Question

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star system 14.4 light-years away (one light-year is the distance lighttravels in one year). Relative to a frame of reference that is fixed with respect to Earth, how longdoes it take Rob to complete the trip?

Answer 1

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

$1Ly =9.4605284*10^{15}m \rightarrow 'Ly'$means  Light Year

Then

$14.4Ly = 1.36231609*10^{17} m$

If we have that

$v= \frac{x}{t} \rightarrow t = \frac{x}{t}$

Where,

v = Velocity

x = Displacement

t = Time

We have that

$t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c$= Speed of light

$t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}$

$t= 454105363 s (\frac{1hour}{3600s})$

$t= 126140 hours(\frac{1day}{24hours})$

$t= 5255.85 days(\frac{1 year}{365days})$

$t = 14.399 years$

Therefore will take 14.399 years