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Non-free-fall Acceleration Problem

  1. Feb 6, 2004 #1
    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."
  2. jcsd
  3. Feb 6, 2004 #2


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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:

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

    (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.
  4. Feb 6, 2004 #3
    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.
  5. Feb 6, 2004 #4


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    No problem at all. The more problems you do, the easier these will get; you may even start having fun doing them.
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