Is it consistent if you are 24 (mi) away?
My teacher made this problem up and I'm having a hard time understanding what I'm supposed to do. We haven't worked with earthquake epicenters at all so I'm confused. The last thing we worked with was linear programming. Anyway here's the problem. Thanks for any help!
If you have ever experienced an earthquake, you realize that it causes a primary wave and also a secondary wave to be created. The primary wave (a logitudinal wave) travels faster, so you feel it first. Shortly afterward you will feel the secondary wave (a transverse wave), which we refer to as an aftershock. For our minor earthquake, the primary wave traveled at 6km/s and the secondary wave traveled at 3km/s. In the late afternoon we felt a small earthquake and 4 seconds later came the aftershock. How far away from the epicenter of the earthquake were we located?
Her hint to me was: Three equations, three unknowns.
Hello iDrum24 km is the correct answer. It's easy to check: the first wave would take 4 sec to travel that distance, and the second 8 sec, arriving 4 sec later.
But if you don't know that's the answer, here's how you could work it out. Suppose that the first wave had been travelling for sec before it hit. Then the second was travelling for sec. So using the formula
Distance = Speed x Timewe have:
The first wave has travelled a distance km in secEach wave travels the same distance. Therefore:
The second wave has travelled a distance km in sec
So each wave travels km.
Grandad
You understand, don't you, that you really don't need to know anything about "earthquakes" or "epicenters" except what is given in this problem. The only "technical" point you need to know is that "speed equals distance divided by time".
Let "t" be the time it takes the first wave to reach you, "d" the distance to the epicenter. Then d/t= 6. Since it you feel the second wave 4 seconds later, its time of travel is t+ 4 and distance is the same: d/(t+ 4)= 3. Two equations for the two unknowns d and t.