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?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Possible duplicates: Why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$? and links therein. $\endgroup$– Qmechanic ♦Commented Aug 2, 2020 at 10:59
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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 $u+\frac12 at$.
Bottom line, any equation that is quoted without a definition of every term (the "wheres) is worse than useless.
$\begingroup$
$\endgroup$
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 $s = ut + at^2/2$.