Why the formula of kinetic energy assumes that the object has started from an initial velocity of zero? According to my physics textbook, the formula of kinetic energy is:
$$
W = \frac{1}{2}mv^2
$$
Where $m$ is is mass of the object and $v$ is the velocity of the object. The equation is calculated from this (according to that book as well): ($a$ is acceleration, $s$ is displacement, $t$ is time and $F$ is force whose value is $ma$)
$$
\begin{align}
W &= Fs \\
&= mas \\
&= ma \cdot \frac{1}{2}at^2
\end{align}
$$
Here comes the problem. According to that book: ($u$ is initial velocity)
$$
s = ut+\frac{1}{2}at^2
$$
But in the equation of kinetic energy, $s$ is replaced by $\frac{1}{2}at^2$, which is only possible when the initial velocity ($u$) is zero ($s = ut+\frac{1}{2}at^2 = 0t+\frac{1}{2}at^2 = \frac{1}{2}at^2$). Why the initial velocity is assumed to be zero here? What if the object has a non-zero initial velocity?
 A: The $s$ in equation $W=mas$ can be replaced from 3rd equation of motion.
$$W=\frac{ma(v^2-u^2)}{2a}$$
$$W=\frac{1}{2}m(v^2-u^2) \rightarrow K.E.=\frac{1}{2}m(v^2-u^2)$$
Or else, if you still want to substitute $s$ from 2nd equation of motion:
$$W=ma(ut+\frac{1}{2}at^2)$$
$$W=ma(ut)+\frac{1}{2}m(a^2t^2)$$
From 1st equation of motion; $t=\frac{(v-u)}{a},\,\, at=v-u$
$$W=mau\cdot \frac{(v-u)}{a}+\frac{1}{2}m(v^2+u^2)-mvu$$
Rearranging and simplifying the terms we get:
$$W=K.E.=\frac{1}{2}m(v^2-u^2)$$
A: Formula $W=Fs$ is the formula for work, that is the energy transferred to the body by the force. It is the difference of kinetic energies between the final and initial state. If you're starting from $v=0$ and assume $E_{kin}(0)=0$, you have
$$
E_{kin}(v)
= E_{kin}(0) + ma \cdot \frac12at^2
= 0 + \frac12 mv^2
$$
If you start from the velocity $u$ and change it to $v=u+\Delta u=u+at$, you have
\begin{align}
E_{kin}(v) &= E_{kin}(u+\Delta u) = \\
&= E_{kin}(u) + ma \cdot (ut+ \frac12 at^2) \\
&= \frac12mu^2 + mu \cdot at + \frac12 m (at)^2 \\
&= \frac12mu^2 + mu\Delta u + \frac12 m (\Delta u)^2 \\
&= \frac12 m(u+\Delta u)^2 = \\
&= \frac12 mv^2
\end{align}
so the formula still holds.
