. One gravity = 32' per second/per second. I need a formula for determining distance travelled at x seconds

Question

How far and how fast would one travel after 1000 seconds if subjected to a constant accelleration of one gravity? What formula was used to arrive at this answer?

Answer 1

s= 1/2 a t^2

s: distance travelled
a: acceleration (32.2 ft/sec^2)
t: time that this acceleraton is maintained

Just be aware that this formula assumes your initial velocity is zero; if not you have to add

s= Vi t + 1.2 a t^2

where Vi is your initial velocity

Answer 2

I don't know what level math you are at because this involves (simple) calculus. Acceleration (gravity in your case) is the second derivative of displacement (distance). It is a constant (32 ft/sec^2) in this problem. So you just integrate twice to get the distance funtion. You can get a general formula (which you can find in the first chapter of any physics book) or you can get a specific solution but you need initial conditions such as initial velocity and initial placement.

Answer 3

Another lazy student. You're going to school to learn this crap. Open your books and figure it out. Ask you teacher for help if you're too damn lazy to pay attention in class. Jeez, you students are giving me hives with your ignorance.....OK, I feel better now.