta v_{y}>0## put the condition ##v_{y}=0## with ##t=0## in

    ##\ln{(-g-\beta v_{y})}=-t\beta +c ##
     
  10. Apr 10, 2016 #9

    Ssnow

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    You can see the value of ##c## also from ##\ln{|-g-\beta v_{y}|}=-t\beta + c##...
     
  11. Apr 10, 2016 #10
    Ssnow, not sure I follow you. From post #1, 3rd line of my attempt, the limits of integration are:

    left side: 0 to v
    (its 0 since initial velocity is given to be 0)
    right side: 0 to t
     
  12. Apr 11, 2016 #11

    Ssnow

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    You forget the constant ##\ln{g}##:

    ##\int_{0}^{v_{y}}\frac{w}{-g-\beta w}=-\frac{1}{\beta}\ln{|-g-\beta v_{y}|}+\frac{1}{\beta}\ln{g}##
     
  13. Apr 11, 2016 #12
    I see now, thanks!
     
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