Why Is My Distance Travelled Overshooting? I have the following pascal code to determine position and velocity from acceleration:
For x:= 1 to 10
Do Begin

    Velocity:=Velocity + Acceleration;
    Position := Position + Velocity;
    Writeln(Position);
    End;

Why am I overshooting?
 A: You are overshooting because your code assumes that the updated velocity was valid  for the entire duration of this tick of the clock.
You get better accuracy if you use (inside the loop)
Begin
  NewVelocity:=Velocity+Acceleration;
  Position:=Position+(Velocity+NewVelocity)/2;
  writeln(Position);
  Velocity:=NewVelocity;
End;

This will be accurate, iff the acceleration is uniform, because it uses the average of the old and updated velocities. If the acceleration is not uniform (say, celestial bodies moving under gravity) then this is still just an approximation.
A: Congratulations, you've just entered the wide world of numerical integration.  You're trying to take a system of ordinary differential equations like this:
$$x'(t) = v(t), \quad v'(t) = a(t)$$
And use numerical data at a fixed interval of $h = 1$ to estimate the true values of $x$ and $v$ at $t=10$.
You can estimate the error inherent in such a scheme from a Taylor series expansion.  Let's look at just $v$:
$$v(t+h) \approx v(t) + a(t) h + \frac{1}{2} v''(t) h^2 + \ldots $$
In particular, notice that because you chose $h=1$, the $h^2$ term could be as large as the order $h$ term if $v'' \geq 2a$.  That naturally jeopardizes the validity of your scheme, as it represents something potentially large that you didn't account for.  One way to attack this problem is to simply use a smaller step $h$.  As you choose a smaller $h$, the quadratic term will be reduced.

Of course, you may not always be able to use better resolution, and even if you can, it is only part of an overall strategy to a better numerical integration.
In the above, I used a simple Taylor series expansion corresponding to the kind of scheme used in the question.  It basically comes down to the idea that we can estimate $v(t+h)$ from $a(t)$ to a certain error.  But what if we used more values of $a$ in the interval $[t, t+h]$ and could reduce the error?  Basically, we use more Taylor expansions to try to eliminate error terms.  For instance, consider
$$\begin{align*}
v(t+h) &= v(t+h/2) + a(t+h/2) \frac{h}{2} + \frac{1}{2} v''(t+h/2) \frac{h^2}{4} + O(h^3)\\
v(t) &= v(t+h/2) - a(t+h/2) \frac{h}{2} + \frac{1}{2} v''(t+h/2) \frac{h^2}{4} + O(h^3)\end{align*}$$
You can use these to derive a scheme that is third-order accurate in the stepsize $h$:
$$v(t+h) = v(t) + a (t+h/2) h + O(h^3)$$
Basically, we use the acceleration at the midpoint of the step, rather than the beginning of the step.  But, this may be impractical if all you have is acceleration data on the actual step values.

At any rate, the method you're using--called an "Euler step"--is very simple, and also not very accurate.  It is hardly surprising to see it over or undershoot significantly.  Consider a smaller step if possible to decrease error, but also consider a more sophisticated method.  I just drew one up above, but even that is very simple. Runge-Kutta methods (look them up) are quite common.
Whatever you use, understand that from a numerical standpoint, you are using an approximation to try to recover $x$ and $v$ at the future times.  Approximations have intrinsic error.  A more sophisticated scheme, or decreasing the step size, can reduce error.  The Euler scheme that you're using is not exact for a general acceleration.
