It is really confusing me. Because the Vehicle is moving at the uniform velocity of 22.22 m/s. So, the object should also moved at the same distance as Vehicle moved. But my calculation gives only 11.11m instead of 22.22 m/s. Where I am wrong? Can you please point out?
thanks,
pmk

An object travelling at 22.22 m/s for one second is going to cover more distance than an object travelling at variable speed between 0 m/s and 22.22 m/s (at constant acceleration). This is normal.
Perhaps what you want to model is something different... something a bit more akin to somebody "picking up" an object and putting it into their car... or a crash. Let's try going through your points and figure things out in a different way.
Originally Posted by SpeedEC 1. A Vehicle running at uniform velocity of 80 KMPH (50 MPH) i.e 22.22 m/s. 
So we have
$\displaystyle u_{car} = 22.22$ m/s
$\displaystyle u_{obj} = 0$ m/s
$\displaystyle m_{car} = ? $ kg
$\displaystyle m_{obj} = 70$ kg
So we have some information relevant for a collision problem.
2. No acceleration since uniform velocity.

This is okay. In collision problems, we can assume that the collision is instantaneous and we just consider conservation of momentum before and after the collision.
3. It picks the object while running.

So we can use the equation of the conservation of momentum:
$\displaystyle m_{car} u_{car} + m_{obj} u_{obj} = (m_{car} + m_{obj}) v$
Consequently, we have
$\displaystyle 22.22 m_{car} = (m_{car} + 70) v$
Solving for v gives:
$\displaystyle v = \frac{22.22 m_{car}}{70 + m_{car}}$
No meaningful solution exists for $\displaystyle v >= 22.22$. Therefore, if the mass of the car is something like 45000 kg, we have
$\displaystyle v = \frac{22.22 \times 45000}{70 + 45000} = 22.185$ m/s
So, in this scenario, the interaction is modelled as if the car picks up the mass and then has to slow down because of the conservation of momentum. I guess this isn't really the kind of information you want though.
If we consider a momentum exchange over a time period (the collision time), then we can use the momentum exchange to find out the force exerted by the object on the car. Let's say we consider a 45000 kg car. The momentum exchange is:
$\displaystyle \Delta p = m_{car} (u_{car}  v)$
$\displaystyle = 45000 \times (22.22  22.185) = 1553$ kg m/s
And using Newton's second law (note: F=ma is the law that can be derived from the rate of change of momentum, which is the actual definition of Newton's second law) with a collision time is one second, we get
$\displaystyle F = \frac{\Delta p}{\Delta t} = \frac{1553}{1} = 1553$ N
That force is rather low, but considering that the collision time is rather long, it makes sense. Note that a collision time of 1 second is much longer than typical collision times of 0.05 to 0.001 seconds. Perhaps the car has a crumple zone
Finally, the acceleration is:
$\displaystyle a = \frac{F}{m_{obj}} = \frac{1553}{70} = 22.186 $ m/s$\displaystyle ^2$
If you absolutely require the acceleration to be 22.22 m/s$\displaystyle ^2$, you will either have to change the collision time or the initial velocity of the car.