1. A student is on a stationary train - Scottish Highers Physics - Question 1 - 2017

To calculate the distance, we can use the formula: $s = ut + \frac{1}{2}at^2$ where:

• $u = 0$ (initial velocity),
• $a = 0.32 \text{ m s}^{-2}$ (acceleration),
• $t = 25 \text{ s}$ (time).

Substituting the values, we find:

$s = 0 \times 25 + \frac{1}{2} \times 0.32 \times (25)^2 = 0 + 0.16 \times 625 = 100 \text{ m}$

To find the speed of the train, we can use the Doppler effect formula:

$f_s = f' \frac{v}{v - v_t}$ Where:

• $f_s = 270 \text{ Hz}$ (source frequency),
• $f' = 290 \text{ Hz}$ (observed frequency),
• $v = 340 \text{ m s}^{-1}$ (speed of sound),
• $v_t$ is the speed of the train.

Rearranging the formula to solve for $v_t$:

$290 = 270 \frac{340}{340 - v_t}$

Cross multiplying gives: $290(340 - v_t) = 270 \times 340$ $290 \times 340 - 290v_t = 91800$ Now, substitute values: $98600 - 290v_t = 91800$ Thus, we can isolate $v_t$: $290v_t = 6800 \implies v_t = \frac{6800}{290} \approx 23.45 \text{ m s}^{-1}$

The frequency of the sound heard decreases due to the Doppler effect. As the train approaches, the sound waves are compressed, increasing the frequency. When the train passes and moves away, the sound waves stretch out, causing a decrease in frequency. This is further illustrated by drawing a diagram that shows the wavelength increasing as the train moves away from the observer.