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To find rocket's acceleration and velocity

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1203.0 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 11.2 s, then the motor stops. The rocket altitude 24.1 s after launch is 2023.0 m. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 11.2 s? What is the rocket's velocity as it passes through a cloud 2023.0 m above the ground?

    2. Relevant equations

    (a)d=Vit + 1/2 at^2...
    (b) Vf= Vi + at...
    (c) d= di + Vf (t) - 1/2 at^2

    3. The attempt at a solution
    So I used the first equation to findthe distance the rocket travelled in the first 11.2 seconds and subtract that with altitude given....and know that the starting speed is 0m/s i tried to determine the speed using vfinal {throughout the first 2 eqns. i used a= -9.80 m/s^2}and then plugged it into the last equation to determine the acceleration..im gettin the wrong answer ..pls help
     
  2. jcsd
  3. May 8, 2009 #2

    diazona

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    Well, first thing to think about: if an object is accelerating at -9.8 m/s^2, and its initial velocity is 0, is it moving down or up? Does that make sense for your problem?

    -9.8 m/s^2 is the acceleration of gravity. But that's not the only acceleration involved - the rocket engine adds some acceleration of its own.
     
  4. May 8, 2009 #3

    LowlyPion

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    They want you to find 2 things. Vfinal and the initial a.

    I'd look to develop 2 equations that are made of the 2 unknowns, and then solve.

    For instance

    Vcutoff = 11.2*a

    and

    Xcutoff = 1/2*a*(11.2)2

    That yields ...

    Vfinal2 = Vcutoff2 - 2*g*(2023 - Xcutoff)

    Substituting into that then should yield your first equation in the 2 unknowns you need to find.

    You just need another equation in the 2 unknowns then and you can solve.
     
  5. May 9, 2009 #4
    so if i use the equation:: Vfinal2 = Vcutoff2 - 2*g*(2023 - Xcutoff) and sub in to it the vcutoff and xcutoff i will still have the 2 unknowns in the same questions..this is a really stupid question but 2 eliminate one of the unknowns i wld have to use the other unknown correct?? but theres no additional info. given in the question to do this....
     
  6. May 9, 2009 #5

    LowlyPion

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    Sure there is.

    Consider how the cutoff velocity is affected by gravity. How does that relate to Vfinal?
     
  7. May 9, 2009 #6
    in the cutoff velocity the acceleration inclds gravity right?? and im not really sure how it relates to vfinal...
     
  8. May 10, 2009 #7

    LowlyPion

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    The statement of the problem - constant acceleration to cutoff - means that gravity was taken into account in determining the acceleration to that point.

    But after cutoff it is being slowed by gravity - there is no more propellant. So ...

    Vfinal = Vcutoff - g*t

    You can figure the time. You know g, so ...
     
  9. May 10, 2009 #8
    and the time wld b the differnce between 11.2 s and 24.1 s where the altitude was measured ..correct??
     
  10. May 10, 2009 #9

    LowlyPion

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    That looks right.
     
  11. May 10, 2009 #10
    ok so i did that and my hwk app. is telling me that the answer im getting is wrong any more hints u can give me wld b really appreciated
     
  12. May 10, 2009 #11

    LowlyPion

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    It look's like a complicated calculation. Looks like you get a quadratic equation don't you? Are you sure you solved the quadratic correctly?

    Maybe show what you get for your quadratic?
     
  13. May 10, 2009 #12

    LowlyPion

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    OK. I see the quadratic in a seems to cancel out the quadratic term., so it is a little simpler.

    Maybe write out your equation with all the substitutions. There seems to be ample opportunity for arithmetic slip ups in this.
     
  14. May 10, 2009 #13
    this is what i get as a quadratic equation:: 125.4a^2 + 1180.48a- 52800.002=0

    and for my acceleration value i got 16.345m/s^2
     
  15. May 10, 2009 #14

    LowlyPion

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    As just mentioned when I worked it out, I lose the quadratic in a. It seems to cancel out.

    Vf2 = (11.2*a)2 - 2*g*(2023 - 1/2*a*11.22)

    Vf = 11.2*a - 12.9*g

    (11.2*a - 12.9*g)2 = (11.2*a)2 - 2*g*(2023 - 1/2*a*11.22)

    The (11.2*a)2 term on both sides cancels. So I don't see an a2 term.
     
  16. May 10, 2009 #15
    i went over my calcs. and i realized i 4got 2 square my Vcutoff but i still didnt get the answer...the gravity will be negative thoughout the calcs right???
     
  17. May 10, 2009 #16

    LowlyPion

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    g is negative, but I accounted for that directly by putting its direction in the equation.
     
  18. May 10, 2009 #17
    thank u so much...:))
     
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