Acceleration.

i know how i calculate how far i can get whit an acceleration.
s=at²/2
but how do i do when the acceleration accelerate?

if the acceleration is not constant in time (which I assume is what you meant), you need to use more advanced math than the simple equation above. if you don't understand what I write below, please tell me and I'll try to explain it in another way.

we know from kinematics of particles that:
a = dv/dt (1)
v = ds/dt (2)
which gives when we substitute dt in both equations:
a*ds = v*dv (3)

if acceleration is constant, we get:
v = v0 + a*t (by integrating both sides in equation (1))
s = s0 + v0*t + a*t^2/2 (by substituting our expression for v into (2) and integrating both sides)
this is the equation you presented above.

if acceleration is a function of time (t), velocity (v) or displacement (s), we can use the three equations above, integrate them and get a result.

what i can see i dont see any where a increase whit time. but lets change the question so it matches what i want more.
Im thinking in forces. a force that is F at R metres from the thing that creates the force, but is 0,5F at 2R. i want to calculate how close we can get to the target body when we have velocity V,the force it creates goes away from the body. would be a negative acceleration

Ok, I assume that you are familiar with Newton's second law? (F = m*a)
First of all, let's cartesian coordinates and say that x=0 is where you start and the velocity is in positive x-direction.
Assuming that the force is linear, its equation would be: Ff(x) = -0.5*F*(1 + x/R)*i where i is the unit vector in the x-direction. Our acceleration would then be: a(x) = -0.5/m*F*(1 + x/R)*i
Let's use the relation named (3) above (a*dx = v*dv)
Insert a into it and we get:
-0.5/m*F*(1+x/R)*dx = v*dv
Now we integrate both sides (which I assume you know how to do) and we get:
v^2/2 = -0.5/m*F*(x+x^2/(2*R))
Insert our values (v(0)=V,v(s)=0) and we get:
-V^2/2 = -0.5/m*F*(s+s^2/(2*R)) or
F/(2*m) * s + F/(4*m*R) * s^2 - V^2/2 = 0
This equation has the solution:

s = -F/(8*m*R) +- sqrt( (F/(8*m*R))^2 + V^2/2)

And there you have your answer.
If you don't understand, just tell me so.

Zelos, you really need to start giving more information in your questions (and making them easier to read...)

A changing acceleration is called "jerk". Perhaps you could give a Clear problem showing more exactly of what your goal is.

its easy... You know a=dv/dt=dx/dt^2 so da/dt that is the rythm of accelerations alteration will be: da/dt=dv/dt^2=dx/dt^3... Which means da/dt=(((x(t))')')'

Just change x(t) with 1/2at^2 and then paragon three times...

wow, what math/science is this? i'm probebley going to have to do it soon so if i know i can study ahead of time
i knew this site would come in handy one day

Well, acceleration is mostly found in Physics--certainly the most worthy of the sciences to study. But without Calculus, you won't get far. Calculus is the Tool through which the Science of Physics is explored.

Originally Posted by Big Bro Davidia

Well, acceleration is mostly found in Physics--certainly the most worthy of the sciences to study. But without Calculus, you won't get far. Calculus is the Tool through which the Science of Physics is explored.

Sorry if this is hijacking the thread, but wha is calculus? i've heared of it before, but have no idea what it means (BTW im studying Physics and Maths as A/S Level)

final speed-start speed= Acceleration, so lets say you are starting at a rate of... 0.
time

yous speed up to 20 km in 5 seconds, or 20km/s.subtract 0 from it, and divide by 5.
20/5 is 4, so it's 4km/s/s. do you understand?