I could use some help in physics!?

Question

A particle lives for a short time before breaking apart into other particles. Suppose it is moving at a speed of 0.984c, and an observer who is stationary in a laboratory measures the particle's lifetime to be 2.50 10-8 s.

(a) What is the lifetime according to a hypothetical person who is riding along with the particle?

(b) According to this hypothetical person, how far does the laboratory move before the particle breaks apart?

Answer 1

It pertains to time dilation (increase in time of two events having been observed by observers having relative speed with relativstic domain.

T = To /{sqrt [1 - v^2/c^2]}

To = is the proper time interval measured by stationary observer having a stationary clock with him.

To = 2.5 * 10^-8 s
V = speed of relative hypothetical observer who is riding the particle moving with 0.984 c

[1 - v^2/c^2] = [1 - (0.984)^2]

[1 - v^2/c^2] = [1 - 0.968256] = 0.031744

T = To /{sqrt [1 - v^2/c^2]}

T = 2.5 * 10^-8 s / {sqrt [0.031744]}

T = 14.031 * 10^-8 sec [this is lifetime for hypothe. obser

b) for this hypotheti person the lab is moving with - V speed
opposite to hypothetical person.

so duting the T (delta t for breaking apart) the lab would have moved (as hypothei will feel)

distance = V * T = 0.984 c *14.031 * 10^-8
= 0.984 * 3*10^8 *14.031 * 10^-8
= 41.4195 meters lab moved