Time difference gives round-trip distance from crack to bottom

Steps:

• The time between echoes is Δt = 5.0 × 10^{-5} s, representing the round trip from crack to bottom and back.
• Round-trip distance = speed × Δt = 5200 m/s × 5.0 × 10^{-5} s = 0.26 m.
• Distance x is half the round-trip distance: x = 0.26 m / 2 = 0.13 m.
• The earlier time (1.0 × 10^{-5} s) locates the crack but is not needed for x.

Why A is correct:

• It equals (speed × time to bottom reflection) / 2, the one-way distance using the full echo time as if from crack.

Why the others are wrong:

• B is the full round-trip distance (2x), not the one-way x.
• C is speed × 10^{-4} s, double the relevant time.
• D is speed × 1.5 × 10^{-4} s, an incorrect time multiple.

Final answer: A