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If astronauts could travel at $v=0.950 c,$ we on Earth would say it takes $(4.20 / 0.950)=4.42$ years to reach Alpha Centauri, 4.20 light-years away. The astronauts disagree. (a) How much time passes on the astronauts' clocks? (b) What is the distance to Alpha Centauri as measured by the astronauts?

a) $\Delta t_{p} \approx 1.38 \mathrm{yr}$

b) $d=1.31$ light-years

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well, speed of astronauts. We is a cool too. Zero point 950 times the speed of light in time taken by astronauts to reach Alka Century with respect to people on Earth. Uh, t it's equal to four point for two years for two years and, ah, distance off Alfa century from Earth Distance D off Alfa Century from our physical too. 4.20 white years. Light years. Well, uh, we know they, um d is equal to proper time divided by squared Rudolph one minus we squared, divided by C Square. Let's splitting the values. Um, here we need to find out the proper time. Well, allow the f d t equals for mind for two is equal to, uh, proper time Divided by square root off one minus we squarely by Macy's. Were we in this case, is it a 0.950 times the speed of light divided by C square and sold ing for proper time? We have ah for point for Do multiply by square. Who does on cedar 0.975 The 0.9 75 and therefore appropriate time. It's a cool too one point 38 years. It is the proper time. 1.38 years. No, let's do part B. Using l is equal to a lot of times our square root one minus we squared, divided by C square. Well, uh l A is equal to we have a lot, which is 4.20 light years into one minus we. Well, we is a 0.950 times the speed of light divided by C square and therefore l is equal to 4.20 I wanted life. I square root off zero point zero 975 975 And finally, we have 1.31 light years, so l is equal to 1.31 light year.