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 $t=780\times {10}^{-9}seconds$

The correct answer is:

$2.72\times {10}^{8}\frac{m}{s}=0.907c$

(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 $t=780\times {10}^{-9}seconds$

The correct answer is:

$2.72\times {10}^{8}\frac{m}{s}=0.907c$