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  1. #1
    Junior Member MathGuy's Avatar
    Join Date
    Dec 2010

    A boat is floating at rest in dense fog near a large cliff...How much...

    ...additional time will it take? Here's the full 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 and the air (340 m/s) simultaneously. The echo in the water takes 0.40 s to return. How much additional time will it take the echo in the air to return?

    This is my attempt:

    First I find the distance by using the ?d = V(?t) and then divide the distance by 2 since it takes twice the distance for the echo to come back, so I have a distance of 65M. Then I would this distance and the given velocity of the sound-wave to find the time it takes to reach where the sound bounces back and have a value ?t = 0.20 seconds. Then I add both, to get a total value of ?t = 0.60 seconds. Is this the answer? Or am I wrong?

  2. #2
    I am seeking the answer to this question as well, I may have mucked it up a bit though here is my attempt:

    If the speed of sound in salt water is 1470m/s (as stated in the lesson I'm taking), then the answer is that it takes 1.32 seconds longer for the sound to travel through the air than the sea water.

    How I got my answer:

    The time, 0.4s, multiplied by the speed of sound in salt/sea water, 1470m/s, is 588m. Now, divide the distance, 588m, by the velocity, 340m/s, and the answer is 1.72s, now minus the time it takes for the sound to travel through the salt water, 0.4s, and the answer is 1.32 s longer.

    I am trying to make sure my answers are correct and my calculations were done correctly, I hope someone comments and corrects us!

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