# Hard equation (for me anyways) help!

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.

Accelaration is rate of change of velocity. Since F=ma, a=F/m=dv/dt
Mass of the rocket varies with time, every second it reduces 2.6kg from 363kg
so m=363-2.6t
a=4990/m=4990/(363-2.6t)=dv/dt
dv=4990 dt/(363-2.6t)
Integrating, you get v=4990/2.6 * ln (Initial mass / final mass)
=4990/2.6*2.4932
=4785m/s

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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thank you very much, that helped a lot.