Calculate Additional Time for Echo in Air to Return | Physics Problem

[Evangeline101]

Homework Statement

A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

Homework Equations

v= Δd/Δt

The Attempt at a Solution

Δd = vΔt = (1470 m/s) (0.40s) = 588 m

The distance for the echo to return is half of this so:
588m /2 = 294 m (I am not so sure about this part)

Δt = Δd / v = 294 m / 340 m/s = 0.86 s

Additional time:
0.86s - 0.40s = 0.46 s

The echo in the air took 0.46s longer to return.

Is this right? Any help will be appreciated.

 

[BvU]

Hello Eve,

The distance for the echo to return is half of this

And why is that ?

 

[Evangeline101]

Well I just thought 588 m would be the total distance for the echo to go their and reflect back so you would divide by two to get the actual distance?

 

[billy_joule]

Well I just thought 588 m would be the total distance for the echo to go their and reflect back so you would divide by two to get the actual distance?

You would divide by two to find the distance to the cliff but you aren't interested in that.

You correctly found the echo in the water travels 588 metres, will the echo in the air travel the same distance?

I think the question is worded poorly, I'm quite sure they are asking for the time between the captain hearing the water echo and the air echo. ie the difference in round trip time, not the difference in 'return' time.
ie
How much additional time will it take the echo in the air to return to the boat?
(incorrectly implying what comes out of the horn is already an echo..)
rather than
How much additional time will it take the echo in the air to return from the cliff?
(correctly implying the echo forms at the cliff)

Although the former is misleading,I think it's the correct interpretation.

 

[Evangeline101]

So is my previous answer wrong? If so I did it again without doing the whole dividing by 2 thing:
v = Δd/Δt

Δd = v Δt = (1470 m/s) (0.40 s) = 588 m

Δt = Δd/v= 588 m / 340 m/s = 1.73 s

Δt = 1.73 s – 0.40 s = 1.33 s

The echo in the air took 1.33 s longer to return.

Is this a better answer?

 

[BvU]

Do you still have to ask ? What do you think yourself ?

Re Billy's ambiguity: same wording for both routes makes the answer correct one way and the other way as well .

 

[bilalsyed25]

so wat is the correct answer, 1.33s or 0.46s?

Pls reply soon!

(I have the same question and I'm having trouble with it..)

 

[haruspex]

so wat is the correct answer, 1.33s or 0.46s?

Pls reply soon!

(I have the same question and I'm having trouble with it..)

No, you have to say what you think the answer is, and why.

 

[bilalsyed25]

The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

relevant equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s

 

[haruspex]

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

It depends how you define d. You can choose to define it as the distance from boat to cliff or as the distance for the return trip (which is, after all the distance traveled by the sound). Either will work as long as you use it consistently.
Think about your own working with respect to that. At each calculation, what is the description of the distance or time you have calculated?

 

[Nestory]

So is my previous answer wrong? If so I did it again without doing the whole dividing by 2 thing:
v = Δd/Δt

Δd = v Δt = (1470 m/s) (0.40 s) = 588 m

Δt = Δd/v= 588 m / 340 m/s = 1.73 s

Δt = 1.73 s – 0.40 s = 1.33 s

The echo in the air took 1.33 s longer to return.

Is this a better answer?

Yes that's so right i also got that answer