Quote by faen
Yes I thought about this too. But the result seems strange to me, because it now predicts that time goes slower in the rest frame and not in the moving frame, just because we changed the location of the experiment.
For example if we had done the experiment in frame A instead of B, t and t' would change place in the equation predicting opposite results, as such:
$$t'=\frac{t}{\sqrt{1\frac{v^2}{c^2}}}.$$.
How is this possible?

According to Special Relativity, time does not go slower in any rest frame you choose, it only goes slower for objects/observers/clocks that are moving in any rest frame you choose. It's no different than saying that the rocket is moving in the ground's rest frame and the ground is moving in the rocket's rest frame. How is that possible?
But it does help to consider that in the ground's rest frame, the clock on the rocket is ticking slower and it is moving away from the clock on the ground. So between each tick of the rocket clock, as it observes the ground clock, it has moved farther away and so the image of the ground clock's ticks have farther to travel resulting in a longer time compared to its own ticks. It turns out that the ratio of the ground clock's tick interval compared to the rocket clock's tick interval as determined by the rocket is the same as the ratio of the rocket clock's tick interval compared to the ground clock's tick as determined by the ground. Work it out, you'll see that this is true.