Special Relativity: Object Motion and Time Dilation

AI Thread Summary
An object traveling at 0.94c for 32 microseconds in a laboratory frame moves approximately 90,177.6 meters. In its rest frame, the time experienced during travel is calculated to be about 10.9 microseconds. The laboratory's length in the rest frame of the object is determined to be 30,716.7 meters. The discussion emphasizes the application of relativistic equations and the concept of simultaneity in different frames. Understanding these principles is crucial for solving problems related to special relativity.
Lucille
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Homework Statement


An object moves at v=0.94c in the frame of a laboratory for 32 us, decaying as it reached the other end.
a) how far did it move in the frame of the laboratory?
b) how long did the object travel in its rest frame?
c) in the rest frame, how long is the laboratory?

Homework Equations


t' = gamma * (-v/c^2*x + t)
x' = gamma* (x-vt)

The Attempt at a Solution


for a, I did:

x=v*t = (0.94c)(32us) =90,177.6 m

b)

t' = gamma * (-v/c^2 * x + t) = 10.9us

c)

x' = gamma * (x - vt) = 0 m

thanks!
 
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Welcome to PF;
Good up to (c) well done.
Recall - inside each inertial frame the regular laws of physics work like normal.
In the rest frame of the particle the laboratory of length d' zips past at speed v, and it does that in time t' ... so how long is d'?
 
would it be x' = vt' = 30,716.7 m? if so, i was wondering why x'=gamma*(x-vt) did not apply/could not work. thanks!
 
The beginning and end of the lab happen in the same place in the rest-frame of the particle.
 
Oh, that makes sense! Thank you so much!
 
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