Bats use echolocation to determine their distance from objects they cannot easil

Question

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 19.8 m/s in a straight line toward a vertical cave wall, makes a single clicking noise and hears the echo 0.07 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)

Explanation / Answer

Assume that the bat, when clicking, is x meters away from the wall. Then the sound has to travel x meters to the wall and then some way back which is *shorter* than the x meters, as the bat flies towards the wall. How much shorter? Well, exactly the distance the bar flew in these 0.07 seconds, which is 19.8*0.07 meters. So the sound, altogether, travels x + (x - 19.8*0.07) meters. For this distance, it takes 0.07 seconds. Therefore, its velocity is x + (x - 19.8*0.07) --------------------------- m/s 0.07 which we know to be equal to 343 m/s. So we have x + (x - 19.8*0.07) --------------------------- = 343 0.12 x + x - 19.8*0.07 = 343*0.07 2x = 0.07 * (343 + 19.8) x = 0.035 * (343 + 19.8) = 12.698 m This is where the bat *starts*. Now, How far away is she from the wall when it hears the click? Well, as we saw Distance = x - 19.8*0.07 which is 0.035 * (343 + 19.8) - 19.8*0.07 = 11.312 m ---------------> Answer She is at 11.312 m close to the wall when she received the echo

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