Find distance using speed of sound

In summary: What I get is that you are expected to compute a number, and the online service is giving you an error message because the number you computed is not the number they have stored in their database. Are you expected to get the right answer, or the answer they have stored in their database?In summary, the sailor strikes the side of his ship just below the waterline and hears the echo of the sound reflected from the ocean floor directly below 2.02 seconds later. Using the formula distance = speed x time, we can calculate the depth of the ocean to be 1515 meters. However, for the second problem, where a rescue plane is searching for a disabled boat and the sound takes 2.17 seconds
  • #1
L1988
11
0
Two questions:
1. A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor directly below 2.02 s later. How deep is the ocean at this point?

I assumed this used distance = speed x time. I knew that speed must equal the speed of sound in water, which I think is 1500 m/s (pretty sure, but not 100%) and I also divided my final distance by 2 so that I only got the distance of depth one way. However, the computer (this is online homework) said I was wrong, so I'm not sure where I went wrong. Any help would probably help me with the next problem too.


2.
A rescue plane flies horizontally at a constant speed searching for a disabled boat. When the plane is directly above the boat, the boat's crew blows a loud horn. By the time the plane's sound detector perceives the horn's sound, the plane has traveled a distance equal to one half its altitude above the ocean. If it takes the sound 2.17 s to reach the plane, determine the altitude of the plane. Take the speed of sound to be 350 m/s.

Just like last time, I assumed that I should use

distance= speed of sound x time

with a new speed (350 m/s), new time, and this time distance doesn't have to be divided by 2. However, that got me the wrong answer.

Then I tried drawing a triangle, (since the plane has moved) and called my final distance r^2 and my x^2 = one half its altitude, so all that was left was y^2. This also was incorrect.

Where am I making a mistake?
 
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  • #2
Well, what WERE your answers? How can we see where you went wrong if you don't show us what you got?
 
  • #3
sorry, I am new to this forum.

for the first one I got 1515m
343m/s x 2.02 s = 3030 m
3030m /2 = 1515 m
 
  • #4
second question:
Note: problem says : Take the speed of sound to be 350 m/s.

attempt 1:
distance = speed x time
d = 350 x 2.17 s = 759.5 m

attempt 2: using a triangle and pythagorean's theorem

759.5 = r^2
x^2 = "distance equal to one half its altitude above the ocean"
x^2= (sqrt (759.5) /2)^2
Y^2 = r^2-x^2
y^2= 759.5-189.9
y^2= 23.9m
 
  • #5
343m/s x 2.02 s = 3030 m

Check your arithmetic here.
 
  • #6
sorry i meant speed of sound
1500 x 2.02
 
  • #7
L1988 said:
second question:
Note: problem says : Take the speed of sound to be 350 m/s.

attempt 1:
distance = speed x time
d = 350 x 2.17 s = 759.5 m

attempt 2: using a triangle and pythagorean's theorem

759.5 = r^2
x^2 = "distance equal to one half its altitude above the ocean"
x^2= (sqrt (759.5) /2)^2
Y^2 = r^2-x^2
y^2= 759.5-189.9
y^2= 23.9m

How did you get the 189.9? If you don't know the altitude, how can you know half the altitude?? You can express the two sides of the triangle in terms of the altitude, h (what you are trying to find). Yes, 759.5 m is the length of the hypotenuse of your triangle, but make sure you are using the Pythagorean theorem correctly (be careful of your algebra).
 
  • #8
ok:

759.5 = (h/2) ^2 + h^2
759.5 = (h^2 / 4) + (h)^2
759.5 = (h)^2 (1+1/4)
759.5 = h^2 (1.25)
759.5 / (1.25) = h^2
h^2 = 607.6
h = 24.6

Still incorrect according to the computer
 
  • #9
Don't forget to square the 759.5 meters.
 
  • #10
awesome! I got it,

final answer = 679.3

Thanks for your help!

I don't think I can use this method for the boat question, right? Do you have any suggestions for that?
 
  • #11
L1988 said:
ok:

759.5 = (h/2) ^2 + h^2
759.5 = (h^2 / 4) + (h)^2
759.5 = (h)^2 (1+1/4)
759.5 = h^2 (1.25)
759.5 / (1.25) = h^2
h^2 = 607.6
h = 24.6

Still incorrect according to the computer

BTW, one of the things you hopefully are learning/practicing is how to check your answers intuitively. When you found the answer, did you stop and ask yourself "does it make sense that the airplane was flying less than 25 metres above the ocean?"
 
  • #12
I still don't understand why

(1500 m/s) x (2.02s) x (.5) = 1515 m

is incorrect for the boat problem
 
  • #13
L1988 said:
I still don't understand why

(1500 m/s) x (2.02s) x (.5) = 1515 m

is incorrect for the boat problem

Frankly, I don't understand how an online service deals with margins of error. What if the correct answer is 1514?
 

1. How does the speed of sound determine distance?

The speed of sound is a constant value that is dependent on the medium through which it travels. By knowing the speed of sound in a specific medium, we can calculate the distance traveled by dividing the time it takes for the sound to travel by the speed of sound.

2. What factors affect the speed of sound?

The speed of sound is affected by the density, temperature, and elasticity of the medium through which it travels. In general, sound travels faster in denser, warmer, and more elastic mediums.

3. Can the speed of sound change?

Yes, the speed of sound can change depending on the medium it is traveling through. For example, sound travels faster in water than in air. It can also change due to changes in temperature and pressure.

4. How do we measure the speed of sound?

The speed of sound can be measured using various methods, such as timing how long it takes for an echo to return or using specialized equipment like a sound wave analyzer. The most common method is to use the time it takes for sound to travel a known distance.

5. What are the practical applications of finding distance using the speed of sound?

Knowing the speed of sound and how to calculate distance using it is essential in various fields, such as navigation, seismology, and engineering. It is also used in everyday life, such as measuring the distance of a lightning strike or estimating the depth of a body of water.

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