An expression for the vertical velocity as a function of time

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Physil
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A rocket of initial mass m0 is launched vertically upwards from the rest. The rocket burns fuel at the constant rate m', in such a way, that, after t seconds, the mass of the rocket is m0-m't. With a constant buoyancy T, the acceleration becomes equal to a=T/(m0-m't) -g. The atmospheric resistance can be neglected, and the gravitational accelereation ,g, is considered a constant for low-level flights. Deduce an expression for the vertical velocity v of the rocket, as a function of time t, before the fuel burns out completely.
 
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We're not here to solve physics problems for you. We're here to help YOU solve them. You have to make an effort towards a solution. If you need a hint to get started, acceleration is the time derivative of velocity.
 
Mister T said:
We're not here to solve physics problems for you. We're here to help YOU solve them. You have to make an effort towards a solution. If you need a hint to get started, acceleration is the time derivative of velocity.
I'm sorry. I new here. That's exactly what I was thinking. I just don't know if I should use definite or indefinite integral.
 
Physil said:
I'm sorry. I new here. That's exactly what I was thinking. I just don't know if I should use definite or indefinite integral.
Why would you use an indefinite integral? Does the integration have a starting point and stopping point?