# If we divide the second equation of motion by time $t$, why don't we get the first equation of motion where has $1/2$ come from? [duplicate]

The first equation of motion is $$v = u + at$$.

The second equation of motion is $$s = ut + \frac{at^2}{2}$$.

If we divide the second equation of motion by time $$t$$, why don't we get the first equation of motion where has $$1/2$$ come from?

The "$$v$$" in the first equation is the instantaneous velocity at the end of the time $$t$$, starting at an instantaneous velocity $$u$$ and accelerating for that time at a rate $$a$$.
On the other hand, the "$$s$$" in the second equation is the distance travelled under the above conditions. So, if you divide the second equation by $$t$$, the left-hand side will be the average velocity over the motion.
Since the object starts at a velocity $$u$$ and adds $$at$$ to its velocity by the end, it's logical that the average velocity is $$u+\frac12 at$$.
Because distance, $$s$$, is the integral of velocity with respect to time.
So, given $$v = u + at$$, you can integrate this to get \begin{align} s &= \int u + at\,dt\\ &= ut + \frac{at^2}{2} + s_0 \end{align} Where $$s_0$$ is a constant of integration. If you further say that $$s = 0$$ when $$t = 0$$ you end up with $$s = ut + at^2/2$$.