Expected 3d position after x seconds, given 3d speed and accel vectors I need to find out where an objects is expected to be, after X units of time has passed.
Say 10 units of time has passed.  The object is at (3,4,1).  The speed vector is (0.01,0.02,0.001).  The acceleration vector is (0.1, 0.01, 0.2).
What is the easiest, quickest, accurate way to find out where the object will be at the end of 10 units of time?
Is there a way to calculate an equation of 3d movement given the speed and acceleration vectors?  then add that to the current position vector?
Any help would be greatly appreciated - thanks!  I am a programmer looking to solve a heavily mathematically problem I do not fully understand :)
 A: This is what physicists call kinematics.  It can get pretty complicated, but thankfully you're looking at a relatively simple example of it.
Since this is all in Cartesian coordinates, and the acceleration is constant, this is fairly straightforward.  Let's consider the $x$-direction first.  You have
$$
\frac{d^2 x}{dt^2} = a_x, 
$$
where $a_x$ is the $x$-component of the acceleration vector.  Integrate this once with respect to $t$ and you get
$$
\frac{dx}{dt} = a_x t + C,
$$
where $C$ is a constant of integration.  Since we want $dx/dt$ to be equal to a known initial velocity $v_{0x}$ at $t = 0$, this implies that $C = v_{0x}$:
$$
\frac{dx}{dt} = a_x t + v_{0x}.
$$
Now we can integrate again with respect to $t$:
$$
x = \frac{1}{2} a_x t^2 + v_{0x} t + D,
$$
where $D$ is another constant of integration.  Since we know the object's position at $t = 0$ to be $x_0$, this implies that $D = x_0$.  All told, then,
$$
\boxed{  x = \frac{1}{2} a_x t^2 + v_{0x} t + x_0, }
$$
where $a_x$ is the acceleration in the $x$-direction, $v_{0x}$ is the object's initial velocity in the $x$-direction, $x_0$ is its initial $x$-coordinate, and $t$ is the amount of time elapsed.
Similar equations hold for the $y$-direction and the $z$-direction;  just replace all instances of "$x$" with $y$ or $z$ in the above equation.
