Rocket velocity/displacement problem, HELP

[UNDEAD WAFFLE]

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 acceleration 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$$\Delta$$t

$$\Delta$$y = Vi$$\Delta$$t + 0.5a($$\Delta$$t)^2

(Vf)^2 = (Vi)^2 + 2a$$\Delta$$y

The Attempt at a Solution

For Part A I calculated displacement with $$\Delta$$y = Vi$$\Delta$$t + 0.5a($$\Delta$$t)^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$$\Delta$$t.

I had a Vi of 0, an acceleration 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$$\Delta$$y
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??

 

[willem2]

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 ${V_f}^2 = {V_i}^2 + 2a \Delta y$