Question

Suppose a particle called a kaon is created by cosmic radiation striking the atmosphere. It moves by you at $0.980c$, and it lives $1.24\times 10^{-8}\textrm{ s}$ when at rest relative to an observer. How long does it live as you observe it?

Final Answer

$6.23\times 10^{-8}\textrm{ s}$

### Solution video

# OpenStax College Physics, Chapter 28, Problem 4 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A kaon particle is moving at 0.980 times the speed of light

*c*; when an observer is at rest with respect to the particle, they measure a lifetime for it of 1.24 times 10 to the minus 8 seconds and because this time is measured at rest with respect to the particle, it's called the proper time,*t naught*. So*Δt*—the time measured by the person watching the things zip by— equals the Lorentz factor*γ*times the proper time. So that's 1 over the square root of 1 minus*v squared*over*c squared*times*Δt naught*and that's 1 over the square root of 1 minus 0.980 times*c*all squared over*c squared*times 1.24 times 10 to the minus 8 seconds and that is 6.23 times 10 to the minus 8 seconds. This is the time measured by an Earth-based observer watching the kaon particle zip past.