# A relativistic train with a rest length of 500.0 m takes 780 ns to pass a stationary observer standing on the train platform, as measured by the stationary observer. (a) What is the speed of the train?

A relativistic train with a rest length of $500.0$ m takes $780$ ns to pass a stationary observer standing on the train platform, as measured by the stationary observer.
(a) What is the speed of the train? (Hint: Remember to account for the Lorentz contraction of the spaceship.)
If $S$ is the inertial frame with respect to the observer and ${S}^{\prime }$ is the inertial frame with respect to the train, then the front of the train and an observer line up at $t={t}^{\prime }=0$ and $x={x}^{\prime }=0$,
Then, in the ${S}^{\prime }$ frame, the front of the train sees itself at ${x}^{\prime }=500m$ when $t=780×{10}^{-9}seconds$
$2.72×{10}^{8}\frac{m}{s}=0.907c$
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Emmanuel Russo
You should know that Lorentz-FitzGerald contraction is (in units of $c=1$):
$L={L}_{0}\sqrt{1-{v}^{2}}\text{,}$
where you are given that ${L}_{0}=500\phantom{\rule{thinmathspace}{0ex}}\text{m}=1.6678\phantom{\rule{thinmathspace}{0ex}}\mu \text{s}$ and that in the inertial frame the contracted length $L$ passes by the observer in $\mathrm{\Delta }t=780\phantom{\rule{thinmathspace}{0ex}}\text{ns}$. Therefore $v=L/\mathrm{\Delta }t$ and:
${L}^{2}={L}_{0}^{2}\left(1-\frac{{L}^{2}}{\mathrm{\Delta }{t}^{2}}\right)\text{.}$
If you solve this for $L$ algebraically and plug in the values that you are given, you should then get $v=0.906$ from the above relationship to $v$. Close enough up to roundoff error.