A passenger on an interplanetary express bus traveling at v=0.99c takes a 5 min
ID: 1311750 • Letter: A
Question
A passenger on an interplanetary express bus traveling at v=0.99c takes a 5 min catnap, according to her watch. How long did her catnap last from the vantage point of an observer on fixed planets ? B) according to the passengers and driver, the bus is 21.34 meter long. What is the length of the bus as seen from the vantage point of the observer on a fixed planet? C) the bus then slows down as it passes earth to 80% of the speed of light. Some passengers get off the bus in an escape pod that moves away from the bus 50% of the speed of light relative to the bus. How fast ( compared to the speed of light) does the escape pod travel relative to earth?
Explanation / Answer
?t1=?t/?[1-(v/c)2]
?t=5min
v=0.99c
then?t1=35.4s
t' = t/gamma
gamma = sq rt(1 - v^2/c^2)
We are dealing with % of speed of light, so we can work in those terms:
V = 0.99 speed of light = 0.99
c= 1,00 speed of light = 1.00
g = sq rt(1- 0.99^2/1)
g = sq rt (1-.99^2)
g = sq rt ( 1 - 0.9801)
g = 0.141
t' = 5 min / 0.141 = 35.46 min
Where:
t' = earth time = 35.46 minutes
t = ship time = 5 minutes
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.