4. Two trains depart from a station at the same time - Scottish Highers Physics - Question 4 - 2023

To find the relative speed, we subtract the speed of train A (and therefore the passenger walking towards the back) from the speed of train B:

$v_{passenger} = v_A - v_B$

Where:

• $v_A = 3.5 \, \text{m s}^{-1}$ (speed of train A)
• the passenger walks at 1.3 m s⁻¹ towards the rear, effectively reducing their speed relative to train A:

The passenger's speed is thus $v_A - 1.3 \, \text{m s}^{-1}$, and then relative to train B:

$v_{passenger} = (3.5 - 1.3) - 4.0 = 1.8 \, \text{m s}^{-1}$

So, the speed of the passenger on train A relative to a passenger seated on train B is 1.8 m s⁻¹.

The speed of the emitted light towards the stationary observer is $3.00 \times 10^8 \, \text{m s}^{-1}$ or $c$, where $c$ represents the speed of light. This speed is invariant and remains constant for all observers regardless of their relative motion.

The reasoning comes from the principles of Einstein's theory of relativity, where the speed of light is always the same in all inertial frames of reference. Hence, the speed measured by the stationary observer will still be $c = 3.00 \times 10^8 \, \text{m s}^{-1}$.

To calculate the contracted length of train A, we use the formula for length contraction:

$L' = L \sqrt{1 - \frac{v^2}{c^2}}$

Where:

• $L$ is the proper length of the train (142 m)
• $v = 0.9c$

This leads to: