Question

An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?

Answer 1

Answer:

The S wave arrives 6 sec after the P wave.

Explanation:

Given that,

Distance of P = 45 km

Speed of p = 5000 m/s

Speed of S = 3000 m/s

We need to calculate the time by the P wave

Using formula of time

$t = \dfrac{D}{v}$

Where, D = distance

v = speed

t = time

Put the value in to the formula

$t_{p} =\dfrac{45\times1000}{5000}$

$t_{p} = 9\ sec$

Now, time for s wave

$t_{s}=\dfrac{45000}{3000}$

$t =15\ sec$

The required time is

$\Delta t=t_{s}-t_{p}$

$\Delta t=15-9$

$\Delta t =6\ sec$

Hence, The S wave arrives 6 sec after the P wave.