Question

A person travels by a car at a speed of 180 km h⁻¹. It takes exactly 10 hours by his wristwatch to go from the station A to the station B. (a) What is the rest distance between the two stations? (b) How much time is taken in the road frame by the car to go from the station A to the station B?

Answer 1

Given:
Speed of car, v = 180 km/hr = 50 m/s
Time, t = 10 h

Let the rest distance be L₀.

Apparent distance is given by
$L\text{'}=L_0\sqrt{1-v^2/c^2}$

Apparent distance, L' = 10 × 180 = 1800 km = 18 × 10⁵ m

Now,
$18\times {10}^5=L_0\sqrt{1-\frac{{\left(50\right)}^2}{{\left(3\times {10}^8\right)}^2}}$
$L_0=\left(18\times {10}^5+25\times {10}^{-9}\right)\mathrm{m}$

Thus, the rest distance is 25 nm more than 1800 km.

(b) Let the time taken by the car to cover the distance in the road frame be t. Then,

$$\begin{gathered} t=\frac{1.8\times {10}^5+25\times {10}^{-9}}{50} \\ =0.36\times {10}^5+5\times {10}^{-8} \\ =10\mathrm{hours}+0.5\mathrm{ns} \end{gathered}$$