solution:-
given data
speed of muon with respect to lab(v)="\\frac{c}{2}"
speed of person with respect to lab(v')="\\frac{c}{3}"
relative velocity can be written as
"v_{rel}=\\frac{v-v'}{1-\\frac{vv'}{c^2}}"
"v_{rel}=\\frac{\\frac{c}{2}-\\frac{c}{3}}{1-\\frac{c^2}{6c^2}}=\\frac{c}{5}"
proper life time for muon can be written
"\\Delta t'=\\Delta t \\sqrt{1-\\frac{v_{rel}^2}{c^2}}"
by putting the value
"\\Delta t=1\\sqrt{1-\\frac{c^2}{25c^2}}=0.9797\\mu s"
distance moved by muon with respect to lab is
"d=v_{rel} \\Delta t'=\\frac{c}{5}\\times(0.9797)\\times10^{-6}"
"=\\frac{3\\times10^8}{5}\\times(0.9797)\\times10^{-6}"
"d=58.78m"
therefore distance moved by muon is 58.78m.
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