An expression for the vertical velocity as a function of time

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The discussion revolves around deriving the vertical velocity of a rocket as a function of time, considering its changing mass due to fuel consumption. The rocket's acceleration is defined as a = T/(m0 - m't) - g, where T is the constant buoyancy and g is the gravitational acceleration. Participants emphasize the importance of understanding the relationship between acceleration and velocity, suggesting that acceleration is the time derivative of velocity. There is a debate on whether to use definite or indefinite integrals for the calculations, with encouragement to try both methods. The overall focus is on guiding individuals to engage actively in solving the physics problem themselves.
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.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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