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 $\displaystyle \frac{d^2h}{dt^2}= \frac{kb}{m_0- bt}$?

[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