91影视

The given data can be listed below as:

The length of the longest stretch limo is ${\mathrm{L}}_{\mathrm{c}}=30.5\text{鈥尘}$.

The length of the car when parked in front of the garage is ${\mathrm{L}}_{\mathrm{g}}=6.00\text{鈥嬧�尘}$.

The value of the attached ${\mathrm{x}}_{\mathrm{c}}$ axis is ${\mathrm{x}}_{\mathrm{c}}=0$.

The value of the attached ${\mathrm{x}}_{\mathrm{g}}$ axis is ${\mathrm{x}}_{\mathrm{g}}=0$.

The velocity of the car is$\mathrm{v}=0.9980\mathrm{c}$ .

The timing of the event is for Garageman and Carman is ${\mathrm{t}}_{\mathrm{g}1}={\mathrm{t}}_{\mathrm{c}1}=0$.

The occurrence of the event is ${\mathrm{x}}_{\mathrm{c}}={\mathrm{x}}_{\mathrm{g}}=0$.

Thespecial relativity mainly shows the relationship between the space and the time. It mainly shows that the physics laws are mainly invariant in the inertial reference frame.

The equation of the length of the limo is expressed as:

$\mathrm{L}={\mathrm{L}}_{\mathrm{c}}\sqrt{1鈭�{\left(\frac{\mathrm{v}}{\mathrm{c}}\right)}^2}$

Here, $\mathrm{L}$ is the length of the limo, ${\mathrm{L}}_{\mathrm{c}}$ is the length of the longest stretch limo,$\mathrm{v}$ is the velocity of the car and$\mathrm{c}$ is the speed of the light.

Substitute the values in the above equation.

$\begin{array}{c}\mathrm{L}=\left(30.5\text{鈥尘}\right)\sqrt{1鈭�{\left(\frac{0.9980\mathrm{c}}{\mathrm{c}}\right)}^2}\\ =\left(30.5\text{鈥尘}\right)\sqrt{1鈭�{\left(0.9980\right)}^2}\\ =\left(30.5\text{鈥尘}\right)\sqrt{1鈭�\left(0.996004\right)}\\ =\left(30.5\text{鈥尘}\right)\left(0.063\right)\\ =1.93\text{鈥尘}\end{array}$

Thus, the length of the limo is $1.93\text{鈥尘}$.

According to the question, the ${\mathrm{x}}_{\mathrm{g}}$ axis is mainly fixed with the garage. Hence, the value of the ${\mathrm{x}}_{\mathrm{g}2}$ will be the length of the car when parked in front of the garage.

Thus, the value of ${\mathrm{x}}_{\mathrm{g}2}$ is $6.00\text{鈥嬧�尘}$ .

The equation of the value of ${\mathrm{t}}_{\mathrm{g}2}$ of event 2 is expressed as:

${\mathrm{t}}_{\mathrm{g}2}={\mathrm{t}}_{\mathrm{g}1}+\frac{{\mathrm{L}}_{\mathrm{g}}鈭�\mathrm{L}}{\mathrm{v}}$

Here, ${\mathrm{t}}_{\mathrm{g}1}$ is the timing of the event is for Garageman and Carman, ${\mathrm{L}}_{\mathrm{g}}$ is the length of the car when parked in front of the garage and $\mathrm{v}$ is the velocity of light.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_{\mathrm{g}2}=0+\frac{6.00\text{鈥尘}鈭�1.93\text{鈥尘}}{0.9980(3脳{10}^8\text{鈥尘/s})}\\ =\frac{4.07\text{鈥尘}}{(2.9脳{10}^8\text{鈥尘/s})}\\ =1.36脳{10}^{鈭�8}\text{鈥塻}\end{array}$

Thus, the value of ${\mathrm{t}}_{\mathrm{g}2}$of event 2 is $1.36脳{10}^{鈭�8}\text{鈥塻}$ .

It has been identified that the lino was inside the garage in the time frame from ${\mathrm{t}}_{\mathrm{g}1}$ and also ${\mathrm{t}}_{\mathrm{g}2}$. Hence, the equation of the time duration is expressed as:

$\mathrm{t}={\mathrm{t}}_{\mathrm{g}2}鈭�{\mathrm{t}}_{\mathrm{g}1}$

Here,$\mathrm{t}$ is described as the time duration.

Substitute the values in the above equation.

$\begin{array}{c}\mathrm{t}=1.36脳{10}^{鈭�8}\text{鈥塻}鈭�0\\ =1.36脳{10}^{鈭�8}\text{鈥塻}\end{array}$

Thus, the limo was temporarily trapped for about $1.36脳{10}^{鈭�8}\text{鈥塻}$.

The equation of the length of the passing garage is expressed as:

${\mathrm{L}}_1={\mathrm{L}}_{\mathrm{g}}\sqrt{1鈭�{\left(\frac{\mathrm{v}}{\mathrm{c}}\right)}^2}$

Here, ${\mathrm{L}}_1$ is the length of the passing garage.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{L}}_1=\left(6.00\text{鈥尘}\right)\sqrt{1鈭�{\left(\frac{0.9980\mathrm{c}}{\mathrm{c}}\right)}^2}\\ =\left(6.00\text{鈥尘}\right)\sqrt{1鈭�{\left(0.9980\right)}^2}\\ =\left(6.00\text{鈥尘}\right)\sqrt{1鈭�\left(0.996004\right)}\\ =\left(6.00\text{鈥尘}\right)\left(0.063\right)\\ =0.379\text{鈥尘}\end{array}$

Thus, the length of the passing garage is $0.379\text{鈥尘}$.

According to the question, the limo is fixed with the ${\mathrm{x}}_{\mathrm{c}}$ axis. Then the value of the $x_{c2}$ will be the value of the length of the car when parked in front of the garage.

Thus, the value of $x_{c2}$ is $30.5\mathrm{m}$.

The equation of the value of ${\mathrm{t}}_{\mathrm{c}2}$of event 2 is expressed as:

${\mathrm{t}}_{\mathrm{c}2}={\mathrm{t}}_{\mathrm{c}1}鈭�\frac{{\mathrm{L}}_{\mathrm{c}}鈭�{\mathrm{L}}_{\mathrm{g}}}{\mathrm{v}}$

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{t}}_{\mathrm{c}2}=0鈭�\frac{30.5\text{鈥尘}鈭�0.379\text{鈥尘}}{0.9980(3脳{10}^8\text{鈥尘/s})}\\ =\frac{30.121\text{鈥尘}}{(2.9脳{10}^8\text{鈥尘/s})}\\ =鈭�1.01脳{10}^{鈭�7}\text{鈥塻}\end{array}$

Thus, the value of ${\mathrm{t}}_{\mathrm{c}2}$ for event 2 is $鈭�1.01脳{10}^{鈭�7}\text{鈥塻}$.

According to the point of view of Carman, the limo was not in the garage. The doors of the limo were also not shut.

Thus, the limo was not in the garage with both doors shut.

According to the question, it has been identified that ${\mathrm{t}}_{\mathrm{c}2}<{\mathrm{t}}_{\mathrm{c}1}$. Hence, from the point of view of the Carman, the event 1 occurs after the event 2.

Thus, the event 2 occurs first.

The event 1 has been sketched below:

Here, in the above diagram, it can be identified that the car is at a certain distance from the door.

The event 2 has been sketched below:

Here, in the above diagram, it can be identified that the car in front of the door.

Thus, and

It has been observed that the limo is not in the garage which has both the doors shut. Hence, if the events were related, then the limo would not have been in the garage.

Thus, the events are not casually related.

It can be observed that both the persons garageman and carman are absolutely correct in their reference frames. Hence, in a certain way, it can be said that both the persons are correct.

Thus, both carman and garageman wins the bet.