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Acceleration of a rocket

  1. Dec 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0s later.

    2. Relevant equations
    [itex]v_x = v_0 + at[/itex]
    [itex]x_f = x_0 + v_0t + 1/2at^2[/itex]
    [itex]v_x^2 = v_0^2 + 2a(Δx)[/itex]

    3. The attempt at a solution
    The biggest part I'm confused about is whether or not the bolt had an initial positive velocity (I am setting down as negative). It says the bolt "falls off the side", implying it just falls down.

    Position of bolt before it falls
    [itex]0m = x + 1/2(-9.8m/s^2)(6.0s)^2[/itex]
    [itex]x_i = 176.4m[/itex]

    Acceleration of rocket
    [itex]176.4m = 1/2(a)(4.0s)^2[/itex]
    [itex]a = 22.05 m/s^2[/itex]

    My answer is incorrect.
     
  2. jcsd
  3. Dec 13, 2014 #2

    Doc Al

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    Staff: Mentor

    The bolt will have the same velocity as the rocket at the moment it falls free. "Falls off the side" just means that it's not shot out or launched, but just let go.
     
  4. Dec 13, 2014 #3

    PeroK

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    Science Advisor
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    Gold Member

    If you were cycling along and you fell off your bike, would you have forward velocity?
     
  5. Dec 13, 2014 #4
    Fair enough. I gotta go now, but I'll attempt the problem later tonight. Conceptually I understand that - but the wording sort of got to me. Thanks for clarifying!
     
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