If the time of flight is 13 μs, what is the reflector depth and the total distance traveled?

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Multiple Choice

If the time of flight is 13 μs, what is the reflector depth and the total distance traveled?

Explanation:
In ultrasound timing, the time of flight is the time for the pulse to travel to the reflector and back. The total distance the pulse covers during that time is v × t, where v is the speed of sound in tissue (about 1540 m/s). The reflector’s depth is half of that round-trip distance, since the pulse travels to the reflector and then the same distance back. Convert the time: 13 μs = 13 × 10^-6 s. Multiply by the speed: v × t ≈ 1540 × 13×10^-6 ≈ 0.02002 m, which is about 2.002 cm for the round-trip distance. Half of that is the depth: ≈ 1.001 cm. So the reflector depth is about 1 cm, and the total distance traveled is about 2 cm.

In ultrasound timing, the time of flight is the time for the pulse to travel to the reflector and back. The total distance the pulse covers during that time is v × t, where v is the speed of sound in tissue (about 1540 m/s). The reflector’s depth is half of that round-trip distance, since the pulse travels to the reflector and then the same distance back.

Convert the time: 13 μs = 13 × 10^-6 s. Multiply by the speed: v × t ≈ 1540 × 13×10^-6 ≈ 0.02002 m, which is about 2.002 cm for the round-trip distance. Half of that is the depth: ≈ 1.001 cm.

So the reflector depth is about 1 cm, and the total distance traveled is about 2 cm.

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