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Math Help - Integration Help!

  1. #1
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    Integration Help!

    we got a homework question about the take off of a rocket, its acceleration proportional to fuel usage.

    we are given acceleration as:

    ((d^2)h) / (d(t^2)) = kb / m(subscript 0) - bt

    where dm/dt = -b refers to (The rate at which the total mass of the rocket changes with time is equal to the fuel consumption rate)

    integrating this function (dm/dt = -b) gives m(t) = -bt + constant, whereby the mass at time (0) = constant = m(subscript 0)

    the initial mass of the rocket is m(subscript 0)
    the initial mass of the fuel is Xm(subscript 0) where X is an unspecified fraction between 0 and 1

    A
    t some point in time, say at t = t*, the fuel runs out and the
    mass of the rocket will be its initial mass minus the initial mass of

    fuel. This is described by
    m(t*) = -bt* + m0
    m(t*) = m0 - X.m0

    k is an undefined constant

    well we start with the original formula for the acceleration:

    ((d^2)h) / (d(t^2)) = kb / m(subscript 0) - bt

    and i need to integrate this once, to find the velocity at a specific time, and then integrate again to find the height at a specific time

    i need to prove:

    h(t*) = (k.m0 / b) ((1-X)ln(1-X) + X(1-(g/2kb).Xm0))
    and then the velocity at the end of the boost phase: v(t*)
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  2. #2
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    Quote Originally Posted by mathsismyfriend View Post
    we got a homework question about the take off of a rocket, its acceleration proportional to fuel usage.

    we are given acceleration as:

    ((d^2)h) / (d(t^2)) = kb / m(subscript 0) - bt[/tex]
    Is all of that in the denominator? Is it \frac{d^2h}{dt^2}= \frac{kb}{m_0- bt}?

    [tex]where dm/dt = -b refers to (The rate at which the total mass of the rocket changes with time is equal to the fuel consumption rate)

    integrating this function (dm/dt = -b) gives m(t) = -bt + constant, whereby the mass at time (0) = constant = m(subscript 0)

    the initial mass of the rocket is m(subscript 0)
    the initial mass of the fuel is Xm(subscript 0) where X is an unspecified fraction between 0 and 1

    A
    t some point in time, say at t = t*, the fuel runs out and the
    mass of the rocket will be its initial mass minus the initial mass of

    fuel. This is described by
    m(t*) = -bt* + m0
    m(t*) = m0 - X.m0

    k is an undefined constant

    well we start with the original formula for the acceleration:

    ((d^2)h) / (d(t^2)) = kb / m(subscript 0) - bt

    and i need to integrate this once, to find the velocity at a specific time, and then integrate again to find the height at a specific time

    Okay, what have you done so far?

    i need to prove:

    h(t*) = (k.m0 / b) ((1-X)ln(1-X) + X(1-(g/2kb).Xm0))
    and then the velocity at the end of the boost phase: v(t*)
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  3. #3
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    i have tried rearranging formula's:

    -bt = Xm0

    m - bt = m0 - Xm0

    so that

    v(t) = integral of (kb / m0 - Xm0) - g .dt

    or i tried:

    v(t) = integral of (-k(dm/dt) / m0 + t(dm/dt)) - g . dt
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  4. #4
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    as a really complex problem i didn't imagine anyone would be able to help

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