Bats use echolocation to determine their distance from objects they cannot easily see in the dark. The time between the emission of a high-frequency sound pulse (a click) and the detection of its echo is used to determine such distances. A bat, flying at a constant speed of 19.3 m/s in a straight line toward a vertical cave wall, makes a single clicking noise and hears the echo 0.12 s later. Assuming that she continued flying at her original speed, how close was she to the wall when she received the echo? (assume the speed of sound is 343 m/s)

## Physics problem dealing with echolocation and a doppler type effect - Please help ASAP

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# 1 Answer

Assume the initial distance from the wall is x. The distance where the echo is heard is x-(19m/s)(0.12 s).

The distance traveled by the sound is (343 m/s)(0.12 s) or 41.16m

Since the sound starts at x and travels back to the point (x- (19m/s)(0.12 s)), we know that

x + (x - (19m/s )(0.12 s))=41.16 m

Simplifying, we get

2x - 2.28 m = 41.16 m

2x = 43.44 m

X = 21.72 m

But x is the initial distance from the wall, so subtract 2.28 m from x to get 19.44 m