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

The proper time measured by the wristwatch in the car is $\Delta \tau = 10 \text{ hours} = 36000 \text{ s}$.

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

The time measured in the road frame is $\Delta t$. According to the time dilation formula:

$\Delta t = \dfrac{\Delta \tau}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Since $v \ll c$, we can use the binomial approximation $(1 - x)^{-1/2} \approx 1 + \dfrac{x}{2}$:

\Delta t \approx \Delta \tau \left(1 +\dfrac{v^2}{2c^2}\right)

The extra time taken in the road frame is:

$\Delta t - \Delta \tau = \Delta \tau$\dfrac{v^2}{2c^2}

Substituting the values:

$\Delta t - \Delta \tau = 36000 \times$\dfrac{1}{2} \times \left(\dfrac{50}{3 \times 10^8}\right)^2$<br><br>$\Delta t - \Delta \tau = 18000 \times $\dfrac{2500}{9 \times 10^{16}}<br><br>$\Delta t - \Delta \tau = 2000 \times $\dfrac{2500}{10^{16}}<br><br>$\Delta t - \Delta \tau = $\dfrac{5 \times 10^6}{10^{16}} =$5 \times 10^{-10} \text{ s}$<br><br>$\Delta t - \Delta \tau = 0.5 \times 10^{-9} \text{ s} = 0.5 \text{ ns}$<br><br>Therefore, the time taken in the road frame is$0.5 \text{ ns}$more than$10 \text{ hours}.