## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $d_0 = 8.5~ly$ $t_0 = 17~years$ (b) $d = 7.36~ly$ $t = 14.72~years$ (c) Both sets of answers are correct according to each frame of reference.
(a) We can find the distance according to earthlings: $d_0 = (2)(4.25~ly) = 8.5~ly$ We can find the time of the journey according to earthlings: $t_0 = \frac{d}{v} = \frac{8.5~ly}{0.500~c} = 17~years$ (b) We can find $\gamma$: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{(0.500~c)^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{0.75}}$ $\gamma = 1.155$ We can find the distance according to the crew: $d = \frac{d_0}{\gamma}$ $d = \frac{8.5~ly}{1.155}$ $d = 7.36~ly$ We can find the time of the journey according to the crew: $t = \frac{t_0}{\gamma}$ $t = \frac{17~years}{1.155}$ $t = 14.72~years$ (c) Both sets of answers are correct according to each frame of reference.