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35E

Expert-verifiedFound in: Page 1

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**A supersonic: - plane travels al ${\mathbf{420}}{\text{\hspace{0.33em}}}{\mathbf{m}}{/}{\mathbf{s}}$. 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 $\text{9.8\hspace{0.33em}ps}$.

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.

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