There are two groups of mutually synchronized clocks K and `K^'` moving relative to each other with a velocity v as shown in Figure. The moment when the clock `A^'` gets opposite the clock A is taken for the beginning of the time count. Draw the approximate position of hands of all the clocks at this moment "in terms of the K clocks", "in terms of the `K^'` clocks".

Explanation

By Lorentz transformation `t^'=(1)/(sqrt(1-v^2/c^2))(t-(vx)/(c^2))`
So at time `t=0`, `t^'=(vx)/(c^2)(1)/(sqrt(1-v^2//c^2))`
If `xgt0t^'lt0`, if `xlt0`, `t^'gt0` and we get the diagram given below "in terms of the K-clock".

The situation in terms of the `K^'` clock is reversed.