Wave race: Help me start off the earthquakes

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SUMMARY

The discussion centers on calculating the distance to an earthquake based on the velocities of transverse and longitudinal waves, specifically 8.9 km/s and 5.1 km/s, respectively. A seismograph records the transverse waves arriving 73 seconds before the longitudinal waves. The final calculated distance to the earthquake is approximately 870 km, derived from the relationship between wave speed, time, and distance. The solution involves setting up equations for both wave types and solving for time and distance using the provided velocities.

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Homework Statement


The velocity of the transverse waves produced by an earthquake is 8.9km/s while that of the longitudinal waves is 5.1km/s. A seismograph records the arrival of the transverse waves 73s before that of the longitudinal waves. How far away was the earthquakes?

Homework Equations


http://tpub.com/content/neets/14191/img/14191_46_1.jpg
displacement= (Initial Velocity)(Time) + (0.5)(a)(time squared)?

The Attempt at a Solution


Given Answer: 8.7x10^2 km=870km=870,000m
I wish I showed you guys the 12/13 problems out of this worksheet that I did to show that I did try and am not trying to mooch..

I scoured the internet and found some similar links..
http://www.glenbrook.k12.il.us/gbssci/phys/p163/ec/u10ec.html
https://www.physicsforums.com/showthread.php?t=88686
http://www.hopperinstitute.com/phys_waves.html

My last link basically showed.. they multiplied the fastest velocity by 98s.. How did they get 98s..I don't know..

Can anyone get me started on this?
==
I've tried..
3.8km (difference in velocity displacement)=73s*x
x=52m/2 .. Then try and use displacement formula.. But I have no acceleration value, so meh..
 
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The waves move at constant speed--no need for accelerated motion, just distance = speed*time.

Try this: Call the distance D; call the time for the transverse wave to arrive T. Now apply the above speed equation for each wave and solve for D.
 
Sorry.. it's not clicking.. How do I find the time for the transverse wave to arrive?
 
You don't need the time, just call it T. Write two equations: one for the transverse wave; one for the longitudinal. You'll be able to eliminate T and solve for D.

Hint: If the time for the transverse wave is T, what's the time for the longitudinal wave?
 
T + 72s? :-)
===
PS, distance = speed*time is (at least in this case) the same as displacement=velocity*time ?
We haven't done much of speed at all.. So I'm more comfortable with delta d:D
===
I've done 5.1 (km/s) + T=delta d=8.9 (km/s) + T + 72s
=5.1 (km/s)=delta d=8.9 (km/s) + 72s..

That can give me 3.8 km/s=72s..
 
Last edited:
pugfug90 said:
T + 72s?
Of course. :smile:
PS, distance = speed*time is (at least in this case) the same as displacement=velocity*time ?
Same thing.
 
kk
So how do I resolve 3.8 km/s = 72s = delta d?
 
Last edited:
I think I did my math wrong.. Ignore post#5..
Hey I think I got it! But this problem is so convoluted I'm going to forget it all meh rawr. PS, want to check my method? (I know the answer)
==
D=(8.9km*T)/s = [(5.1km)(T+72s)]/s
They both have same denominator so that cancels..
8.9km*T=5.1km*T + 367.2km*s
3.8km*T=367.2km*s
Isolate T..
T=(367.2km*s)/(3.8km)=~96.6s
That means it takes the transverse (faster), 96.6s to get there, and ~170s for the other..
I can use each's respective time and velocity to find distance.. Thanks Doc Al! I think I got it.
 

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