A person travelling in a car moving at 100 km h^ -1 observes that...

The time taken by the car to travel from tower $A$ to tower $B$ in the Earth's frame is given by $t = \dfrac{d}{v}$.

Substituting $d = 1000 \text{ km} and$v = 100 \text{ km h}^{-1}$, we get$t = \dfrac{1000}{100} = 10 \text{ h} = 36000 \text{ s}$.<br><br>The speed of the car in$\text{m s}^{-1}$is$v = 100 \times \dfrac{5}{18} = \dfrac{250}{9} \text{ m s}^{-1}$.<br><br>According to the time dilation formula, the proper time$t_0$measured by the wristwatch in the car is related to the time$t$in the Earth's frame by$t_0 = t \sqrt{1 - \dfrac{v^2}{c^2}}$.<br><br>Since$v \ll c$, using the binomial expansion, we can approximate this as$t_0 \approx t \left(1 - \dfrac{v^2}{2c^2}\right).<br><br>The time difference between the clock on towerB$and the wristwatch is$\Delta t = t - t_0 = t $\dfrac{v^2}{2c^2}.<br><br>Substituting the values, we get$\Delta t = 36000 \times $\dfrac{1}{2} \times \left(\dfrac{250}{9 \times 3 \times 10^8}\right)^2$.

$\Delta t = 18000 \times \left(\dfrac{25}{27 \times 10^7}\right)^2 = 18000 \times$\dfrac{625}{729 \times 10^{14}} = \dfrac{11250000}{729 \times 10^{14}} \text{ s}.

$\Delta t =$\dfrac{1125}{729} \times$10^{-10} \text{ s} \approx 1.543 \times 10^{-10} \text{ s} = 0.154 \text{ ns}$.

Since $t_0 < t$, the wristwatch shows less time than the clock on tower $B$. Therefore, the wristwatch lags the clock on tower $B$ by $0.154 \text{ ns}$.