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Related Rates and acceleration

  1. Dec 31, 2009 #1
    I have a question about a related rates, i already solved it, but i would like someone to confirm it to me .

    * The relation between distance s and velocity v is given by v=[tex]\frac{150s}{3+s}[/tex]. Find the acceleration in terms of s.



    As you can see my final answer is a in terms of v and s, not s only .. .but am not sure what to do to getrid of this v..

    Thanks
     
  2. jcsd
  3. Dec 31, 2009 #2
    There's a mistake between the third and forth lines. Also, what substitution can you make to get rid of v?
     
    Last edited: Dec 31, 2009
  4. Jan 1, 2010 #3

    thanks, the 4th line should be [tex]\frac{dv}{dt}[/tex]= [tex]\frac{450v}{(3+s)^2}[/tex]

    am not sure what sub. get rid of v.
    but v' will give me a instead of v, but is that possible, i mean, if i want to get V' i will have to take thderivative for the whole thing.
     
  5. Jan 1, 2010 #4

    kreil

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    Gold Member

    Your original velocity equation expresses v in terms of s; use that.
     
  6. Jan 1, 2010 #5
    sorry but i don't know what you mean .

    The Question asked me to find a in terms of s , not v in terms of s
     
  7. Jan 3, 2010 #6

    kreil

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    Gold Member

    your expression for a is correct, but it contains a 'v'. in order to express a completely in terms of s, you need to use the equation [tex]v=\frac{150s}{3+s}[/tex] and plug this into your equation for a.
     
  8. Jan 3, 2010 #7
    Ah, that make sense, thank you :D
     
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