Actually, this is basically the same in 2D and 3D and it all follows directly from the definition of velocity.
velocity = distance / time, time = distance / velocity. All you need is the actual distance, which can be computed using vector subtraction (x2-x, y2-y) using the Pythagorean theorem: distance = sqrt(a^2+b^2). For 3D, you add the third dimension and are done.
have another question, how do i get X2, Y2 position depending of the given time, velocity, and distance? Do i need the slope of the line?
http://en.wikipedia.org/wiki/Slope
float2 startingPos = given;
float2 velocity = given;
float time = given; // Time since t=0. At t=0, the object is at position startingPos.
return startingPos + time * velocity;
Well... distance or time and velocity, you don't need all of them. You have to know the direction of motion (as an angle or just give the velocity as a vector). In the first case, you have to use trigonometry (posX = posXold + velocity*time*sin(angle), posY = posYold + velocity*time*cos(angle)), in the second case it's just pos = oldpos + time * velocity (pos and velocity as a vector).
But seriously, that are really basic questions usually covered in high school maths and physics, I strongly recommend you read up on basic algebra, geometry and kinematics.
