Thread: 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*)

Originally Posted by mathsismyfriend

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 ?

[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*)

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

as a really complex problem i didn't imagine anyone would be able to help