Solve Physics Echo Problem: Air Echo Returns 1.33s Later

[SpicyWassabi101]

Homework Statement

A boat is floating at rest in dense fog near a 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

First Attempt:
v = Δd/Δt

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

Δt = Δd/v = 588m / 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.

Second Attempt:
I was thinking that 1.73 s would be the total time, so then I would have to divide 1.73 s by 2 to get the return time (0.865 s). Then I would subtract 0.40s from 0.865 s to get the additional time it took the echo to return, which would be 0.465 s. Like this:

v = Δd/Δt

Find total time:
Δd = v Δt = (1470 m/s) (0.40 s) = 588 m

Δt = Δd/v = 588m / 340 m/s = 1.73 s

Then find return time:
1.73 s/2 = 0.865 s
Then find additional time:
0.865 s - 0.40s = 0.465 s

The echo in the air took 0.465 s longer to return.

Please help me figure out which process is correct, if they even are correct :)

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[mgkii]

pretty sure your 1st answer is correct. The question is not asking about the return journey, but total journey time - 0.4s in total for the underwater echo, so the question is looking for the difference in total time

 

[Evangeline101]

I am doing a similar qts

 

[SpicyWassabi101]

Thanks so much for your help, mgkii :)