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Math Help - earthquake

  1. #1
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    earthquake

    When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.03 km/s and the secondary, or S, wave has a speed of about 5.63 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.9 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?
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  2. #2
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    Quote Originally Posted by lovinhockey26 View Post
    When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.03 km/s and the secondary, or S, wave has a speed of about 5.63 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.9 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?
    You know that the distance is calculated by:

    d = v \cdot t

    Both waves travel the same distance

    d = v_P \cdot t_P and

    d = v_S \cdot t_S. Use the fact that t_S - t_p = 77.9\ s :

    v_P \cdot t_P = v_S \cdot (t_P + 77.9) . After a few steps you'll get:

    t_P=\frac{v_S \cdot 77.9}{v_P - v_S}\approx 182.7\ s

    And now you can calculate the distance to the earthquake.
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  3. #3
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    Hello, lovinhockey26!

    When an earthquake occurs, two types of sound waves are generated
    and travel through the earth.
    The primary or P wave has a speed of about 8.03 km/s
    and the secondary or S wave has a speed of about 5.63 km/s.

    A seismograph, located some distance away, records the arrival of the P wave
    and then, 77.9 s later, records the arrival of the S wave.

    Assuming that the waves travel in a straight line,
    how far is the seismograph from the earthquake?

    The P wave travels for t seconds at 8.03 km/sec.
    . . Its distance is: . 8.03t feet.

    The S wave travels for (t+77.9) seconds at 5.03 km/sec.
    . . Its distance is: . 5.03(t+77.9) feet.

    These distances are equal: . 8.03t \;=\;5.63(t + 77.9)

    Go fot it!

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