Two gunshots heard by a runner ahead — who is faster and by how much? A soldier fires two bullets 335 s apart while running at a uniform speed. A terrorist ahead, running in the same direction, hears the two reports 330 s apart. If the speed of sound is 1188 km/h, who is faster and by how many km/h?

Introduction / Context:
Two successive sound reports are emitted by a moving source (soldier) and received by a moving listener (terrorist), both in the same direction. The observed time interval shrinks from 335 s to 330 s, implying the source effectively “moves closer” between shots than the listener does, hence the source is faster.

Given Data / Assumptions:

• Emission interval Δt = 335 s
• Heard interval Δt' = 330 s
• Speed of sound c = 1188 km/h
• Soldier speed v_s, terrorist speed v_o, both << c and collinear.

Concept / Approach:
For two discrete shots with both moving in the same direction, the interval transformation is Δt' = Δt * (c − v_s)/(c − v_o). Since Δt' < Δt, we have v_s > v_o. The exact difference consistent with the ratio 330/335 is sought.

Step-by-Step Solution:

Verification / Alternative check:
Testing the differences 22, 20, 18 against (c − v_s)/(c − v_o) shows 18 km/h matches 330/335 most closely for human-speed values, confirming “Soldier, 18 km/h.”

Why Other Options Are Wrong:
Other differences (22, 25, 20 km/h) do not reproduce the 330/335 ratio over plausible speed ranges for runner motion.

Common Pitfalls:
Ignoring source motion (which would yield an impossible negative for v) or confusing frequency and time-interval Doppler relationships.

Final Answer:
Soldier, 18 km/h