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A runner

  1. Nov 9, 2004 #1
    A runner hopes to complete the 10,000 m run in exactly 30 minutes. After exactly 27 minutes, there are still 1100 meters to go. The runner must accelerate at .20 m/s^2 for how many seconds in order to achieve the desired time? Assume the runner ran with constant velocity for the first 27 minutes.

    Here's what I did:

    8900 m/27 minutes

    8900 m = .5(V0 + Vf)1620 s
    Vo + Vf = 10.9 m/s
    Vf= 5.5 m/s

    1100 m = 5.5t + .5(.2 m/s^2)t^2
    t= 80.9 seconds

    Is this right?
     
  2. jcsd
  3. Nov 9, 2004 #2

    Galileo

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    Looks good to me...
     
  4. Nov 9, 2004 #3

    BobG

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    No, that's not right. You calculated how long it would take him to run the remaining 1100 meters if he accelerated .2 m/sec^2 for the entire 1100 meters. (That would be a pretty impressive final kick - he'd be running over 48 mph by time he hit the finish line).

    He has to complete the final 1100 meters in 180 seconds. He's currently running too slow to do that. He has to accelerate from 5.5 m/s to a higher speed, but every second it takes for him to accelerate increases the final speed he has to achieve in order to complete the 1100 meters within 180 seconds.

    He has to average a little over 6.1 m/s for the final 1100 meters. But the 3 seconds it takes to accelerate from 5.5 m/s to 6.1 m/s means he still falls short (now he has to average about 6.2 m/s over the final 1083 meters).

    In other words, your problem's a little more complicated.
     
  5. Nov 9, 2004 #4
    riiiighhttt...I understand why its wrong. But, now, I have NO CLUE how to fix it. This is due tomorrow, please help me ASAP.
     
  6. Nov 9, 2004 #5

    BobG

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    Your position will change due to accleration for awhile, then it changes due to a constant velocity for awhile. Both need to be taken into account.

    So, how about this:

    [tex]s_f=s_i+v_it+\frac{1}{2}at^2+v_f(180-t)[/tex]

    final position is 10000 m
    initial position is 8900 m (you have 1100 m to go)
    initial velocity is 5.5 m/s
    acceleration is .2 m/s^2
    t is the time that you're accelerating (you have 180 sec total to reach your destination)

    The big problem is that you don't know your final velocity. But you do know how to get it. It's equal to your initial velocity plus (acceleration times time).

    [tex]v_f=v_i+at[/tex]

    That way, you get your equation down to one unknown: time. Pretty ugly looking by time you're done. You must be going over related rates and, ideally, they would have a method that didn't look quite so ugly. Maybe. Some problems are just pretty ugly and you have to take a look at them from a few different angles, kind of like lining up a putt.
     
  7. Nov 9, 2004 #6
    Is the anser 3.08 seconds?

    I added all the areas and set it equal to 1100

    .1t^2 -36t + 110 = 0
     
  8. Nov 9, 2004 #7

    BobG

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    Yes, that's the right answer.
     
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