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Physics problem about echoes

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    v= Δd/Δt

    3. 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.
     
  2. jcsd
  3. Feb 23, 2016 #2

    BvU

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    Hello Eve, :welcome:

    And why is that ?
     
  4. Feb 23, 2016 #3
    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?
     
  5. Feb 24, 2016 #4

    billy_joule

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    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.
     
  6. Feb 24, 2016 #5
    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?
     
  7. Feb 24, 2016 #6

    BvU

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    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 :smile: .
     
  8. Jan 15, 2017 #7
    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..)
     
  9. Jan 15, 2017 #8

    haruspex

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    No, you have to say what you think the answer is, and why.
     
  10. Jan 15, 2017 #9
    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?

    Relevent 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
     
  11. Jan 15, 2017 #10

    haruspex

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    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 travelled 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?
     
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