Echo timing matches round-trip sound travel Steps:

• Sound travels distance 2x (to wall and back) for each echo.
• Time between claps equals round-trip time T = 2x / v.
• N claps in time t means (N-1) intervals, so t ≈ N T for large N.
• Thus, v ≈ 2 N x / t.

Why B is correct:

• Speed of sound v equals total round-trip distance for N cycles (≈ 2 N x) divided by total time t, per the definition of speed as distance over time in the approximation for many claps.

Why the others are wrong:

• A: Uses one-way distance x instead of round trip 2x.
• C: Uses one-way distance without accounting for clap count N.
• D: Ignores multiple claps, treating t as single round-trip time.

Final answer: B