- 02 Apr '11 15:37I 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?
- 04 Apr '11 09:47

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.*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....** - 05 Apr '11 02:11

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*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....**

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 - 05 Apr '11 04:25

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.*Originally posted by mtthw***Relative to what? Are we ignoring the velocity of the Earth?**