Question

A subatomic particle has a 410 ns lifetime in its own rest frame. If it moves through the lab at 0.980 c, how far does it travel before decaying, as measured in the lab?

Answer 1

As we know that

Time dilation is dT = dt/sqrt(1 - (V/C)^2) where in your case V = .98 C and C is light speed. If you want more precision, look up the entire C on the web and plug that in instead of the 299E6 mps I used. dT is the decay period as measured by the observer and V is the speed relative to the same observer.

In its own frame dt = 410E-9 seconds is its period of decay.

So as seen by the guy in the white coat who's looking through the detection device on the LHC, the decay period is dT = 410E-9/sqrt(1 - .98^2) = 2.060327521E-06 seconds.

In which time the little guy goes S = dT .98 C = 2.060327521E-06 * .98 * 299E6 = 605.7362913 m. ANS.