Question

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

28.82 minutes

Step-by-step explanation:

This problem is an example of time dilatation by acceleration. The formula is:

`\Delta t=\gamma \Delta t_0=\frac{\Delta t_0}{\sqrt{1-(v^2)/(c^2) }}`

`\Delta t` is the time measured by the observer on a fixed position (in this case in the fixed planet),
`\Delta t_0` is the time measured by the person moving away from the fixed observer (the one on an interplanetary express bus),
`c` is the speed of light and
`v` the velocity of the observer that is moving away (the velocity of the ship).

`\Delta t_0=9 min=540s\\\Delta t=\frac{540}{\sqrt{1-((0.95c)^2)/(c^2)}}=\frac{540}{\sqrt{1-(0.95^2c^2)/(c^2)}}=(540)/(√(1-0.95^2))\\\Delta t=(540)/(√(1-0.9025))=(540)/(√(0.0975))=(540)/(0.3122)=1729.38s\\\Delta t=28.82minutes`