22117.09.11 (0.99)γβ===−−Thus, ∆t0= (26.26 y)/(7.09) = 3.705 y. 15. (a) The speed of the traveler is v= 0.99c, which may be equivalently expressed as 0.99 ly/y. Let dbe the distance traveled. Then, the time for the trip as measured in the frame of Earth is ∆t = d/v= (26 ly)/(0.99 ly/y) = 26.26 y.
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