91Ó°ÊÓ

The given data can be listed below as:

The velocity of the ship is $\stackrel{→}{\mathrm{v}}=0.950\mathrm{c}\widehat{\mathrm{i}}$.

The velocity of the proton is ${\mathrm{v}}_1=0.980\mathrm{c}$.

The ship’s length is $\mathrm{l}=760\text{″¾}$.

The relativity mainly refers to the behaviour of a particular object is time and space. This theory is also beneficial for predicting the existence of a particle and light’s bending.

The equation of the temporal separation of a passenger is expressed as:

${\mathrm{t}}_1=\frac{\mathrm{l}}{{\mathrm{v}}_1}$

Here, ${\mathrm{t}}_1$ is the temporal separation of a passenger, $l$ is the ship’s length and ${\mathrm{v}}_1$ is the velocity of the proton.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_1=\frac{760\text{″¾}}{0.980(3×{10}^8\text{″¾/s})}\\ =\frac{760\text{″¾}}{(2.9×{10}^8\text{″¾/s})}\\ =2.6×{10}^{−6}\text{ s}\end{array}$

Thus, the temporal separation of a passenger in the ship is $2.6×{10}^{−6}\text{ s}$.

As in our case, the length of the ship is taken negative. The equation of our temporal separation is expressed as:

${\mathrm{t}}_2=\frac{1}{\sqrt{1−{\left(\frac{{\mathrm{v}}_1}{\mathrm{c}}\right)}^2}}\left({\mathrm{t}}_1+\frac{{\mathrm{v}}_1(−\mathrm{l})}{{\mathrm{c}}^2}\right)$

Here, $t_2$is the temporal separation of us and ${\mathrm{v}}_1$is the velocity of the proton.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_2=\frac{1}{\sqrt{1−{\left(\frac{0.980\mathrm{c}}{\mathrm{c}}\right)}^2}}\left(2.6×{10}^{−6}\text{ s}+\frac{\left(0.980\mathrm{c}\right)(−760\text{″¾})}{{\mathrm{c}}^2}\right)\\ =\frac{1}{\sqrt{1−{\left(0.980\right)}^2}}\left(2.6×{10}^{−6}\text{ s-}\frac{\left(744.8\text{″¾}\right)}{3×{10}^8\text{″¾/s}}\right)\\ =\frac{1}{\sqrt{1−0.9604}}(2.6×{10}^{−6}\text{ s-2.4}×{\text{10}}^{−6}\text{ s})\\ =\frac{1}{0.19}(0.2×{10}^{−6}\text{ s})\\ =1.05×{10}^{−6}\text{ s}\end{array}$

Thus, our temporal separation is $1.05×{10}^{−6}\text{ s}$.

According to the observers of ship, there is no difference of firing proton from front to back that significantly makes no difference. Hence, the temporal separation of the passenger is the same as previous.

Thus, the temporal separation of a passenger in the ship is $2.6×{10}^{−6}\text{ s}$.

The equation of our temporal separation is expressed as:

${\mathrm{t}}_3=\frac{1}{\sqrt{1−{\left(\frac{{\mathrm{v}}_1}{\mathrm{c}}\right)}^2}}\left({\mathrm{t}}_1+\frac{{\mathrm{v}}_1\left(\mathrm{l}\right)}{{\mathrm{c}}^2}\right)$

Here, ${\mathrm{t}}_3$ is the temporal separation of us.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_3=\frac{1}{\sqrt{1−{\left(\frac{0.980\mathrm{c}}{\mathrm{c}}\right)}^2}}\left(2.6×{10}^{−6}\text{ s}+\frac{\left(0.980\mathrm{c}\right)\left(760\text{″¾}\right)}{{\mathrm{c}}^2}\right)\\ =\frac{1}{\sqrt{1−{\left(0.980\right)}^2}}\left(2.6×{10}^{−6}\text{ s+}\frac{\left(744.8\text{″¾}\right)}{3×{10}^8\text{″¾/s}}\right)\\ =\frac{1}{\sqrt{1−0.9604}}(2.6×{10}^{−6}\text{ s+2.4}×{\text{10}}^{−6}\text{ s})\\ =\frac{1}{0.19}(5×{10}^{−6}\text{ s})\\ =2.63×{10}^{−5}\text{ s}\end{array}$

Thus, our temporal separation is $2.63×{10}^{−5}\text{ s}$.