Wave race: Help me start off the earthquakes

  • Thread starter pugfug90
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  • #1
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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 [Broken]
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 [Broken]
https://www.physicsforums.com/showthread.php?t=88686
http://www.hopperinstitute.com/phys_waves.html [Broken]

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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Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
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Sorry.. it's not clicking.. How do I find the time for the transverse wave to arrive?
 
  • #4
Doc Al
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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?
 
  • #5
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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..
 
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  • #6
Doc Al
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T + 72s?
Of course. :smile:
PS, distance = speed*time is (at least in this case) the same as displacement=velocity*time ?
Same thing.
 
  • #7
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kk
So how do I resolve 3.8 km/s = 72s = delta d?
 
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  • #8
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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, wanna 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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