## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Let $t_p$ be the time that the P-wave took to reach the seismograph. Let $t_s$ be the time that the S-wave took to reach the seismograph. Note that $t_s = t_p + 120~s$ We can use the P-wave to write an expression for the distance. $d = v_p~t_p$ $d = (8000~m/s)~t_p$ We can use the S-wave to write an expression for the distance. $d = v_s~t_s$ $d = (4500~m/s)~(t_p+120~s)$ $d = (4500~m/s)~t_p+540,000~m$ We can equate the two expressions for the distance and solve for $t_p$: $(8000~m/s)~t_p = (4500~m/s)~t_p+540,000~m$ $(3500~m/s)~t_p = 540,000~m$ $t_p = \frac{540,000~m}{3500~m/s}$ $t_p = 154.3~s$ We can use the P-wave to find the distance. $d = (8000~m/s)~t_p$ $d = (8000~m/s)~(154.3~s)$ $d = 1.2\times 10^6~m = 1200~km$ The earthquake was 1200 km away.