Echo timing measures round-trip distance Steps:

• The student claps 11 times, with each subsequent clap timed to the echo of the previous one, so the time between claps equals one round trip: $t = \frac{2x}{v}$.
• Total time from first to eleventh clap covers 11 round trips: $11t = 9.4$ s.
• Solve for $t$: $t = \frac{9.4}{11} \approx 0.8545$ s.
• Distance: $x = \frac{v t}{2} = \frac{340 \times 0.8545}{2} \approx 145$ m (150 m).

Why B is correct:

• Matches calculation of $x = \frac{340 \times 9.4}{2 \times 11} \approx 150$ m using round-trip time formula $t = \frac{2x}{v}$.

Why the others are wrong:

• A: Assumes 12 round trips ($9.4/12 \approx 0.783$ s, $x \approx 133$ m).
• C: Assumes 9 round trips ($9.4/9 \approx 1.044$ s, $x \approx 178$ m).
• D: Assumes 5 round trips ($9.4/5 = 1.88$ s, $x \approx 320$ m).

Final answer: B