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Wave race: Help me start off the earthquakes

  1. Mar 13, 2007 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    [​IMG]
    displacement= (Initial Velocity)(Time) + (0.5)(a)(time squared)????

    3. 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..
     
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2

    Doc Al

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    Staff: Mentor

    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.
     
  4. Mar 13, 2007 #3
    Sorry.. it's not clicking.. How do I find the time for the transverse wave to arrive?
     
  5. Mar 13, 2007 #4

    Doc Al

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    Staff: Mentor

    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?
     
  6. Mar 13, 2007 #5
    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: Mar 13, 2007
  7. Mar 13, 2007 #6

    Doc Al

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    Staff: Mentor

    Of course. :smile:
    Same thing.
     
  8. Mar 13, 2007 #7
    kk
    So how do I resolve 3.8 km/s = 72s = delta d?
     
    Last edited: Mar 13, 2007
  9. Mar 13, 2007 #8
    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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