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Runner and a bird heading towards finish line

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A runner is jogging at a steady velocity of 3.4 km/h. When the runner is 4.1km from the finish line, a bird begins flying from the runner to the finish line at velocity 10.2km/h. When the bird reaches the finish line it turns around and flies back to the runner. Assume the birds length equals zero... How far does the bird travel?

    3. The attempt at a solution

    i solved that it would take the bird .4019h to get to the finish line from that point...

    at that time the jogger is 2.733 km from the finish

    what are my next steps to calculate how far the bird goes to reach the runner... since the runner would be moving as the bird moves back towards it??
     
  2. jcsd
  3. Sep 13, 2007 #2
    since hte bird travels 3 times faster than the jogger and we know at the point the bird is at the finish line.... that the distance away is 2.733333 then can we divide that by 4 and then multiply by 3 to get the distance tehy are at when they intersect again? since if the bird travels 3 times faster then the distance would be 3 times further it travels at this point and would be 2.049975 of the possible 2.7333 distance after reaching the finish line

    therefore i add 2.049975 and 4.1 km to get my final answer for how far the bird traveled?
     
  4. Sep 13, 2007 #3
    so the distance the bird traveled was 6.149975 km

    now for part two of the problem PLEASE HELP

    After the first encounter the bird then turns and heads back to the finish line, turns around again and heads back to the runner. then repeats that until the runner reaches the finish line.... how far does the bird travel including the 6.149975 distance from the first part of the answer. answer in units of km

    please help as i am stuck on this one... is there a formula to use
     
  5. Sep 13, 2007 #4

    learningphysics

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    You've calculated that when the bird hits the finish line... the runner is 2.733 km away...

    Suppose t seconds beyond that point the runner and the bird meet... what is the distance the runner travelled in terms of t... what is the distance the bird ran in terms of t... what is the sum...


    EDIT: did you finish the first part? For the second part... they want the total distance from when the runner was at 4.1km... from that point... how long does it take the runner to hit the finish line?
     
    Last edited: Sep 13, 2007
  6. Sep 13, 2007 #5
    i have done that,... do i just keep repeating those steps over and over again until the runner is 0km away
    ???

    seems like alot of work to get that answer
     
  7. Sep 13, 2007 #6

    learningphysics

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    No... when from when the runner is 4.1km away... how long does it take the runner to hit the finnish line?
     
  8. Sep 13, 2007 #7
    if the runner is 4.1 km away.... it takes him 1.205882353 hours to cross the finish line....

    so if that is the total time it takes him to cross the line do i just multiply that by the velocity of the bird to get its total distance flown?
     
  9. Sep 13, 2007 #8

    learningphysics

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    yup.
     
  10. Sep 13, 2007 #9
    dude u are a genius at this stuff

    i have 2 more iam struggling with... should i post them in here or in a diff thread... mind helpin me out a lil?
     
  11. Sep 13, 2007 #10

    learningphysics

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    you can post here.
     
  12. Sep 13, 2007 #11
    the velocity of the transverse waves produced by an earthquake is 4.93 km/s while the longitudinal waves are 8.8247 km/s

    a seismograph records teh arrival of the transverse waves 51.4s after that of the longitudinal waves

    how far away was the earthquake

    i calculated the longitudinal waves travel 1.79 times faster than the transverse waves
     
  13. Sep 13, 2007 #12
    does knowing how many times faster one travels than the other help me out at all?
     
  14. Sep 13, 2007 #13

    learningphysics

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    just think of it as two runners... they leave the start position at the same time... one finishes the race 51.4 s after the other... use t as time taken by one wave... what's the time taken by the other wave... what's the distance in terms of t?
     
  15. Sep 13, 2007 #14
    im not sure how to do this one without knowing the distance... it is really jumbled in my head

    so runner 1 goes 8.8247 km/s

    runner 2 goes 4.93 km/s
    and 51.4s slower

    i feel like i need one other number to make the next step
     
  16. Sep 13, 2007 #15

    learningphysics

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    runner 1 takes t seconds to finish the race. What distance does it travel in terms of t? If runner 1 takes t seconds how many seconds does runner 2 take? What distance does does runner 2 travel in terms of t?
     
  17. Sep 13, 2007 #16
    ok so runner 1 goes 8.8247 km/s and he takes t seconds to finish therefore his distance is 8.8247/t

    runner 2 goes 4.93km/s and he takes t+51.4 seconds to finish nad his disctance is 4.93/t+51.4

    correct?

    so where do i go from there?
     
  18. Sep 13, 2007 #17

    learningphysics

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    you mean multiply... not divide... multiply the velocity and time... can you relate the two distances? then you can solve for t.
     
  19. Sep 13, 2007 #18
    what do mean by relate the 2 distances

    if runner 1 ----- d = 8.8247 * t

    if runner 2 ------ d = 4.93 * (t+51.4)

    how would i relate those 2 now???
     
  20. Sep 13, 2007 #19

    learningphysics

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    Both runners are running the same race. :wink:
     
  21. Sep 13, 2007 #20
    so 4.93 * ( t+51.4) = 8.8247 * t

    can i then deduce 4.93t + 253.402 = 8.8247t
     
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