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Math Help - Finding Velocity with Respect to Time (IS FRUSTRATINLGY HARD)

  1. #1
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    Exclamation Finding Velocity with Respect to Time (IS FRUSTRATINLGY HARD)

    hey this is a question i have for Year 12 Maths C in Australia. It is irritating me no end. <line of text deleted by CaptainBlack>
    here we go...

    A particle of mass one unit moves in a straight line in the horizontal plane against a resistance v+av^2 , where a is a positive constant. If v is the velocity of the particle at time t , then dv/dt=-(v+av^2) represents the particle’s motion.
    Given this particle is released from the origin, O, with an initial velocity of u, find the velocity of the particle v as a function of the time t .

    As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?

    PLEASE HELP! I WILL BE FOREVER IN YOUR DEBT. EMAIL ME OR RESPOND HERE, I WILL CHECK IT EVERY 30 MINUTES

    THANKS
    Last edited by CaptainBlack; August 27th 2007 at 10:22 PM.
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  2. #2
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    Quote Originally Posted by Ned Hunter View Post
    hey this is a question i have for Year 12 Maths C in Australia. It is irritating me no end. <line of text deleted by CaptainBlack>
    here we go...

    A particle of mass one unit moves in a straight line in the horizontal plane against a resistance v+av^2 , where a is a positive constant. If v is the velocity of the particle at time t , then dv/dt=-(v+av^2) represents the particle’s motion.
    Given this particle is released from the origin, O, with an initial velocity of u, find the velocity of the particle v as a function of the time t .

    As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?

    PLEASE HELP! I WILL BE FOREVER IN YOUR DEBT. EMAIL ME OR RESPOND HERE, I WILL CHECK IT EVERY 30 MINUTES

    THANKS
    The ODE

    <br />
\frac{dv}{dt}=-(v+av^2)<br />

    is of variables seperable type so has solution:

    <br />
\int_{v=u}^{v(t)} \frac{1}{v+a v^2} dv = -\int_0^t dt<br />

    The first integral is done by resolving the integrand into partial fractions,
    and the second is -t.

    That will give you t as a function of v which
    you will need to rearrange to give v as a function of t.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Ned Hunter View Post

    As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?
    If the initial velocity u is positive the acceleration is negative and so the particle slows down and as t \to \infty,\ \ v \to 0.

    If the initial velocity is negative, the velocity will change until the acceleration is zero, that is v+av^2=0, that is v=0 or v=-1/a. The first of these will not occur as all trajectories starting with negative velocity are repelled by this solution. The second is an attractor, that is all trajectories of v which start negative more towards it.

    RonL
    Last edited by CaptainBlack; August 28th 2007 at 05:17 AM.
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    Quote Originally Posted by CaptainBlack View Post
    The ODE

    <br />
\frac{dv}{dt}=-(v+av^2)<br />

    is of variables seperable type so has solution:

    <br />
\int_{v=u}^{v(t)} \frac{1}{v+a v^2} dv = -\int_0^t dt<br />

    The first integral is done by resolving the integrand into partial fractions,
    and the second is -t.

    That will give you t as a function of v which
    you will need to rearrange to give v as a function of t.

    RonL
    hey erm i dont quite get what you are saying here? do u think you could type the steps you took to achieve this? that would be a huge help. thanks.

    Ned
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  5. #5
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    Quote Originally Posted by Ned Hunter View Post
    hey erm i dont quite get what you are saying here? do u think you could type the steps you took to achieve this? that would be a huge help. thanks.

    Ned
    Hello,

    as the Captain pointed out you have to get the variable t at one side of the equation and the variable v at the other side:

    \frac{dv}{dt}=-(v+av^2) . Multiply by dt

    dv=-(v+av^2) \cdot dt. Now divide by (v+av^2)

    \frac{1}{(v+av^2)} \cdot dv=-dt. Now integrate both sides. You'll get the equation which CaptainBlack gave you.
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  6. #6
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    thanks

    hey thanks heaps for that. erm hate to be an irritating twerp but can you type it out? i dont get how to rearrange it to get v(t)

    cheers,
    ned
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  7. #7
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    Quote Originally Posted by Ned Hunter View Post
    hey thanks heaps for that. ... i dont get how to rearrange it to get v(t)
    ...
    Hello,

    CaptainBlack wrote:"The first integral is done by resolving the integrand into partial fractions, and the second is -t."
    I'm going to show you what he meant:

    \int_{v=u}^{v(t)} \frac{1}{v+a v^2} dv = -\int_0^t dt. I assume that the integral of the right side isn't a problem to you.

    Take \frac{1}{v+a v^2} = \frac{N}{v} + \frac{M}{1+av} = \frac{N+Nav+Mv}{v(1+av)}. Thus you know that
    N = 1 \text{ and } Nav+Mv = 0 (There isn't a summand in the numerator of the fraction which contains the variable v). Solve for N and M:
    N = 1 and M = -a. Therefore your integral becomes:

    \int_{v=u}^{v(t)} \frac{1}{v} dv - \int_{v=u}^{v(t)} \frac{a}{1+a v} dv= -\int_0^t dt

    1. \int \frac{1}{v} dv = \ln(v)

    2. \int \frac{a}{1+a v} dv = \ln(1+av) Use integration by substitution here: u = 1+av; u' = a

    3.  -\int dt = -t

    Without considering the bounds you have now:

    \ln(v) - \ln(1+av) = -t

    \ln \left(\frac{v}{1+av} \right) = -t. Now de-logarithmize(?) using the base e:

    \frac{v}{1+av} = e^{-t}. Multiply both sides by the denominator:

    v = e^{-t} \cdot (1+av)~\Longrightarrow~ v = e^{-t} + a \cdot e^{-t} \cdot v

    v(1-a\cdot e^{-t})=e^{-t}

    v=\frac{e^{-t}}{1+a\cdot e^{-t}} =\frac{1}{e^t (1+a\cdot e^{-t})}= \frac{1}{e^t-a}

    I'll leave the rest for you
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