Observer $\mathit{S}$reports that an even occurred on the$\mathit{x}$axis of his reference frame at $\mathit{x}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{m}$at time $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{s}$. Observer${\mathit{S}}^{\mathbf{\text{'}}}$and her frame are moving in the positive direction of the $\mathit{x}$axis at a speed of$\mathbf{0}\mathbf{.}\mathbf{400}\mathit{c}$. Further$\mathit{x}\mathbf{=}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$at$\mathit{t}\mathbf{=}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$. What are the (a) spatial and (b) temporal coordinate of the event according to ${\mathit{S}}^{\mathbf{\text{'}}}$? If ${\mathit{S}}^{\mathbf{\text{'}}}$were, instead, moving in the negative direction of the$\mathit{x}$axis, what would be that (c) spatial and (d) temporal coordinate of the event according to${\mathit{S}}^{\mathbf{\text{'}}}$?

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:

$$\begin{gathered} x-vt=x^{\text{'}}\sqrt{1-{\left(\frac{v}{c}\right)}^2} \\ 3.00\times {10}^8-0.400.3\times {10}^8.2.5=x^{\text{'}}\sqrt{1-{\left(\frac{0.400c}{c}\right)}^2} \\ {x}^{\text{'}}=0 \end{gathered}$$

Thus, the spatial coordinate of the event according to $S^{\text{'}}$is$0m$ .

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 of inertial observer.

Substitute the given values in the above equation:

$$\begin{gathered} t^{\text{'}}=\frac{t-{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}} \\ {t}^{\text{'}}=\frac{2.5-{\displaystyle \frac{3.00\times {10}^8}{0.400.3\times {10}^8}}}{\sqrt{1-{\left({\displaystyle \frac{0.400}{c}}\right)}^2}} \\ {t}^{\text{'}}=2.29s \end{gathered}$$

Thus, the temporal coordinate of the event according to $S^{\text{'}}$is$2.29s$ .

Here, the observer is traveling in the opposite direction. So, the length contraction equation becomes $x+vt=x^{\text{'}}\sqrt{1-{\left(\frac{v}{c}\right)}^2}$.

Substitute the given values in the above equation:

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

$$\begin{gathered} 3.00\times {10}^8+0.400.3\times {10}^8.2.5=x^{\text{'}}\sqrt{1-{\left(\frac{0.400c}{c}\right)}^2} \\ {x}^{\text{'}}=6.54\times {10}^{8}m \end{gathered}$$

Thus, the spatial coordinate according to moving in the negative direction of $x$axis is$6.54\times {10}^{8}m$ .

Here, the observer is traveling in the opposite direction. So, the time dilation equation becomes $t^{\text{'}}=\frac{t+{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$$t^{\text{'}}=\frac{t+{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$.

Substitute the given values in the above equation:

$$\begin{gathered} t^{\text{'}}=\frac{t+{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}} \\ {t}^{\text{'}}=\frac{2.5+{\displaystyle \frac{3.00\times {10}^8}{0.400.3\times {10}^8}}}{\sqrt{1-{\left({\displaystyle \frac{0.400c}{c}}\right)}^2}} \\ {t}^{\text{'}}=3.16s \end{gathered}$$

Thus, the temporal coordinate according to moving in the negative direction of axis is 3.16s.