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Math Help - difficult integration

  1. #1
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    difficult integration

    well atleast i think this problem involves integration. i also duno how to derive this into a formula.

    ok, so this rocket is taken out to space so there are no external forces acting on it (ie if the rocket moves at 2m/s it will continue to move at that speed FOREVER as there is no other forces acting on it).

    total mass of the rocket is 363kg, fuel in the rocket is 333kg. (wiehgt of actual rocket itself is 363-333=30kg). the rate at which the fuel is burnt is 2.6kg/s therefore the engine can run for 128seconds because 333/2.6=128. the thrust force of the engine is 4990N (N = Newtons). now force=acceleration x mass therefoer acceleration=force/mass.

    when the rocket starts up the velocity is zero, the acceleration initially is 13.74m/s and the total weight of the rocket is 363kg. and just before all the fuel has ran out the acceleration is now 166.8m/s the total weight of the rocket is 30kg.

    ok so thats all the information you can use (you dont have to use all the information). the QUESTION is "what is the maximum speed when all of the fuel has been used". obviously thoughout the whole time (128seconds) the force comming out of the engine is 4990N and as the fuel is used up, the mass decreases and therefore acceleration increases so the rate of velocity will increase because there is no other external forces on the system except for the thrust force.

    ive been trying to do this question for a long time but unsuccessful. the only way i think this can be done is by integration but i have no idea how to derive this into a formula. yes it is a physics question but scince i have given all information needed it can be done using maths. much help will be appreciated (as theres a few of us trying to do this same question) thanks.
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  2. #2
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    Quote Originally Posted by phlik View Post
    well atleast i think this problem involves integration. i also duno how to derive this into a formula.

    ok, so this rocket is taken out to space so there are no external forces acting on it (ie if the rocket moves at 2m/s it will continue to move at that speed FOREVER as there is no other forces acting on it).

    total mass of the rocket is 363kg, fuel in the rocket is 333kg. (wiehgt of actual rocket itself is 363-333=30kg). the rate at which the fuel is burnt is 2.6kg/s therefore the engine can run for 128seconds because 333/2.6=128. the thrust force of the engine is 4990N (N = Newtons). now force=acceleration x mass therefoer acceleration=force/mass.

    when the rocket starts up the velocity is zero, the acceleration initially is 13.74m/s and the total weight of the rocket is 363kg. and just before all the fuel has ran out the acceleration is now 166.8m/s the total weight of the rocket is 30kg.

    ok so thats all the information you can use (you dont have to use all the information). the QUESTION is "what is the maximum speed when all of the fuel has been used". obviously thoughout the whole time (128seconds) the force comming out of the engine is 4990N and as the fuel is used up, the mass decreases and therefore acceleration increases so the rate of velocity will increase because there is no other external forces on the system except for the thrust force.

    ive been trying to do this question for a long time but unsuccessful. the only way i think this can be done is by integration but i have no idea how to derive this into a formula. yes it is a physics question but scince i have given all information needed it can be done using maths. much help will be appreciated (as theres a few of us trying to do this same question) thanks.
    This problem does deal with Newton's 2nd Law, but possibly in a way that isn't familiar to you.

    We need to know the relative velocity of the ejected material from the rocket. Using the impulse-momentum theorem we get:
    SumF(ext)*(Delta t) = p2(tot) - p1(tot)
    where (Delta t) is a small time interval, p2(tot) is the total moment of the rocket at time t + (Delta t) and p1(tot) is the total momentum of the rocket at time t.

    Let v denote the velocity of the rocket and V the relative velocity of the ejected fuel. Let (Delta m) represent the mass loss of the rocket over a time (Delta t). Then

    SumF(ext)*(Delta t) = (m + (Delta m))(v + (Delta v)) - (mv+ (Delta m)(v + V))
    ((Delta v) is the change in velocity of the rocket over a time (Delta t).)

    SumF(ext)*(Delta t) = m*(Delta v) + (Delta m)(Delta v) - V*(Delta m)

    Dividing through by (Delta t) and taking the limit as (Delta t) goes to 0, we get:

    SumF(ext) = m*dv/dt - V*dm/dt

    Since there are no external forces on the rocket SumF(ext) = 0, so
    m*dv/dt = V*dm/dt

    We may "cancel" the dt's, so:
    m dv = V dm

    V is a constant (presumably) so this equation separates:
    dv = V*dm/m

    Int[dv, 0, v(t)] = Int[V*(1/m) dm, m0, m(t)]
    where m0 is the initial mass of the rocket + fuel and m(t) is the mass of the rocket + fuel at time t.

    v(t) = V*ln(m0/m(t)) <-- This is a variation on the "rocket equation."

    We need to know V.


    We know the thrust is a constant F = 4990 N. So:
    F = V*dm/dt
    (I'm taking magnitudes here, but note that V would typically be in a negative sense, and dm/dt is negative so this works as vectors as well.)

    V = F/(dm/dt) = 1919.23 m/s


    So
    v(t) = (1919.23)*ln(363/m(t)) m/s

    When t = 128 s the rocket is out of fuel and the rocket has attained its maximum speed. The mass at this point is 30 kg, so
    v(128) = (1919.23)*ln(363/30) m/s

    v(128) = 4785.04 m/s

    -Dan
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  3. #3
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    that was exactly what i was seeking for. thank you very much, it helped me a lot.
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