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Rocket acceleration/displacement problem

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    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.



    2. Relevant equations
    Vf = Vi + a[tex]\Delta[/tex]t

    [tex]\Delta[/tex]y = Vi[tex]\Delta[/tex]t + 0.5a([tex]\Delta[/tex]t)^2

    (Vf)^2 = (Vi)^2 + 2a[tex]\Delta[/tex]y



    3. The attempt at a solution

    For Part A I calculated displacement with [tex]\Delta[/tex]y = Vi[tex]\Delta[/tex]t + 0.5a([tex]\Delta[/tex]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[tex]\Delta[/tex]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[tex]\Delta[/tex]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??
     
  2. jcsd
  3. Sep 21, 2009 #2
    Hi,

    You need to now how much time the second stage is burning.
     
  4. Sep 23, 2009 #3
    Okay, lets sort it all out.

    So you are calculating from the starting point,

    First stage net acceleration - [tex]4m/s^{2} [/tex]
    First stage burning time - [tex]10s[/tex]
    Second stage net accelaration - [tex]6m/s^{2}[/tex]

    When I talk about net acceleration, I am assuming those accelerations you are providing already accounted for acceleration due to gravity [tex] \stackrel{\rightarrow}{g} = -9.8 m/s^{2}[/tex].

    Equations that we are going to use,

    [tex] \stackrel{\rightarrow}{v_{f}} = \stackrel{\rightarrow}{v_{i}} + \stackrel{\rightarrow}{a}}\Delta t [/tex]

    [tex] \stackrel{\rightarrow}{\Delta y} = 0.5 \stackrel{\rightarrow}{a}} \Delta t^{2} + \stackrel{\rightarrow}{v_{i}}\Delta t[/tex]

    [tex] \stackrel{\rightarrow}{v_{f}^{2}} = \stackrel{\rightarrow}{v_{i}^{2}} + 2 (\stackrel{\rightarrow}{a}})(\stackrel{\rightarrow}{\Delta y})[/tex]

    Checking at your work for part A and part B, I found no problem with those.

    So you have [tex]\stackrel{\rightarrow}{y_{first}} = 200m [/tex] and [tex] \stackrel{\rightarrow}{v_{first}} = 40m/s [/tex].

    Here is the problem, as JasonGodbout has pointed out, you don't have the time for the second stage. In order to reach its highest point it needs to burn at a fixed time so that [tex]g[/tex] can acts on it and slows it down. Otherwise the rocket is just going to keep moving at a constant acceleration and it never stops. Unless your second stage acceleration is not net acceleration (I doubt it).

    Generally your strategy would be to solve for the displacement covered by the second stage acceleration. Then solve for the displacement covered after second stage acceleration has ended ([tex]g[/tex] is now your third acceleration) with a final velocity of [tex]0m/s[/tex]. Finally add the displacement from first stage to the two displacement findings above to solve for part 3.

    I'll leave part 4 to you then.
     
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