Kinematics problem- seems easy but having trouble

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The discussion revolves around a kinematics problem involving a rocket's acceleration and a bolt that falls off after four seconds. The initial velocity of the rocket is zero, and the challenge lies in determining the rocket's acceleration based on the bolt's fall, which takes six seconds. Participants clarify that while the bolt initially moves upward with the rocket, its subsequent motion is influenced by gravity. The equations of motion for both the rocket and the bolt are discussed, emphasizing the importance of correctly applying kinematic equations. The conversation concludes with a consensus that the bolt's acceleration is not zero, but its upward motion ceases once it detaches from the rocket.
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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:
 
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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.
 
Just for kicks, how did you come up with the notion that the acceleration of the screw is zero? I am 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.
 
DaMastaofFisix said:
Just for kicks, how did you come up with the notion that the acceleration of the screw is zero? I am 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.
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.
 
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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