A submarine can use sonar (sound traveling through water) to determine its dista

Question

A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a "ping") and the detection of its echo carn be used to determine such distances. Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine's speed may be determined by comparing the time between pings. Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is known to be 1522 m/s. If you sent out pings every 4.90 s, and your apparatus receives echoes reflected from an undersea cliff every 4.88 s, how fast is your submarine approaching the cliff? 3.11 Your goal is to find the time between successive pulses received by the submarine, ot received. Let t-0 be when the submarine passes one of two successive pulses that approach it. After passing the first pulse, the next sound pulse moves toward the submarine at....V sound and the submarine moves toward the pulse at speed V sub- The distance between successive pulses may be divided into the distance the sub travels and the distance the second sound pulse travels in the time interval received. m/s eBook

Explanation / Answer

x= vsoundtemitted. We need to find the time treceivedbetween successive pulses received by the submarine.
We start our clock, as it were, when the submarine passes one of two successive pulses that approach it,
separated by the distance x. After passing the first pulse, the next sound pulse moves toward the
submarine atvsoundand the submarine moves toward the pulse at speed vsub. The distance between successive
pulses xmay be divided into xsuband xsound, which are equal to vsubtreceivedand vsubtreceived, respectively

The distance b/w successive pulses is given by

= x = x sub +x recv

substituting for all three term in this equatin yield

= Vsound (t)emitted = (x)sub(t)recv + Vsound(t)recv

solve for Vsub to obtain = Vsub = Vsound ((t)emitted - (t)recv / (t)recv

substitute numerical values and evaluate Vsub:

= Vsub = (1522)*(4.90-4.88)/(4.90) = 6.212m/s

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