1. Heard Duration at Station P ($\tau_P$)

The train moves towards P with velocity $v = 90 \text{ km/h} = 25 \text{ m/s}$.

• Start of Honk ($t=0$): Signal emitted at distance $x$. Arrives at $t_{start} = x/c$.
• End of Honk ($t=\tau$): Train has moved closer by $d = v\tau$. New distance is $x – v\tau$. Signal emitted at $t=\tau$ arrives at $t_{end} = \tau + \frac{x – v\tau}{c}$.

The duration $\tau_P$ is the difference in arrival times:

$$ \tau_P = t_{end} – t_{start} = \left(\tau + \frac{x – v\tau}{c}\right) – \frac{x}{c} $$ $$ \tau_P = \tau – \frac{v\tau}{c} = \tau \left(1 – \frac{v}{c}\right) $$

Substituting values ($c = 350 \text{ m/s}, v = 25 \text{ m/s}, \tau = 44 \text{ s}$):

$$ \tau_P = 44 \left(1 – \frac{25}{350}\right) = 44 \left(1 – \frac{1}{14}\right) = 44 \left(\frac{13}{14}\right) \approx 40.86 \text{ s} $$

2. Heard Duration at Village Q ($\tau_Q$)

Village Q is located perpendicular to the track at distance $y = 1.2 \text{ km}$. We must calculate the path length of sound for the start and end of the honk.

Start of Honk:

• Train position: $x_1 = 1.6 \text{ km} = 1600 \text{ m}$ from P.
• Distance to Q (Hypotenuse): $d_1 = \sqrt{x_1^2 + y^2} = \sqrt{1600^2 + 1200^2} = 2000 \text{ m}$.
• Arrival time: $t_1 = \frac{d_1}{c} = \frac{2000}{350}$.

End of Honk ($t = 44 \text{ s}$):

• Train travels distance $s_{train} = v \times \tau = 25 \times 44 = 1100 \text{ m}$ towards P.
• New position from P: $x_2 = 1600 – 1100 = 500 \text{ m}$.
• New Distance to Q: $d_2 = \sqrt{x_2^2 + y^2} = \sqrt{500^2 + 1200^2} = 1300 \text{ m}$.
• Signal emitted at $t=44$ arrives at: $t_2 = 44 + \frac{d_2}{c} = 44 + \frac{1300}{350}$.

Duration Calculation:

$$ \tau_Q = t_2 – t_1 = 44 + \frac{1300}{350} – \frac{2000}{350} $$ $$ \tau_Q = 44 + \frac{1300 – 2000}{350} = 44 – \frac{700}{350} $$ $$ \tau_Q = 44 – 2 = 42 \text{ s} $$