A passenger on an interplanetary express bus traveling at = 0.95 takes a 9.0-minute catnap, according to her watch.How long does her catnap from the vantage point of a fixed planet last?

Answer 1

You need to tell us what the units of velocity are. I assume that the "bus" is travelling at 0.95c, or 95% of the speed of light, with respect to a "fixed" planet - presumably the home planet where the passenger came from.

Someone is apparently trying to teach you the theory of relativity. That person should know that it is measingless to say something is "fixed" unless a frame of reference is specified.

You need to use the "time dilation" formula of Special Relativity. It is probably explained in your textbook or class notes. A tutorial on the subject can easily be found with Google.

The nine minute time interval measured on the watch of the traveler will appear longer on the planet, by a factor
1/sqrt[1 - (0.95)^2] = 3.20
That means the nap will be 28.8 minutes long as measured on the home planet.