A rocket is at rest on the moon, where g=2 m/s^2. It has a total mass of 1500kg, 1000kg of which is fuel. Fuel is propelled out of the bottom of the rocket at 400m/s, at a rate of 10kg/s. Derive an equation to show its upwards acceleration over time.

The first step is to construct a force diagram of the rocket, which has two forces acting on it; its weight (W) and thrust (T) in opposite directions. T can be calculated by combining the equations F=ma and v=u+at. Rearranging the equations (and removing "u" as it is equal to zero) gives us a=F/m and a=v/t. These can then be combined to give F/m=v/t and rearranged to present F=mv/t. We can now substitute in the values given to us; F=10*400/1= 4000N of thrust. We must now find the weight component. As the rocket is getting constantly lighter (it is burning fuel), its weight will change and so the force opposing its upward motion will decrease. We must therefore introduce a time variable "t". We can now represent the weight of the rocket as (1500-10t)*2= 3000-20t N (0<t<100). The resultant upwards force on the rocket can now be seen to be 4000-(3000-20t) = 1000+20t N. Going back to F=ma, we have derived the force on the rocket and its mass, so can substitute those in to the equation; 1000+20t=(1500-10t)*a. This is now rearranged to give "a" in terms of "t" (so that we answer the question) to give a=(1000+20t)/(1500-10t) (0<t<100).