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Thread: 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 $\displaystyle 20m/s$ to $\displaystyle 10m/s$ in a distance of $\displaystyle 100m$. Find the average speed (w.r.t. time) during this period.

    Would the equation be

    $\displaystyle \dfrac{dv}{dt}=-kv^2$ where $\displaystyle k$ is a constant of proportionality
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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 $\displaystyle 20m/s$ to $\displaystyle 10m/s$ in a distance of $\displaystyle 100m$. Find the average speed (w.r.t. time) during this period.

    Would the equation be

    $\displaystyle \dfrac{dv}{dt}=-kv^2$ where $\displaystyle k$ 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.

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

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

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

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

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

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

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

    $\displaystyle \dfrac{1}{v}=kt+C$
    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.

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

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

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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.

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

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

    $\displaystyle \dfrac{dv}{dx}=-kv$

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

    $\displaystyle \ln v=-kx+C$

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

    $\displaystyle \ln v =-kx+\ln 20$

    When $\displaystyle x=100, v=10$

    $\displaystyle \ln 10=-100k+\ln 20$

    $\displaystyle 100k = \ln 20 - \ln 10$

    $\displaystyle 100k = \ln 2$

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

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

    The answer in the book is $\displaystyle 20\ln 2$
    Last edited by acevipa; Sep 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.

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

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

    $\displaystyle \dfrac{dv}{dx}=-kv$

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

    $\displaystyle \ln v=-kx+C$

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

    $\displaystyle \ln v =-kx+\ln 20$

    When $\displaystyle x=100, v=10$

    $\displaystyle \ln 10=-100k+\ln 20$

    $\displaystyle 100k = \ln 20 - \ln 10$

    $\displaystyle 100k = \ln 2$

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

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

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

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