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Homework Help: Physics Problem

  1. May 22, 2005 #1

    VietDao29

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    Hi,
    I just want to ask you guys a problem I had at the very beginning of grade 10th. I am gonna go to grade 11th next year. But I still cannot figure this out. (So it's obviously not homework). Any help will be highly appreciated.
    A canoe (mass 0.5kg) has an initial velocity of 10 m/s. As it moves along the river, a force F = 0.5v acts on it (to make it decelerate). Calculate the distance the canoe will go when:
    a/ Its velocity is 5 m/s.
    b/ The canoe stops.
    (I'm sorry if my translation of the problem is not clear).
    The book gives the answer for a is 5m, and b is 10m.
    -----
    So what I have so far is a = -v. And I know that to calculate the distance, I must have a function v with respect to t. But... how can I have it? :confused:
    Thanks a lot.
    Viet Dao,
     
  2. jcsd
  3. May 22, 2005 #2

    dextercioby

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    Can u integrate that ODE...?It's separable.

    Daniel.
     
  4. May 22, 2005 #3

    VietDao29

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    Can you give me a little hint (just a little bit to clear my mind)? I am stuck...
    Viet Dao,
     
  5. May 22, 2005 #4

    dextercioby

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    [tex] 0.5\frac{dv}{dt}=-0.5 v(t) [/tex]

    is the separable I-st order ODE that needs to be solved,using the initial condition the problem's giving.

    Daniel.
     
  6. May 22, 2005 #5

    VietDao29

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    Okay, I'm not sure. But here's what I get:
    [tex]\frac{dv}{dt} = -v(t)[/tex]
    And I want
    [tex]s = \int^\varepsilon_\xi v(t)dt[/tex]
    Plug that in the function and I have:
    [tex]s = \int^\alpha_\beta -1 dv[/tex]
    [tex]s = \int^\beta_\alpha 1 dv[/tex]
    For a/
    [tex]s = \int^{10}_{5} 1 dv = 5[/tex]
    For b/
    [tex]s = \int^{10}_{0} 1 dv = 10[/tex]
    Is it correct? Is there another way?
    Thanks,
    Viet Dao,
     
  7. May 22, 2005 #6

    dextercioby

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    It's not correct.There's another way.It's called separation of variables.

    [tex] \frac{dv}{v}=-dt [/tex] (1)

    [tex] \int \frac{dv}{v}=-\int dt [/tex] (2)

    [tex] \ln v(t)=-t+C [/tex] (3)

    [tex] v(t)=\bar{C}e^{-t} [/tex] (4)

    ,where [itex] \bar{C}=e^{C} [/itex] (5)

    (4) is the solution to your problem.You need to find the integration constant by imposing the initial condition.

    Daniel.
     
  8. May 23, 2005 #7

    VietDao29

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    Thanks very much. Finally, I know how to solve this problem. :smile:
    Viet Dao,
     
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