Why do two formulas $d = v_0 t + \frac{at^2}{2}$ and $d=vt$ yield different results? All of you are sure to be familiar with these two equation:
$d = v_0 t + \frac{at^2}{2}$ and
   $d=\overline vt$
Given the same initial and final velocities, and time and acceleration. With the second equation I need not use the acceleration. But the distance found using these equations are different!
For this who say that the second equation assume a = 0, it isn't always true. 
I use the formula for graphs of uniformly accelerated motion to find the distance travelled at the end of certain time. I use initial and final velocities to find the average velocity and multiply the velocity by the time in second on the graph.
Wait, I just tested another uniformly accelerated motion problem  with both formulas. The result are the same! What? This is only for some specific problems. 
 A: If you have a uniform acceleration the average velocity $\bar{v}$ is just
$$
\bar{v}=v_0+\frac{at}{2}.
$$
because the final velocity is $v_{initial}=v_0$ and $v_{final}=v_0+at$, i.e.
$$
\bar{v}=\frac{1}{2}(v_{initial}+v_{final})=v_0+\frac{at}{2}.
$$
Then you get
$$
d=\bar{v}t=v_0 t+\frac{at^2}{2}.
$$
A: The second formula is a special case of the first one where the acceleration is zero. If you substitute $a=0$ into the first formula you get the second one, as is expected
A: Don't get too stuck on formulas.
More generally:
(1) velocity is the time rate of change of position and
(2) acceleration is the time rate of change of velocity
So given any (1 dimensional) function of x(t), v(t) or a(t) to begin with, the calculus constrains all remaining relationships.
The formulas you wrote are for special situations: the first assumes constant acceleration and an initial velocity. The second position with presumably variable velocity.
A: Actually, a more correct full formula would be:
$$
x = x_0 + v_0t + \frac{1}{2}at^2
$$
Where I have used $x$ instead of $d$.  Note that $x_0$ and $v_0$ are fixed constant values.
A: D=vt is used when acceleration is zero but D=ut+1/2 at^2 is used when there is a constant acceleration.
