(a) 492.2 ns
The lifetime of the particle in the frame of reference of the laboratory is given by
[tex]T= \frac{T_0}{\sqrt{1-(\frac{v}{c})^2}}[/tex]
where
[tex]T_0[/tex] is the proper lifetime
v is the speed of the particle
c is the speed of the light
For this particle we have:
[tex]T_0 = 161 ns[/tex]
[tex]v=0.945c[/tex]
Substituting into the equation, we get:
[tex]T= \frac{161}{\sqrt{1-(0.945)^2}}=492.2 ns[/tex]
(b) 139.5 m
In the reference frame of the laboratory, the distance travelled by the particle is given by
[tex]L = v T[/tex]
where
v is the speed of the particle
T is the lifetime of the particle in the laboratory frame of reference (found in part a)
The speed of the particle is
[tex]v=0.945c=0.945(3\cdot 10^8)=2.84\cdot 10^8 m/s[/tex]
The lifetime of the particle in the laboratory's frame of reference is
[tex]T=492.2 ns = 492.2\cdot 10^{-9}s[/tex]
And substituting into the equation, we find:
[tex]L=(2.84\cdot 10^8)(492.2\cdot 10^{-9})=139.5 m[/tex]
(c) 426.5 m
Here we have to calculate the distance travelled by the particle in its frame of reference. This can be calculated by using the equation
[tex]L=L_0 \sqrt{1-(\frac{v}{c})^2}[/tex]
where
[tex]L_0[/tex] is the distance travelled measured by an observer moving with the particle
L is the distance travelled measured by an observer in the laboratory
We already know that
L = 139.5 m
So, solving the formula for [tex]L_0[/tex],
[tex]L_0 = \frac{L}{\sqrt{1-(\frac{v}{c})^2}}=\frac{139.5}{\sqrt{1-(0.945)^2}}=426.5 m[/tex]
