An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seismograph at some distance from the earthquake records the arrival of the S waves 2.0 min after the arrival of the P waves. How far away was the earthquake? You can assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.

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adelekeademolu adelekeademolu · Answer 1 · 2020-02-11T02:53:42+01:00

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be [tex]d[/tex], the time taken by P waves be [tex]t_P[/tex] and the time taken by the S waves be [tex]t_S[/tex].

[tex]\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}[/tex]

[tex]\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}[/tex]

For the P waves,

[tex]t_P=\dfrac{d}{8000}[/tex]

[tex]d=8000t_P[/tex]

For the S waves,

[tex]t_S=\dfrac{d}{4500}[/tex]

[tex]d=4500t_S[/tex]

Equating the [tex]d[/tex],

[tex]8000t_P=4500t_S[/tex]

Divide both sides of the equation by 500 to reduce the terms.

[tex]16t_P=9t_S[/tex]

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

[tex]t_S-t_P=120[/tex]

[tex]t_S=120+t_P[/tex]

Substitute this in the equation of the distance.

[tex]16t_P=9(t_P+120)[/tex]

[tex]16t_P=9t_P+1080[/tex]

[tex]7t_P=1080[/tex]

[tex]t_P=\dfrac{1080}{7}[/tex]

Substitute this in the equation for [tex]d[/tex] involving [tex]t_P[/tex].

[tex]d=8000t_P[/tex]

[tex]d=8000\times\dfrac{1080}{7}[/tex]

[tex]d=1234285.7 \text{ m }= 1234.3 \text{ km}[/tex]