1. Subscribersonhouse
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    02 Apr '11 15:37
    I am accelerating at 1.8 G's away from Earth, taking off from the ground, I happen to be at the north pole so I don't get the advantage of the spinning Earth, where you launch on the equator to give a 1.6 odd Km/sec freebie. So I go straight up, I can maintain that 1.8 g's right through the atmosphere, powerful engines crank up to keep the craft at exactly 1.8 g's regardless of drag. So how fast am I going after one hour of that accel and how far have I traveled?
  2. SubscriberAThousandYoung
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    03 Apr '11 00:001 edit
    1.8 * 9.8 m/s^2 * 1 hour * 60 min/hour * 60 sec/min = velocity.

    velocity * time (360 s) = distance

    EDIT - Engines will need to crank DOWN as drag and gravity decrease over time.
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    03 Apr '11 11:45
    AVERAGE velocity * time = distance
  4. Subscriberjoe shmo
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    03 Apr '11 14:521 edit
    (Vf^2-Vo^2)/(2a)= distance

    or

    distance = Vo*t + 1/2*a*t^2
  5. Subscribersonhouse
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    03 Apr '11 16:14
    Originally posted by AThousandYoung
    1.8 * 9.8 m/s^2 * 1 hour * 60 min/hour * 60 sec/min = velocity.

    velocity * time (360 s) = distance

    EDIT - Engines will need to crank DOWN as drag and gravity decrease over time.
    I think you are forgetting one thing.
  6. SubscriberAThousandYoung
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    03 Apr '11 21:40
    Originally posted by sonhouse
    I think you are forgetting one thing.
    You're right...average velocity times time.
  7. Subscribersonhouse
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    04 Apr '11 02:53
    Originally posted by AThousandYoung
    You're right...average velocity times time.
    Something else.
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    04 Apr '11 04:321 edit
    Originally posted by joe shmo
    (Vf^2-Vo^2)/(2a)= distance

    or

    distance = Vo*t + 1/2*a*t^2
    Agree with the second equation, but what were you getting at with the first equation? Distance travlled seems to go down as acceleration goes up....
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    04 Apr '11 09:43
    Originally posted by sonhouse
    So how fast am I going after one hour of that accel and how far have I traveled?
    Relative to what? Are we ignoring the velocity of the Earth?
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    04 Apr '11 09:47
    Originally posted by iamatiger
    Agree with the second equation, but what were you getting at with the first equation? Distance travlled seems to go down as acceleration goes up....
    No, because Vf is also a function of acceleration. That equation is what you get if you take the last equation, and use Vf = Vo + at to eliminate t.
  11. Subscriberjoe shmo
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    05 Apr '11 02:11
    Originally posted by iamatiger
    Agree with the second equation, but what were you getting at with the first equation? Distance travlled seems to go down as acceleration goes up....
    as well as what mtthw said, you can also think about it a a statement of conservation of energy, negating all other forms with the exception of kinetic, and work

    multiply both sides of the equation by mass do a little rearranging and you come to:

    1/2*m*(Vf^2 - Vo^2) = m*a*(Xf -Xo) = F*d = Work
  12. Subscribersonhouse
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    05 Apr '11 04:25
    Originally posted by mtthw
    Relative to what? Are we ignoring the velocity of the Earth?
    No, we are not ignoring the velocity of Earth. Just the velocity of the sun around the galaxy and the 200 mps or so velocity of the whole galaxy. However, that makes for a vector velocity. Remember where it took off.
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