An astronaut leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth.The astronaut's destination is a star system 14.4 light-years away.(One light-year is the distance light travels in one year.)According to the astronaut,how long does the trip take?
A) 14.4 y
B) 22.7 y
C) 9.34 y
D) 15.0 y
E) 4.20 y

Question

Quizplus · Accepted Answer

To find the time experienced by the astronaut, we use the time dilation formula in special relativity: $ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} $, where $ t' $ is the time experienced by the astronaut, $ t $ is the time in the Earth frame, $ v $ is the velocity of the spaceship, and $ c $ is the speed of light. First, calculate the time it would take at this speed in the Earth frame: $ t = \frac{distance}{speed} = \frac{14.4 \, \text{light-years}}{0.960c} = 15 \, \text{years} $. Then, apply the time dilation formula: $ t' = \frac{15}{\sqrt{1 - 0.960^2}} \approx 4.20 \, \text{years} $.

An Astronaut Leaves Earth in a Spaceship at a Speed