91Ó°ÊÓ

1. The speed of the S wave is$v_s=4.\text{5km/s}$
2. The speed of the P wave is$v_p=4.\text{5km/s}$
3. The time delay is$t=3.0\text{min}=3.0\text{min}×60\text{s}=1\text{80s}$

By finding the equation for the time delayand using the given speed of waves, we can find the distance from the location of the earthquake to the seismograph.

Formulae are as follow:

1. The time t is, $t=\frac{d}{v}$.
2. The time delay $∆tis∆t=t_s-t_p$.

where, tis time, d is distance and vis velocity.

If d is the distance from the location of the earthquake to the seismograph and v_s is the speed of the S waves, then the time for this wave to reach the seismograph is,

$t_s=\frac{d}{v_s}$

Similarly, the time taken by waves to reach the seismograph will be,

$t_p=\frac{d}{v_p}$

Therefore, the time delay is,

$$\begin{gathered} ∆t=t_s-t_p \\ =d\left(\frac{1}{v_s}-\frac{1}{v_p}\right) \\ =d\left(\frac{v_p-v_s}{v_s{v}_p}\right) \end{gathered}$$

Therefore, the distance from the location of the earthquake to the seismograph is,

$$\begin{gathered} d=\frac{t×v_s{v}_p}{\left(v_p-v_s\right)} \\ =\frac{180\text{s}×4.5×8.0}{\left(8.0-4.5\right)} \\ =1.85×{10}^{\text{3}}\text{m} \end{gathered}$$

Hence,the distance from the location of the earthquake to the seismograph is

$1.85×{10}^{\text{3}}\text{km}$

Thus, the earthquake originated at a distance of from the seismograph.