Non-free-fall Acceleration Problem

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The discussion centers on a student's confusion regarding a physics problem involving a rocket's acceleration. The student initially used the wrong formula and calculated the speed incorrectly as 3000 m/s. A respondent clarified that the correct approach is to use the formula v = v0 + at, given the constant acceleration and initial velocity of zero. They emphasized the importance of recognizing the problem's structure and provided key kinematic equations for future reference. The student expressed gratitude for the assistance and acknowledged the need to practice more problems for better understanding.
holly
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Sorry to have so many questions. Obviously, I am just not understanding the material. Our professor does not provide keys to problem sets he gives...

The Question: If a rocket initially at rest accelerates at a rate of 50m/s/s for 1 minute, its speed will be: ? I used d=1/2 at**2 (evidently the only formula I can remember, ha ha)...and got 3000 m/s. Correct?

Thank you...
"If I have seen less than others, it is because some giant's shoulders are always in the way."
 
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Your answer is correct, but it looks like you made it more complicated than necessary; if you write out exactly how you got the answer you did, we may be able to clear that up a little.

This problem is exactly the same as your last problem. You are given a constant acceleration, a duration (time) for that constant acceleration, and an initial velocity (starts from rest; v0 = 0). Using v = v0 + at will give you the velocity at a given time.

Here are the main kinematic equations that you should keep in mind for constant acceleration problems:

v = v_0 + at
v^2 = v_0^2 + 2a\Delta x
x = x_0 + v_0 t + \frac{at^2}{2}

(They all come from the definitions of velocity, acceleration, and position, but that's only important if you're interested.)

I hope those all look familiar. You should notice that the first equation gives velocity as a function of time, the second one gives velocity as a function of distance, and the last one gives position as a function of time. (They all are for constant acceleration.) Try to get a feel for what you should use based on what information is given in your problem and what is asked in the problem. Hope that clears things up a bit.
 
Thx for Help

Thank you for the help and especially for the advice about getting a feel for what it is I am being asked in the questions. I'm mortified that I did not recognize that it was basically the same problem as the one you answered before. And thank you for the kinematic equations, I will share them with my classmates. Thank you again.
 
No problem at all. The more problems you do, the easier these will get; you may even start having fun doing them.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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