A person travels by a car at a speed of 180 km h^ -1 . By his...

Speed of the car, $v = 180 \text{ km h}^{-1} = 180 \times \dfrac{5}{18} \text{ m s}^{-1} = 50 \text{ m s}^{-1}$.

Time measured by the wristwatch of the person in the car is the proper time, $\Delta t_0 = 10 \text{ h}$.

Distance measured in the car's frame is $L = v \Delta t_0 = 180 \text{ km h}^{-1} \times 10 \text{ h} = 1800 \text{ km}$.

The rest distance between the two stations is the proper length $L_0$. In the car's frame, the distance between the stations appears contracted, so $L = L_0 \sqrt{1 - \dfrac{v^2}{c^2}}$.

Therefore, $L_0 = \dfrac{L}{\sqrt{1 - \dfrac{v^2}{c^2}}}$.

Using the binomial expansion for $v \ll c$:

$L_0 \approx L \left(1 + \dfrac{v^2}{2c^2}\right) = L + \dfrac{L v^2}{2c^2}$

The additional distance is $\Delta L = \dfrac{L v^2}{2c^2}$.

Substituting the values $L = 1800 \text{ km} = 1.8 \times 10^6 \text{ m}$ and $c = 3 \times 10^8 \text{ m s}^{-1}$:

$\Delta L = \dfrac{1.8 \times 10^6 \times (50)^2}{2 \times (3 \times 10^8)^2}$

$\Delta L = \dfrac{1.8 \times 10^6 \times 2500}{18 \times 10^{16}} = 25 \times 10^{-9} \text{ m} = 25 \text{ nm}$

Thus, the rest distance is $25 \text{ nm}$ more than $1800 \text{ km}$.