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Whenever a particle moves with a specific speed in a particular direction, then the particle would cover the distance that varies with the values of time and speed. The relation between distance and time taken by the particle is linear.

The given data can be listed below as,

• The typical speed of\(P\)wave is\({v_1} = \left( {8.5\;{\rm{km/s}}\; \times \frac{{{{10}^3}\;{\rm{m/s}}}}{{1\;{\rm{km/s}}}}} \right) = 8.5 \times {10^3}\;{\rm{m/s}}\).
• The typical speed of\(S\)wave is\({v_2} = \left( {5.5\;{\rm{km/s}}\; \times \frac{{{{10}^3}\;{\rm{m/s}}}}{{1\;{\rm{km/s}}}}} \right) = 5.5 \times {10^3}\;{\rm{m/s}}\).
• The time lag between the arrival of the two waves is \(t = 1.5\;\min \).

(a) Since, the distance traveled by the both waves is same, then the expression of the distance traveled by the both waves can be represented as,

\(\begin{aligned}{c}\Delta {x_1} = \Delta {x_2}\\{v_1}{t_1} = {v_2}{t_2}\end{aligned}\)

Here,\({t_1}\)and\({t_2}\)are the time taken by both waves\(\left( P \right)\)and\(\left( S \right)\)respectively.

As the time taken by the\(S\)wave will be more when compared to that of the\(P\)wave as the speed of the\(S\)wave is less when compared to that of the\(P\)wave then the time lag between the arrival of the two waves can be represented as,

\(\begin{aligned}{c}{t_2} - {t_1} &= \left( {1.5\;\min \; \times \frac{{60.0\;\sec }}{{1\;\min }}} \right)\\{t_2} - {t_1} &= 90.0\;\sec \\{t_2} &= \left( {{t_1} + 90.0\;\sec } \right)\end{aligned}\)

Substitute all the known values in the above expression of distance travelled by both waves.

\(\begin{aligned}{c}\left( {8.5 \times {{10}^3}\;{\rm{m/s}}} \right){t_1} &= \left( {5.5 \times {{10}^3}\;{\rm{m/s}}} \right)\left( {90.0\;{\rm{s}} + {t_1}} \right)\\8.5{t_1} &= 495 + 5.5{t_1}\\3{t_1} &= 495\;{\rm{s}}\\{t_1} &= 165\;{\rm{s}}\end{aligned}\)

The value of \({t_2}\) can be calculated with the help of known values as,

\(\begin{aligned}{c}{t_2} &= \left( {{t_1} + 90\;{\rm{s}}} \right)\\ &= \left( {165\;{\rm{s}}\;{\rm{ + }}\;{\rm{90}}\;{\rm{s}}} \right)\\ &= 255\;{\rm{s}}\end{aligned}\)

The expression of the distance travelled by the waves can be calculated as,

\(\begin{aligned}{c}\Delta {x_1} &= {v_1}{t_1}\\ &= \left( {8.5 \times {{10}^3}\;{\rm{m}}/{\rm{s}}} \right)(165\;{\rm{s}})\\ &\approx \left( {1.40 \times {{10}^6}\;{\rm{m}}} \right)\left( {\frac{{1.0\;{\rm{km}}}}{{{{10}^3}\;{\rm{m}}}}} \right)\\ &\approx 1400\;{\rm{km}}\end{aligned}\)

Thus, the distance travelled by the both the waves is \(1400\;{\rm{km}}\).

(b) No, one seismic station is not enough in order to find the epicenter. Only distance can be known through one seismic station, and one cannot identify the direction of an earthquake. So, data from two to three other stations is necessary for the correct position of the epicenter.