Question

You are in a bus travelling on a straight road at $\mathbf{20}\mathbf{m}/\mathbf{s}$. As you pass a gas station, your clock and a clock in station read precisely 0. You pass another gas station $\mathbf{900}\mathbf{m}$ farther down the road. (in the frame of reference of the gas stations., all gas station clocks synchronized.) (a) As you pass the: second station, do you find its clock to be ahead of, or: behind your own clock and (b) by how much?

Short answer

(a) The clock of second station is ahead clock in bus.

(b) The duration by which clock of second gas station is ahead of bus clock is $1\times {10}^{-13}\text{\hspace{0.33em}s}$.

Step-by-step solution

Identification of given data

The speed of bus is $\mathrm{u}=20\text{\hspace{0.33em}m/s}$

The distance between gas stations is $\mathrm{L}=900\text{\hspace{0.33em}m}$.

Definition of Time Dilation

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.

Determine whether clock of second station is ahead or behind bus clock

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.

Determination of time by which the clock is ahead your clock

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.

$\begin{array}{l}t=\frac{900\text{\hspace{0.33em}m}}{20\text{\hspace{0.33em}m}/\text{s}}\\ t=45\text{\hspace{0.33em}s}\end{array}$

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

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

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

Substitute all the values in the above equation.

$\begin{array}{l}t-t_1=t\frac{v^2}{2c^2}\\ t-t_1=\left(45\text{\hspace{0.33em}s}\right)\left(\frac{{\left(20\text{\hspace{0.33em}m}/\text{s}\right)}^2}{2{\left(3\times {10}^8\text{\hspace{0.33em}m}/\text{s}\right)}^2}\right)\\ t-t_1=1\times {10}^{-13}\text{\hspace{0.33em}s}\end{array}$

Therefore, the duration by which clock of second gas station is ahead of bus clock is $1\times {10}^{-13}\text{\hspace{0.33em}s}$.