Sound interval with a moving listener — hunting scene: A hunter fires two shots 76 s apart. A tiger running away hears the two reports 83 s apart. If the speed of sound is 1195.2 km/h, find the speed of the tiger.

Introduction / Context:With a stationary source (hunter) and a listener moving directly away (tiger), the time interval between the arrivals of discrete sounds lengthens. This is a time-domain Doppler reasoning using speeds rather than frequencies.

Given Data / Assumptions:

• Emission interval Δt = 76 s
• Heard interval Δt' = 83 s
• Speed of sound c = 1195.2 km/h
• Tiger runs straight away from the hunter at speed v (km/h).

Concept / Approach:For a moving listener and stationary source, the ratio of heard interval to emitted interval is Δt' / Δt = c / (c − v). Rearranging gives v = c * (1 − Δt/Δt').

Step-by-Step Solution:

Verification / Alternative check:Plug back: Δt' = Δt * c/(c − v) = 76 * 1195.2/(1195.2 − 100.8) = 76 * 1195.2/1094.4 ≈ 83 s, confirming consistency.

Why Other Options Are Wrong:112.8, 80.16, 96.0 km/h mismatch Δt' when substituted into c/(c − v). “None of these” is unnecessary since a precise match exists.

Common Pitfalls:Using frequency Doppler formulas directly or forgetting that time intervals scale by the inverse of the relative “closing” rate of sound minus listener speed.

Final Answer:100.8 km/h