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Math Help - Modelling with differential equation

  1. #1
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    Modelling with differential equation

    A cyclist freewheeling on a level road experiences a negative acceleration which is proportional to the square of his speed. His speed is reduced from 20m/s to 10m/s in a distance of 100m. Find the average speed (w.r.t. time) during this period.

    Would the equation be

    \dfrac{dv}{dt}=-kv^2 where k is a constant of proportionality
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  2. #2
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    Quote Originally Posted by acevipa View Post
    A cyclist freewheeling on a level road experiences a negative acceleration which is proportional to the square of his speed. His speed is reduced from 20m/s to 10m/s in a distance of 100m. Find the average speed (w.r.t. time) during this period.

    Would the equation be

    \dfrac{dv}{dt}=-kv^2 where k is a constant of proportionality
    yes.
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    yes.
    Thanks. So I'm a little unsure of what to do next.

    \dfrac{dv}{dt}=-kv^2

    -\dfrac{dv}{v^2}=k \ dt

    \displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt

    \dfrac{1}{v}=kt+C
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  4. #4
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    Quote Originally Posted by acevipa View Post
    Thanks. So I'm a little unsure of what to do next.

    \dfrac{dv}{dt}=-kv^2

    -\dfrac{dv}{v^2}=k \ dt

    \displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt

    \dfrac{1}{v}=kt+C
    You have to set up a differential equation which involves the displacement and velocity.

    Hint: Notice that a=dv/dx=(dv/dx) x (dx/dt) = v (dv/dx)
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  5. #5
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    Quote Originally Posted by acevipa View Post
    Thanks. So I'm a little unsure of what to do next.

    \dfrac{dv}{dt}=-kv^2

    -\dfrac{dv}{v^2}=k \ dt

    \displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt

    \dfrac{1}{v}=kt+C
    ... all right!... the solution is then...

    \displaystyle v(t)= \frac{1}{k t + c} (1)

    ... where of course is c= \frac{1}{v(0)}. Now suppose that v(0) is negative, for example v(0) = -1, so that the solution is...

    \displaystyle v(t)= \frac{1}{k t -1} (2)

    A little question: what does if happend at the time t=\frac{1}{k} ? ...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by mathaddict View Post
    You have to set up a differential equation which involves the displacement and velocity.

    Hint: Notice that a=dv/dx=(dv/dx) x (dx/dt) = v (dv/dx)
    Thanks that makes a lot of sense.

    a=-kv^2 \Longrightarrow \dfrac{dv}{dt}=-kv^2

    v\dfrac{dv}{dx}=-kv^2

    \dfrac{dv}{dx}=-kv

    \displaystyle\int\dfrac{dv}{v}=\int -k \ dx

    \ln v=-kx+C

    When x=0, v=20 \Longrightarrow C = \ln 20

    \ln v =-kx+\ln 20

    When x=100, v=10

    \ln 10=-100k+\ln 20

    100k = \ln 20 - \ln 10

    100k = \ln 2

    k = \dfrac{\ln 2}{100}

    Have I done everything right? If so, I'm not too sure what to do next.

    The answer in the book is 20\ln 2
    Last edited by acevipa; September 8th 2010 at 05:37 AM.
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  7. #7
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    Quote Originally Posted by acevipa View Post
    Thanks that makes a lot of sense.

    a=-kv^2 \Longrightarrow \dfrac{dv}{dt}=-kv^2

    v\dfrac{dv}{dx}=-kv^2

    \dfrac{dv}{dx}=-kv

    \displaystyle\int\dfrac{dv}{v}=\int -k \ dx

    \ln v=-kx+C

    When x=0, v=20 \Longrightarrow C = \ln 20

    \ln v =-kx+\ln 20

    When x=100, v=10

    \ln 10=-100k+\ln 20

    100k = \ln 20 - \ln 10

    100k = \ln 2

    k = \dfrac{\ln 2}{100}

    Have I done everything right? If so, I'm not too sure what to do next.

    The answer in the book is 20\ln 2
    That's correct. Then, work from \dfrac{dv}{dt}=-kv^2

    For velocity, integrate from 20 to 10 and for time, integrate from 0 to t.
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