accelerating and non accelerating duality of matter Kinetic energy of any object , assumes accelerating nature of the object as it is derived from equations of motion for uniformly accelerated objects and the work-energy theorem which deals with accelerating objects $but$ momentum assumes non accelerating nature of the same object . Also we use momentum and kinetic energy equations simultaneously for calculation of velocity of object after or before collsion . So , does that mean that matter when in motion accelerates and not accelerates the same time ? That doesnt make sense .Please correct me.
Also kinetic energy cannot be a virtual concept , if thats the answer , because energy gets exchanged in collisions.
 A: If the velocity of a mass $m$ at some moment of time is $v$, then the kinetic energy and momentum are:
$$\begin{align}
E &= \tfrac{1}{2}mv^2 \\
p &= mv
\end{align}$$
If the velocity is changing with time, i.e. it is a function of time $v(t)$, then the kinetic energy and momentum will also be functions of time:
$$\begin{align}
E(t) &= \tfrac{1}{2}mv^2(t) \\
p(t) &= mv(t)
\end{align}$$
These equations are correct no matter how the velocity is changing with time.
The equation for the kinetic energy can be derived in lots of ways. For students it is common to derive it from the equation for the work done (i.e. force times distance) assuming a constant force and therefore constant acceleration. However this does not mean kinetic energy is only defined for constantly accelerating objects. Likewise, the momentum is defined for any object regardless of how its velocity is changing. Both kinetic energy and momentum are simply given by the equations above no matter how velocity is changing with time.
You'll need to clarify your question about kinetic energy being a virtual concept as it isn't clear what you mean by this.
A: To compliment John's answer I'll give you an example: the kinetic energy of a harmonic oscillator. First we need to determine the velocity 
$ v=\frac{dx}{dt}=\frac{d(Asen(\omega t +\phi_0))}{dt}=A\omega cos(\omega t+\phi_0)$. 
Because kinetic energy is $\frac{1mv^2}{2}$ we substitute $v$
$KE=\frac{1}{2}mA^2\omega^2cos^2(\omega t+\phi_0)$ 
Because $m\omega^2$ is just the spring constant $k$, the usual formula for the KE of a harmonic oscillator (in the case of a spring) is given by
$KE=\frac{1}{2}KA^2cos^2(\omega t+\phi_o)$ 
As you see, kinetic energy can have $v$ as a function of time.
