An expression for the vertical velocity as a function of time

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Homework Help Overview

The problem involves a rocket launched vertically, with its mass decreasing over time due to fuel consumption. The focus is on deriving an expression for the vertical velocity as a function of time, considering factors like buoyancy and gravitational acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity, with some suggesting the need to consider integration methods. Questions arise about whether to use definite or indefinite integrals and the implications of each choice.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to integration and questioning the appropriate methods to apply. Some guidance has been offered regarding the relationship between acceleration and velocity, but no consensus has been reached on the integration approach.

Contextual Notes

Participants express uncertainty about the integration process, specifically regarding the starting and stopping points for the integral. There is an emphasis on the need for individual effort in solving the problem.

Physil
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New user has been reminded to show their work on schoolwork questions
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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