Suppose a small single-stage rocket of total mass m(t) is launched vertically, the positive direction is upward

Suppose a small single-stage rocket of total mass m(t) is launched vertically, the positive direction is upward, the air resistance is linear, and the rocket consumes its fuel at a constant rate.

A mathematical model for the velocity v(t) of the rocket is given by

$\f[\frac{dv}{dt} + \frac{k- \lambda}{m_0-\lambda t}v=-g +\frac{R}{m_0-\lambda t}\f]$

where k is the air resistance constant of proportionality, $\f[\lambda\f]$ is the constant rate at which fuel is consumed, R is the thrust of the rocket, $\f[m(t)=m_0-\lambda t\f]$ , $\f[m_0\f]$ is the total mass of the rocket at t=0, and g is the acceleration due to gravity.

I will show how to solve for part a.

(a) Find the velocity v(t) of the rocket if

$\f[m_0=200kg\f]$

$\f[R=2000N\f]$

$\f[\lambda=1kg/s\f]$

$\f[g=9.8m/s^2\f]$

$\f[k=3kg/s\f]$

and

$\f[v(0)=0\f]$

I looks like a super daunting equation at first but at least all of these inputs are in matching units, so we should be able to plug and a play to start breaking down our answer. and the IVP they have set up starts at the origin (0,0) so any constants of integration should be fairly simple to come up with at the end as well.

So starting off with filling in values

$\f[\frac{dv}{dt}+\frac{3kg/s-1kg/s}{200kg-1kg/s\cdot t}v=-9.8m/s^2 + \frac{2000N}{200kg-1kg/s\cdot t}\f]$

This can be simplified of course.

$\f[\frac{dv}{dt}+\frac{2}{200- t}v=-9.8 + \frac{2000}{200-t}\f]$

Now even though the integrating factor could be written out by hand, I thought I’d try my hand at coding a handle that would work for matlab to produce an answer.

I came up with the following:

simplify(dsolve(‘Dv+(2/(200-t))*v=-9.8+2000/(200-t)’,’v(0)=0′,’t’))

matlab returned…

ans =

(t*(3*t + 25))/125

while this answer is exact, it’s not how we usually write a velocity function, so the proper expansion is

$\f[v(t)=\frac{3}{125}t^2+\frac{1}{5}t\f]$