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GR9677 #50
Problem
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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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 TypT2018-10-26 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 mpdude82012-04-20 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 d2011-09-16 14:40:56 <3 spacetime intervalReply to this comment

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$