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Classical Mechanics,motorboat

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    When engine was turned off,boat with mass m was moving with speed v_0.
    The force of friction F=-[itex]\alpha[/itex][itex]\nu[/itex]-[itex]\beta[/itex]v^2.
    How long it would take to drop speed of boat 3 times?
    Find the distance which the boat will travel in this time?


    2. Relevant equations



    3. The attempt at a solution
    I think i should try to solve differental equation in form of
    m[itex]\dot{v_0}[/itex]=-[itex]\alpha[/itex][itex]\nu[/itex]-[itex]\beta[/itex]v^2

    But I realy dont know what to do next,could someone,please help me with some steps or tips?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2013 #2
    Ok,now Im pretty sure that i have to solve DE :
    [itex]\int(\frac{dv}{ \alpha v+\beta v^2})[/itex]=[itex]\int dt[/itex]

    And integration bondaries from v=v_0 to v_0/3 ?
    And for other intergral offcourse its fromt=0 to t_0

    Either way, how can I find the distance?

    Please help .
     
  4. Oct 9, 2013 #3

    vela

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    Hint: Use
    $$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}.$$ This is just the chain rule.
     
  5. Oct 10, 2013 #4

    haruspex

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    You dropped a minus sign (as well as a factor m). You could fix the sign by changing the v limits to being from v0/3 to v0.
    I note that in the OP you gave the force as being ##-\alpha \nu -\beta v^2##. I assume you meant ##-\alpha v -\beta v^2##
     
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