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Math Help - Problem solving question about distance

  1. #1
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    May 2009
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    Problem solving question about distance

    Hello, I'm not sure how to solve this question:

    The velocity of a shock wave in water is 1500 m/s, while the velocity of sound in air is 330 m/s. The shock wave from a depth charge just beneath the surface of the water is felt by a ship 5.5 s before the sound is heard by the crew. How far is the ship from the explosion.

    How I approach it is like this:

    5.5 s delay x 330 m/s = 1815 m (how far the sound travels in 5.5 seconds to reach the ship)

    Then I divide 1815 m by 1500 m/s = 1.21 s, meaning it took the shock wave 1.21 s to move the distance the sound travelled in 5.5 seconds, meaning the sound travelled for an additional 1.21 s prior to the 5.5 s delay, since it took the shock wave 1.21 s to first reach the ship.

    So this would mean that the sound travelled for 6.71 s to reach the boat, which is 2214.2 m.

    The problem is that there is another answer, and I don't know what is wrong with my approach, or what the other approach is, because I'm having trouble conceptualizing this. Any explanations and steps would be much appreciated. Thank you

    Edit: oh and the answer given in the text is 2327 m.
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  2. #2
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    Joined
    May 2009
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    Ok I figured out a way to do this, but I'm sure there is a faster way, so any advice is greatly appreciated and wanted

    The velocity of a shock wave in water is 1500 m/s, while the velocity of sound in air is 330 m/s. The shock wave from a depth charge just beneath the surface of the water is felt by a ship 5.5 s before the sound is heard by the crew. How far is the ship from the explosion.

    Here's what I did:

    I determined the displacement of the shock wave from the boat, as if it continued to move 1500 m/s for the 5.5 s it took for the sound to reach the ship. So 5.5 seconds times 1500 m/s, which equals 8250 m.

    So now I take 8250 m as both the distance of the shockwave from the ship, but also the distance between the shockwave and the sound.

    Now I work out a pattern, with a hypothetical Time sequence:

    Sound:
    0 seconds = travelled 0 meters
    1s=330m
    2s=660m
    3s=990m, etc.

    Shockwave:
    0 seconds = travelled 0 meters
    1s=1500m
    2s=3000m
    3s=4500m

    Displacement between the two:
    0 seconds = 0 meters displaced
    1s=1170m
    2s=1340m
    3s=3510m

    This means that every second, the shockwave and sound are displaced by 1170m. Therefore, I construct a formula from this:

    Displacement=time(in seconds) x 1170 meters or d=t(1170m)

    Now, since at the final time, at 5.5 seconds after the shockwave, when the sound is heard at the ship, I've calculated the displacement of the sound and shockwave to be 8250 m, I use the formula:

    8250=t x 1170 meters
    Therefore, t=8250m/1170m
    Therefore, t= 7.051282051... seconds

    Now, since I have the time required for that length of displacement, I can answer the question: what is the distance that the ship is from the blast.

    So I can do either:

    [(7.051282051...)seconds x 1500m/s]-8250m (since the shockwave is 8250m away from the boat)

    or

    (7.051282051...)seconds x 330m/s

    Both equal: 2326.923077 meters, or 2327m (as stated in the text book)

    Any tips? Thanks for your time
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  3. #3
    Newbie findmehere.genius's Avatar
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    India, UP
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    Sure..your method was good, but I do have a problem in understanding the question....its that, the siesmic wave and and the sound wave both, are from one source in the water...and we have speed of sound in air. so how can we calculate distance travelled by sound in a particular "in water".
    if i get the question I will try to answer you a bit more better.
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