Question

A neutron lives 900 s when at rest relative to an observer. How fast is the neutron moving relative to an observer who measures its life span to be 2065 s?

Final Answer

$2.70\times 10^{8}\textrm{ m/s}$

### Solution video

# OpenStax College Physics, Chapter 28, Problem 6 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A neutron at rest with respect to an observer has a lifetime of 900 seconds so this time is called

*t naught*— the proper time— since it's measured at rest with respect to the neutron; in a different experiment when the neutron is zipping past the observer, its lifetime is measured to be 2065 seconds and this*Δt*—the time measured by someone watching it zip past— equals the Lorentz factor*γ*times the proper time and we will use this relation to figure out the speed with which the particle is moving in this scenario here. So*γ*is 1 over the square root of 1 minus*v squared*over*c squared*times*Δt naught*and we have to do some algebra to solve for*v*. So first we'll multiply both sides by this Lorentz factor 1 minus*v squared*over*c squared*and we'll also divide both sides by*Δt*. So we have square root 1 minus*v squared*over*c squared*equals*Δt naught*over*Δt*then we square both sides and I have 1 minus*v squared*over*c squared*equals*Δt naught*over*Δt*squared and then add*v squared*over*c squared*to both sides and then subtract*Δt naught*over*Δt*squared from both sides and then switch the sides around and we have*v squared*over*c squared*equals 1 minus*Δt naught*over*Δt*squared then multiply both sides by*c squared*and then you get this line here and take the square root of both sides to finally solve for*v*. So*v*then is*c*multiplied by the square root of 1 minus*Δt naught*over*Δt*squared. So that's 2.998 times 10 to the 8 meters per second times the square root of 1 minus 900 seconds divided by 2065 seconds squared and this is 2.70 times 10 to the 8 meters per second.