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### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# A supersonic: - plane travels al ${\mathbf{420}}{\text{ }}{\mathbf{m}}{/}{\mathbf{s}}$. 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 ${\mathbf{10}}{\mathbf{s}}$?

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

See the step by step solution

## Write the given data from the question.

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

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

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

## 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 $\left(3×{10}^{8}\right)$ for ${c}^{2}$ into equation (1).

$\begin{array}{c}\Delta {t}_{0}=10\sqrt{1-\frac{{\left(420\right)}^{2}}{{\left(3×{10}^{8}\right)}^{2}}}\\ =9.8×{10}^{-12}\\ =9.8\text{ ps}\end{array}$

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

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