Question

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 $\mathbf{10}\mathbf{s}$?

Short answer

The value of time difference in the plane’s clock is $\text{9.8\hspace{0.33em}ps}$.

Step-by-step solution

Write the given data from the question.

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 $\text{4200\hspace{0.33em}m}$.

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

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.

Determine the value of time difference in the plane’s clock.

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.