A supersonic: - plane travels al $\mathbf{420}\mathbf{m}/\mathbf{s}$. As this plane passes two markers a distance of $\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{km}$ apart on the ground, how will the time interval registered on a very precise clock onboard me plane differ from $\mathbf{10}\mathbf{s}$?

Consider a time difference is$\mathrm{\Delta t}=10\text{\hspace{0.33em}s}$.

Consider a speed of the plane is$\mathrm{\upsilon }=420\frac{\text{m}}{\text{s}}$.

Consider a distance is $\text{4200\hspace{0.33em}m}$.

Write the formula of time difference in the plane’s clock.

${\mathrm{\Delta t}}_0=\mathrm{\Delta t}\sqrt{\mathbf{1}-\frac{{\mathrm{\upsilon }}^2}{{\mathbf{c}}^2}}$ …… (1)

Here, $\Delta \mathbf{t}$ is time difference, $\upsilon$ speed of the plane and $\mathbf{c}$ speed of light.

Due to relativistic effects, time relative to an object moves more quickly when it is moving very quickly. The time-dilation equation helps explain this:

$\Delta t=\frac{\Delta t_0}{\sqrt{1-\frac{{\upsilon }^2}{c^2}}}$

Arrive at the following expression for the time difference in the plane's clock using the time-dilation equation:

$\begin{array}{c}\Delta t=\frac{\Delta t_0}{\sqrt{1-\frac{{\upsilon }^2}{c^2}}}\\ \Delta t_0=\Delta t\sqrt{1-\frac{{\upsilon }^2}{c^2}}\end{array}$

Calculate the time difference in the clock of the aircraft using the formula for${\mathrm{\Delta t}}_0$:

Determine the time difference in the plane’s clock.

Substitute $10$ for$\Delta t$, $420$ for $\upsilon$ and $\left(3\times {10}^8\right)$ for $c^2$ into equation (1).

$\begin{array}{c}\Delta t_0=10\sqrt{1-\frac{{\left(420\right)}^2}{{\left(3\times {10}^8\right)}^2}}\\ =9.8\times {10}^{-12}\\ =9.8\text{\hspace{0.33em}ps}\end{array}$

According to the findings, the plane's clock would display the time $\text{9.8\hspace{0.33em}ps}$ earlier.