Rocket velocity/displacement problem, HELP

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The discussion revolves around solving a two-stage rocket problem involving acceleration and displacement. For Part A, the user calculated the vertical displacement of the first stage as 200 meters using the appropriate kinematic equation. In Part B, the final speed after 10 seconds was determined to be +40 m/s. The user seeks clarification on Part C, specifically how to calculate the total height after the second stage is activated, noting confusion with negative signs in their calculations. The conversation emphasizes the need to account for the second stage's burn time and the transition to freefall after the rocket's engines stop.
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Homework Statement


A two stage rocket is launched with an average acceleration of +4 m/s/s. After 10 seconds, a second stage is activated and the rocket's acceration is now +6 m/s/s.

Part A: Find the vertical displacement of stage one of the rocket, before accleration changes to 6.

Part B: Find the final speed after 10 seconds of motion.

Part C: The second stage is activated, find the total height the rocket ascends (its highest point) before it starts to travel back to earth.

Part D: Find the displacement traveled by the second stage only.



Homework Equations


Vf = Vi + a\Deltat

\Deltay = Vi\Deltat + 0.5a(\Deltat)^2

(Vf)^2 = (Vi)^2 + 2a\Deltay



The Attempt at a Solution



For Part A I calculated displacement with \Deltay = Vi\Deltat + 0.5a(\Deltat)^2.
y was my unknown variable and my Vi was 0, my t 10 seconds, and my a +4.
I got 200meters.


For Part B I calculated the first stage's final speed by using the equation Vf = Vi + a\Deltat.

I had a Vi of 0, an accleration of +4 and a t of 10. My velocity final was +40 m/s.

As far as part c goes I plugged everything into
(Vf)^2 = (Vi)^2 + 2a\Deltay
this time I had a velocity initial as +40 [part b] and i used 0 as my velocity final because i was solving for the peak. However, my negative signs got messed up. Here's where I need help. Am I on the right track??
 
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you need to know how long the second stage burns. When the second stage burns
a = 6 m/s^-2 and you can compute the final speed and height as you did in part A and B.
After the second stage stops burning the rocket is in freefall, but it will still move up for some time and you can use {V_f}^2 = {V_i}^2 + 2a \Delta y
 
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