91Ó°ÊÓ

The length contraction equation states that $x-vt=x^{\text{'}}\sqrt{1-{\left(\frac{v}{c}\right)}^2}$, where the expression$x-vt$ is equal to the length measured in the inertial reference frame.

By using the above given formula, we can calculate the spatial coordinate, $x^{\text{'}}$, of the event in the moving observer’s time frame.

Substitute the given values in the above formula:

$x-vt=x^{\text{'}}\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

$$\begin{gathered} 100×{10}^3-0.950.3×{10}^8.200×{10}^{-6}=x^{\text{'}}\sqrt{1-{\left(\frac{0.950c}{c}\right)}^2} \\ {x}^{\text{'}}=1.38×{10}^{5}m \\ {x}^{\text{'}}=138km \end{gathered}$$

Thus, the spatial coordinate in $S^{\text{'}}$ is $138km$.

Here, we use time dilation equation, $t^{\text{'}}=\frac{t-{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$, where $t-\frac{x}{v}$ shows that time measured in the reference frame.

Substitute the given values in the above equation:

$$\begin{gathered} t^{\text{'}}=\frac{t-{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}} \\ {t}^{\text{'}}=\frac{200×{10}^{-6}-{\displaystyle \frac{100×{10}^3}{0.950.3×{10}^8}}}{\sqrt{1-{\left({\displaystyle \frac{0.950c}{c}}\right)}^2}} \\ {t}^{\text{'}}=-3.74×{10}^{-4}s \\ {t}^{\text{'}}=-374\mathrm{μ}s \end{gathered}$$

Thus, the temporal coordinate in $S^{\text{'}}$ is $-374\mathrm{μ}s$.