Can anyone tell me the correct equation thats something like X= Xo + Vt + 1/2at^2 or something like that?

Question

Can anyone tell me the correct equation thats something like X= Xo + Vt + 1/2at^2 or something like that?

Answer 1

That's correct, assuming X = position, Xo = original position. V = original velocity, a = constant acceleration, t = time.

Answer 2

Hi,

I am assuming you meant the formula used for Constant-Acceleration Problems (well it's obvious, hehehe). Anyway, to help you out there are 4 fundamental equations to answering Constant-Acceleration Problems and the one you just showed is actually correct. In any case if ever you would need them, here are the 4:

1.) To look for the velocity: v=vi + at; where (v) is the velocity, (vi) is the initial velocity and (a)(t) is acceleration multiplied to time.
2.) To look for the displacement in terms of velocity: d-di = (1/2)(vi+v)(t); where d is the displacement and (di) is the initial displacement.
3.) To look for displacement in terms of a: d-di=(vi)(t) + (1/2)(a)(t^2); where (t^2) is time squared (which is the equation you are refering to.)
and last:
4.) the velocity in terms of a and d: (v^2)=(vi^2) + 2a(d-di)

Hope this helps!

Answer 3

this equation is for mechanics and projectiles.

to determine (x max) = horizontal range of projectile
to determine (y max) = maximum point of projectile
(y)=any point of y at certain (t) time
(x)=any point of y at certain (t) time

x = V(initial) cos (alpha) (t)
y = V(initial) sin (alpha) (t) -1/2gt^2

ymax = V(inital) cos(alpha) -1/2gt^2