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Homework Help: Solve 1-D dynamics problem using integration and ?

  1. Dec 15, 2013 #1
    Solve 1-D dynamics problem using integration and ???

    1. The problem statement, all variables and given/known data

    A 75kg rocket is launched directly upward. The force on the rocket due to its engine decreases from 5000N to 0N as a quadratic function of time. The rocket reaches a speed of 200m/s during the time interval that its engine fires.

    2. Relevant equations

    Ft = 5000 - At2
    a = -1/75At2 + 66.6 (the force of thrust minus force of gravity divided by mass)

    3. The attempt at a solution

    I tried integrating acceleration and end up with both a time and a variable A left that I can't seem to get rid of. With no given time or position beyond the start, I'm at a loss as to how to solve this? I even tried impulse-momentum and comparing it to the integrated acceleration (velocity), but I ended up getting a negative time for my answer. Also, this question is in a part of our textbook that is only analyzing kinematics and dynamics, so it seems as if they aren't even expecting you to use conservation laws on it. If anyone can help point me in the right direction, I'd appreciate it.
    Last edited: Dec 15, 2013
  2. jcsd
  3. Dec 15, 2013 #2


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    You haven't made clear what it is you're supposed to work out.

    In any case, there is a big assumption here - that the rocket's mass doesn't change (from fuel burn). Is this justifiable?

    Assuming it is, there still seems to be an error in your expression for the acceleration because the 4265 term also needs to be divided by the mass.
  4. Dec 15, 2013 #3


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    After you've made Curious3141's correction, I think you are given that the maximum velocity your rocket reaches is 200m/s, right? At that point the acceleration is 0. Use that to solve for the time when that happens in terms of A. Then put that time into your integrated acceleration expression, which must equal 200m/s at that time. That should let you solve for A.
    Last edited: Dec 15, 2013
  5. Dec 15, 2013 #4
    Apologies about that. I corrected the value in the problem and as far as what we are solving for, it is for the time where it reaches the apex.
  6. Dec 15, 2013 #5
    Ah, thank you. I was definitely over-complicating this one. Tried it out and got the right value. I really appreciate the help.
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