91影视

The speed of bus is $\mathrm{u}=20\text{鈥刴/蝉}$

The distance between gas stations is $\mathrm{L}=900\text{鈥刴}$.

The difference in the observed time and actual time of event due to the relative movement between observer and frame is called the time dilation.

As of initially clocks in bus reference frame and gas station reference frame reads 0 in the clock. To maintain the reading of clock of bus with second gas station clock, the second bus station clock should be ahead of bus clock.

Therefore, the clock of second station is ahead clock in bus.

The time reading of bus clock at second gas station is given as:

$t=\frac{L}{u}$

Substitute all the values in the above equation.

$\begin{array}{l}t=\frac{900\text{鈥刴}}{20\text{鈥刴}/\text{s}}\\ t=45\text{鈥剆}\end{array}$

The duration by which clock of second gas station is ahead of bus clock is given as:

$\begin{array}{l}{t}_1=t{\left(1鈭�\frac{v^2}{c^2}\right)}^{1/2}\\ t_1鈮�t\left(1鈭�\frac{v^2}{2c^2}\right)\end{array}$

$\begin{array}{c}{t}_1=t鈭�t\frac{v^2}{2c^2}\\ t鈭�t_1=t\frac{v^2}{2c^2}\end{array}$

Substitute all the values in the above equation.

$\begin{array}{l}t鈭�t_1=t\frac{v^2}{2c^2}\\ t鈭�t_1=\left(45\text{鈥剆}\right)\left(\frac{{\left(20\text{鈥刴}/\text{s}\right)}^2}{2{\left(3脳{10}^8\text{鈥刴}/\text{s}\right)}^2}\right)\\ t鈭�t_1=1脳{10}^{鈭�13}\text{鈥剆}\end{array}$

Therefore, the duration by which clock of second gas station is ahead of bus clock is $1脳{10}^{鈭�13}\text{鈥剆}$.