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

Answer:

28.82 minutes

Explanation:

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

$\Delta t=\gamma \Delta t_0=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2} }}$

$\Delta t$ 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).

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