Video: Finding a Distance by the Timing of a Reflection

A sonar echo returns to a submarine 1.20 s after being emitted. Find the distance to the object creating the echo. Use a value of 1540 m/s for the speed of sound in sea water.

06:17

Video Transcript

A sonar echo returns to a submarine 1.20 seconds after being emitted. Find the distance to the object creating the echo. Use a value of 1540 meters per second for the speed of sound in sea water.

So the first thing we’ll do, let’s draw a picture of what’s going on here with this submarine that is emitting some sonar waves. So the sonar that’s a series of sound waves that come out the nose of the submarine. And let’s imagine that we have an object some distance away from the submarine, and these waves are moving out through the sea water, and eventually they come in contact with this object.

And when they do that, they bounce back off the object. They reflect backward and make their way back to the submarine, eventually reaching the source of the sound waves. And if we were to draw the journey at this take, the sound waves move out, they reflect off our object, and they come back. And we’re told that all that happens in a time — we’ll call it 𝑡 — a time of 1.20 seconds.

Now you can see from our diagram that the submarine is some distance away from this object. And if we call that distance 𝑑, that’s what we’re trying to solve for, that distance between the submarine and our object detected by sonar, and we’re told further that the speed of sound in water, in sea water that we’re gonna work with, is 1540 meters per second, which just as a side note is several times faster than sound travels in air. So that’s interesting that sound travels so much faster in sea water than in air.

We can write the speed of sound in sea water and represent it with a symbol, alright 𝑣 sub 𝑠 for the speed of sound in this scenario, and we’re given again that that’s 1540 meters per second.

Now given all this information, we can start to make headway by remembering a relationship between speed, distance, and time. And you may recall that the speed of an object, which we can represent as 𝑣, is defined as the distance that object travels divided by the time it takes that object to travel that distance.

Now let’s take this definition of speed being defined as distance over time and apply it to our situation in particular. Instead of simply a 𝑣 for speed, what we’ve been given is a 𝑣 sub 𝑠 for the speed of sound in sea water.

That speed of sound is equal to the distance that our sound waves travel. Now this is a total distance, so rather than being 𝑑, which would be, say, from the submarine to the object, we actually have a total distance of two times 𝑑 because, look again at our diagram, we see that the sound wave indeed travels a distance of 𝑑 out to the object and then a distance of 𝑑 back.

So that tells us why we would use two 𝑑 rather than 𝑑 in this equation. And finally, we’re given the time that this whole process takes, and again that value of time is 1.20 seconds.

Okay great, now we want to isolate the 𝑑 in this equation all by itself, so we’re gonna do that by algebraically rearranging this equation. As a first step, we can multiply both sides of the equation by 𝑡, the time that this whole process takes. And when we do that, you can see on the right side we cancel out the time.

Our next step is to multiply both sides of the equation by one-half. When we multiply our right side by one-half, you may notice that we again get cancellation: the two in the numerator and the two in the denominator both drop out.

So what we are left with is an equation that says 𝑑, our distance between the submarine and the object, is equal to the speed of sound in sea water multiplied by the time divided by two.

From here, we plug in the variable values we’ve been given. So remember that 𝑣 sub 𝑠 is 1540 meters per second and our time value is 1.20 seconds, and all of this is divided by two.

Just as a quick side note, we can actually check the units of our answer before we calculate it out. We can see here that we’ve got meters per second multiplied by seconds, so the second units cancels. And we’re left just with meters in our final answer. That makes sense because our final answer is a distance.

We can now move into finding the final answer by multiplying these values together on our calculator, and we find that the distance between the submarine and the object is 1848 meters divided by two, which equals 924 meters.

Now notice that this answer has three significant figures, which agrees with the values we were given to start the problem off. So again, the distance between the submarine and the object in this example is 924 meters.

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