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    A recording station observed that there was an interval of 68s between the reception of P (push or primary) waves and S (shake or secondary waves) from an underground nuclear test explosion.
    The speed of P and S waves in the Earths crust are 7800 ms^-1 and 4200 ms^-1, respectively.

    Calculate the distance of the test site from the explosion

    Completely stuck here... any advice?
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    Use a modified s = vt
    Tell me if you need more detail.
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    Speed P * (Time S - 68) = Speed S * Time S

    Find Time S with this equation, and then use distance = speed x time for S.
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    (Original post by morgan8002)
    Use a modified s = vt
    Tell me if you need more detail.
    we dont know the time for the individual waves, just the difference is 68..

    so v= s x (t-68) ?
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    (Original post by tjthedj)
    we dont know the time for the individual waves, just the difference is 68..

    so v= s x (t-68) ?
    As the post above said, you can create two s = vt equations and equate them.

    s = v_{p}(t)
    s = v_{s}(t + 68)
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    (Original post by tjthedj)
    we dont know the time for the individual waves, just the difference is 68..

    so v= s x (t-68) ?
    I just told you how to do it.

    The difference is 68 seconds

    So Time(1) = Time(2)-68


    They travel the same distance, so in the equation s=vt, we have vt=vt


    Speed (1) * Time(2)-68 = Speed (2) * Time (2)

    Solve for Time (2).
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    (Original post by Kolasinac138)
    I just told you how to do it.

    The difference is 68 seconds

    So Time(1) = Time(2)-68


    They travel the same distance, so in the equation s=vt, we have vt=vt


    Speed (1) * Time(2)-68 = Speed (2) * Time (2)

    Solve for Time (2).

    sorry completely missed this! i get it now, thanks so much
 
 
 
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