This, I know might perfectly be a very stupid question. But trying to make sense of it in terms of a very simple explanation. For distance $d=vt$. Velocity might be another function of time $v=at$. The formula states that (if $v_0=d_0=0$) $d=\frac{1}{2}at^2$. I have seen the mean speed theorem, but do not get why simple substitution would give an incorrect answer ($d=at^2$).
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$v=at$ is valid for constant acceleration.
$d=vt$ is valid for constant velocity. But if there is acceleration, then the velocity is not constant.
For constant acceleration, starting from a standstill, velocity at time $t$ is $v=at$. As the velocity varies linearly with time, it follows that the average velocity forms a standstill until time $t$ is $1/2 \, at$ (plot $v$ against $t$ on an $x-y$ graph, and it will be obvious). The distance travelled over the time period from standstill to time $t$ is the average velocity multiplied by the time, which gives $d=1/2 \, at^2$.
