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

 

[tomcue]

First of all, well done on your attempt at solving this problem! You have used the correct equations for calculating displacement and final velocity. However, there are a few things that need to be corrected in your solution.

For Part A, you have correctly used the equation \Deltay = Vi\Deltat + 0.5a(\Deltat)^2. However, the value of Vi should not be 0. In fact, Vi is the initial velocity of the rocket, which is not given in the problem. So, you cannot solve for the displacement of the first stage using this equation.

For Part B, you have correctly used the equation Vf = Vi + a\Deltat to find the final velocity of the first stage. However, the value of Vi should not be 0. It should be the initial velocity of the rocket, which is not given in the problem. So, you cannot solve for the final velocity of the first stage using this equation.

For Part C, you have used the correct equation (Vf)^2 = (Vi)^2 + 2a\Deltay. However, you have used the wrong values for Vi and Vf. As I mentioned earlier, Vi is the initial velocity of the rocket, which is not given in the problem. So, you cannot solve for the displacement of the second stage using this equation.

To solve this problem, you need to use the equations of motion for constant acceleration, which are:

Vf = Vi + at

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

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

Where Vi is the initial velocity, Vf is the final velocity, a is the acceleration, t is the time, and \Deltay is the displacement.

For Part A, you can use the equation Vf = Vi + at to find the final velocity of the first stage after 10 seconds. Since the acceleration is +4 m/s/s, the final velocity of the first stage will be 40 m/s.

For Part B, you can use the equation \Deltay = Vi\Deltat + 0.5a(\Deltat)^2 to find the displacement of the first stage. Since the initial velocity is not given, you can assume it to be 0. Therefore, the displacement of the first stage will be 200 meters