GREPhysics.NET: GR9677 #50

GR9677 #50

Problem

GREPhysics.NET Official Solution

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Special Relativity $\Rightarrow$ }Spacetime Interval

The spacetime interval is defined by the metric that negates spatial and time variables as $dS^2 = (cdt)^2 - (dx)^2$ . $dS$ is invariant. One has thus $dS^2 = dS'^2 \Rightarrow (3c)^2=(5c)^2-(ct)^2\Rightarrow ct = 4 c$ minutes, as in choice (C).

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Comments

TypT 2018-10-26 22:10:37	Quick, dumb way I did it without having to figure out what exactly the question was asking:\r\n\r\nSpace time intervals have a Pythagorean relationship, the two numbers we are given are 3 and 5, to complete the Pythagorean triple we need a 4 so we get (C). It even has the trap answer of $4^2 = 16$ (E), to give a little more confidence in (C). Reply to this comment
mpdude8 2012-04-20 15:12:22	Man, I was kind of surprised that only 1/3 of test-takers got this one correct. Reply to this comment
pam d 2011-09-16 14:40:56	<3 spacetime interval Reply to this comment

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