Earthquakes are essentially sound waves—called seismic

waves—traveling through the earth. Because the earth is solid, it can support both longitudinal and transverse seismic waves. The speed of longitudinal waves, called P waves, is 8000 m/s. Transverse waves, called S waves, travel at a slower 4500 m/s. A seismograph records the two waves from a distant earthquake. If the S wave arrives 2.0 min after the P wave, how far away was the earthquake? You can assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.

The speed of longitudinal wave (P-wave) is, $v_p=8000m/s.$

The speed of transverse wave is (S-wave),$v_s=4500m/s.$

Consider the arrival time of longitudinal wave is, $t_p$and the arrival time of transverse wave is, $t_s.$

According to the question,

$$\begin{gathered} t_s=t_p+2.0min \\ {t}_s=\left(t_p+120s\right) \end{gathered}$$

We take, the earthquake is at a distance of $d$

So consider the P-wave,

role="math" localid="1648834069033" $$\begin{gathered} d=v_p{t}_p \\ d=\left(8000m/s\right)t_p...\left(1\right) \end{gathered}$$

For the S-wave,

role="math" localid="1648833989432" $$\begin{gathered} d=v_s{t}_s \\ d=\left(4500m/s\right)t_s \\ d=\left(4500m/s\right)\left(t_p+120\right) \\ d=\left(4500m/s\right)t_p+540000m...\left(2\right) \end{gathered}$$

Assuming the waves travel at straight lines, then we can write from the above equations,

role="math" localid="1648833831997" $$\begin{gathered} \left(8000m/s\right)t_p=\left(4500m/s\right)t_p+540000m \\ \left(8000-4500\right)m/s\times t_p=540000m \\ 3500m/s\times t_p=540000m \\ {t}_p=\frac{540000m}{3500m/s} \\ {t}_p=154.3s \end{gathered}$$

Substituting the value of $t_p$in equation $\left(1\right)$, we get,

$$\begin{gathered} d=\left(8000m/s\right)t_p \\ d=\left(8000m/s\right)\left(154.3s\right) \\ d=1.2\times {10}^{6}m \\ d=1200km \end{gathered}$$