How Fast Must The Particle With Lifespan 30.0 ns At Rest, Must Move To Travel 10.0 Meters?

Question

The particle has an average lifetime of 30.0 ns when at rest. In order to travel 10.0 meters, how fast must it move?

It's a bit confusing to me since there might be Length Contraction, as well as Time Dialation.

Δt = Δt. / (1-(v^2/c^2)^(.5))

L = L. (1-(v^2/c^2)^(.5))

Answer 1

I think you can assume that a "particle" has no dimension, so you don't have to be concerned with length contraction.  Time dilation, however, is another matter.

Let:

d = distance to cross (10.0 m)

t = time (in laboratory frame) to cross the distance

t′ = particle's proper time during crossing (30 ns)

v = particle's speed

Then we have:

t = d/v

also:

t = t′ / sqrt(1-v²/c²)

So:

d/v = t′ / sqrt(1-v²/c²)

Solve the above for v.  If I have made no mistakes, this is:

v = dc / sqrt(c²t′² + d²)

Answer 2

Skip the length contraction, it doesn't apply.

Without time contraction, the speed turns out faster than light, so we have to apply it.

t' = t/√(1-v²/c²)

v = d/t = 10/t/√(1-v²/c²) = (10/t)√(1-v²/c²)

you have to square both sides and solve for v.

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