Length contraction of an accelerating rod Let us assume a rod of length, $4l$, initially at rest. The coordinates are defined s.t - (assuming uniform density)
$$X_{back-end}(0) = -2l$$
$$X_{front-end}(0) = +2l$$
$$X_{mid-point}(0) = 0$$
Let it accelerate in a short period and then move with velocity s.t it contracts to length $2l$ and at time $t$ mid point of the body is at 
$$X_{mid-point}(t) = Y$$ 
Now where do I expect the back end and front end of the rod to be? 
It seems to me that, because of the symmetry in the problem, -
$$X_{back-end}(t) = Y-l$$
$$X_{front-end}(0) = Y+l$$
However, this would mean -
$$(V_{front-end})_{avg} = \frac{Y-l}{t}$$
$$(V_{back-end})_{avg} = \frac{Y+l}{t}$$
$$(V_{mid-point})_{avg} = \frac{Y}{t}$$
This is completely un-intiuitive. Why do I see a non-uniform acceleration for different points in the body when I had uniform acceleration? What is source of this contradiction?
Is something wrong with my vizualisation of length contraction?
EDIT :
After some digging I found a few related questions. 


*

*Length contraction, front middle or back 

*An accelerating and shrinking train in special relativity

*https://en.wikipedia.org/wiki/Rindler_coordinates#A_.22paradoxical.22_property
But none of these explain what is the source of this? i.e If there is a non-uniform acceleration then there is non uniform Force which doesn't make sense because my whole problem is with constant uniform force. Why?
 A: You built in the asymmetry yourself when you declared that you're going to  accelerate the rod in such a way that it shrinks.  If it's going to shrink (in your frame) then (in  your  frame) the back has to start moving forward before the front does.  There's your asymmetry.  (Or you could have a more complicated but equally antisymmetric motion, involving something like the front moving backward for a while while the back moves forward.)
You then tried to impose symmetry after the fact by implicitly assuming in your 
 calculation that the front and back ends both started moving at the same time.  But this contradicts what you've already done.
If you assume that the front and back ends of the rod both start moving simultaneously (and mimic each others' acceelerations thereafter) in your own frame, then the length of the rod can't change in your own frame.  But in the moving frame, the front starts moving before the back does, and the rod stretches so that (as you expect) it's always longer in its own frame than it is in yours.
