# "Easy" Physics Problem

Discussion in 'Community Discussion' started by ipodtouchy333, Sep 5, 2009.

1. ### ipodtouchy333 macrumors 65816

Joined:
Nov 15, 2007
Location:
US
#1
I am really confused on how to set this physics problem up. Any help would be appreciated! Thanks!

The velocity of the transverse waves produced
by an earthquake is 6.76 km/s, while that of
the longitudinal waves is 12.3708 km/s. A
seismograph records the arrival of the trans-
verse waves 53.7 s after that of the longitudi-
nal waves.
How far away was the earthquake?

2. ### mkrishnan Moderator emeritus

Joined:
Jan 9, 2004
Location:
Grand Rapids, MI, USA
#2
Unless I'm missing something obvious, if you use the x = vt formula....

x = v1 * t (for the fast wave)
x = v2 * (t+53.7) (for the slow wave)

Now x and t are the same in the two equations. Two equations, two unknowns, get to work.

Joined:
Dec 5, 2007
4. ### CalBoy macrumors 604

Joined:
May 21, 2007
#4
You really should be a good physics student and use the proper formula, so don't follow my example.

I honestly couldn't remember what formula to use so I improvised a tad.

The first thing you need to do is figure out the difference in velocity between the two waves (12.3708-6.76=5.6108 km/s).

Next, figure out how far the second wave had to travel during those 53.7 seconds. Since we know that (53.7*6.76) is 363.012 km, we can then figure out the total time that Wave 1 had to travel by using the difference in velocity of the waves.

363.012/5.6108=64.698795 seconds. This represents the total time Wave 1 must have had to travel in order to leave Wave 2 53.7 seconds behind it. With a little multiplication (64.698795*12.3708) we see that Wave 1 traveled a total of 800.376 km before passing the station.

You can check your work by adding 53.7 seconds and then multiplying by 6.76, or by dividing 800.376 km by 6.76 and subtracting 53.7.

But like I said, you really should use a proper formula if this is for an assignment.