An expression for the vertical velocity as a function of time

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SUMMARY

The discussion focuses on deriving an expression for the vertical velocity of a rocket as a function of time, considering its initial mass (m0), fuel burn rate (m'), and constant buoyancy (T). The acceleration is defined as a = T/(m0 - m't) - g, where g represents gravitational acceleration. Participants emphasize the importance of understanding the relationship between acceleration and velocity, specifically noting that acceleration is the time derivative of velocity. The conversation encourages users to attempt both definite and indefinite integrals to find the solution.

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  • Familiarity with calculus, specifically integration techniques.
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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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What are your ideas about the problem?
 
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?
 
Physil said:
I just don't know if I should use definite or indefinite integral.
Try it both ways and see. It doesn't take that much time.
 

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