Special theory of relativity: Time dilation in an airplane

Question:
Physicists conduct an experiment to measure time dilation of an atomic clock in a flying airplane. If the velocity of the airplane is 960 km/h, find the time dilated of the atomic clock from the laboratory flame. (You do not need to consider the effect from the general theory of relativity.)

Answer:
We consider one direction of motion, say, $x$ direction. The Lorentz transformations for the displacement and time are
\begin{eqnarray}
x' &=& \frac{-vt+x}{\sqrt{1-\beta^2}}  \\
t' &=& \frac{t-(v/c^2)x}{\sqrt{1-\beta^2}}
\end{eqnarray}
where $\beta=v^2/c^2$. Suppose that the clock is placed at the origin of moving flame, $S'$. Then, one measures the time in laboratory flame, $S$. The origin in $S'$ indicates that $x'=0$. Thus, we have
\begin{equation}
0 = \frac{-vt+x}{\sqrt{1-\beta^2}} \quad \rightarrow \quad x = vt
\end{equation}
Plug it in the transformation for time.
\begin{equation}
t' = \frac{t-(v/c^2)(vt)}{\sqrt{1-\beta^2}}=\frac{t(1-\beta^2)}{\sqrt{1-\beta^2}}=\sqrt{1-\beta^2}t
\end{equation}
Since $v\ll c \rightarrow \beta \ll 1$, we can use the approximated expression with a Taylor expansion;
\[
t' = \left(1-\frac{1}{2}\beta^2\right)t
\]
Now, convert the velocity of the airplane into m/s. Note that 1 km = 1000 m and 1 hour = 3600 s.
\[
960 \ \mathrm{km/h} = 960 \times 1000 \div 3600 = 266.7 \ \mathrm{m/s}
\]
Therefore, we have the time dilation compared with the lab frame.
\[
t'= \left\{1-\frac{1}{2}\left(\frac{266.7}{3.00\times 10^8}\right)^2\right\}t=(1-3.95\times 10^{-13})t
\]
This means: While one second elapses in lab frame, the clock in the airplane only elapses $1-3.95\times 10^{-13}$ seconds.

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