Observer a, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c and determines its lifetime to be τa=129 ns. observer a places markers in the laboratory at the locations where the particle is produced and where it decays. how far apart are those markers in the laboratory?

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: