Answer to Question #294468 in Classical Mechanics for Jly Miyavi

Question #294468

Assuming that you are moving with a constant speed of 0.65c when you passed by a man on


the sidewalk who started looking at his watch. From your observation, it took him 5 seconds to


look at his watch then look up. How much time elapsed in the man’s frame?

1
Expert's answer
2022-02-07T08:44:49-0500

Solution

The time elapsed in the man’s frame

Is given by

t=t1v2c2t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}

Putting all values

t=51(0.65c)2c2t=50.58t=50.76=6.85sect'=\frac{5}{\sqrt{1-\frac{(0.65c)^2}{c^2}}}\\t'=\frac{5}{\sqrt{0.58}}\\t'=\frac{5}{0.76}\\=6.85sec




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