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Kinematics problem- seems easy but having trouble

  1. Sep 21, 2005 #1

    G01

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    So the problem is this: A rocket is launched straight up with a constant acceleration. Four seconds after liftoff a bolt falls off the rocket and hits the ground 6 seconds later. What was the rockets acceleration?

    Well the initial velocity of the rocket is 0m/s and the velocity of the rocket at four seconds is equal to the initial velocity of the screw. The acceleration of the screw is -g and at first I assumed the distance the bolt fell is the same as the distance the rocket went up but i didn't take into account the fact the the bolt would continue to move up for a tiny bit before it started to fall. Thats all I can get , can someone point me in the right direction here. I know I know how to solve it but i can't find a formula or group of formulas to use.

    Thanks Alot :confused:
     
  2. jcsd
  3. Sep 21, 2005 #2
    The 1D kinematic equations for a constant acceleration is:

    x(t) = x_o + v_o*t + at^2/2, v(t) = dx/dt = v_o + at

    Let r(t) = the position of the rocket, and s(t') = the position of the screw, where t' = t-4. It's easy to find the kinematic equation for the rocket. As for the screw, its initial position is r(4) and its initial velocity is v_r(4), where v_r(t) is the velocity of the rocket. Obviously, the acceleration of the screw is zero. Figure out what s(t'=6) is and solve for a.
     
  4. Sep 22, 2005 #3
    Just for kicks, how did you come up with the notion that the acceleration of the screw is zero? Im 100 percent sure that there is not only no need for calculus, but that the acceleration is never zero. Unless I too skipped something in this problem, I found the soltion wih just some simple algebraic substitution. If you both are interested, let me know.
     
  5. Sep 22, 2005 #4
    Nope, you're right, the acceleration of the screw is never zero. What I meant to say is that the contribution to the acceleration by the rocket for the screw is zero for any time after four seconds.
     
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