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Kinematics runner problem

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data
    A runner hopes to complete the 10,000-m run in less than 30.0 min. After exactly 27.0 min, there are still 1100 m to go. The runner must then accelerate at 0.20 m/s/s for how many seconds in order to achieve the desired time?


    2. Relevant equations

    Kinematic equations.

    3. The attempt at a solution

    Well I figured out at his current pace, he would finish the 1100 remaining meters in 200 seconds. However, he needs to complete it in 180 seconds. I'm not sure how to find out how long he has to accelerate at 0.20 m/s/s to complete it in 180 seconds.

    Thanks.
     
  2. jcsd
  3. Sep 9, 2010 #2

    Redbelly98

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    The runner accelerates at 0.20 m/s2 for some time, then runs at constant speed for the remaining time.

    Those two times must add up to ____?

    Use the kinematic equations to figure out how far he travels during the accelerating portion, and also how far during the constant speed portion.

    Those two distances must add up to _____?.
     
  4. Sep 10, 2010 #3
    I only have one time and one distance to input into the equations. The time is 180 seconds and the distance is 1100 meters. If I put those in to find his final velocity, it'll tell me the velocity he'll be at if he accelerates at .2 m/s/s for the entire 180 seconds. I'm not sure how to do it without knowing how long he accelerates. I know it's possible and I can see how I should have enough information to get the right answer, but the intuition for it just isn't coming to me.

    Thanks for the response.
     
  5. Sep 10, 2010 #4

    Redbelly98

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    That is the tricky part. The key is to say he accelerates for t seconds, and proceed to write out the equations.

    So, how far does the runner go if he accelerates at 0.2 m/s2 for t seconds? (To answer this, you'll need the runner's initial velocity, at the moment he begins accelerating.)
     
  6. Sep 10, 2010 #5
    How do I incorporate him accelerating for t seconds into the other equations? Also, I seem to have two unknowns no matter what equation I put my known information in. I know his velocity when he first starts accelerating is 5.49 m/s. I know that he accelerates at .4 m/s/s. I know that the remaining distance is 1100 meters and the time he has left to make it to the finish is 180 seconds. But I can't figure out how to make use of that information, because it's not the distance or time he's accelerating, it's the total distance and total time left. Should I put my known information into one of the equations, get one variable alone on one side of the equation and then plug the entire other side of the equation into another equation in place of the variable?
     
  7. Sep 10, 2010 #6

    Redbelly98

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    Yes, there are two unknowns: the time t and the distance d over which he accelerates. Since there are two unknowns, you will need two equations.
    Correct, he does not accelerate for the entire 180 seconds, and he does not travel the entire 1100 meters while accelerating. He accelerates for an unknown time t, and for an unknown distance d.
    Yes. Can you write that first equation now? Just use "t" for the time, and use "d" for the distance. Which of the kinematic equations has distance and time, as well as the known quantities a=0.2m/s2 and vo=5.49m/s? Write that equation out.
     
  8. Sep 12, 2010 #7
    Well this is the best one I've come up with and it still doesn't make sense. d = 5.49t + .1t^2.
    To get that I had to use two different equations.
    If I plug that into another equation I get 5.49t + .1t^2 = 5.49t + .1t^2 and that doesn't help me.
    I've been doing physics problems in this book for a while. I had to think a little for each of them, but now all of a sudden I reach one that I'm mentally incapable of doing. The other ones took me a maximum of 20 minutes. I've been on this one for about a week now. Did I suddenly get brain damage? There's nothing more frustrating than this.
     
  9. Sep 12, 2010 #8

    Redbelly98

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    This one is a more advanced difficulty level, so don't be too disheartened! :smile:

    That equation is correct. I'm surprised you had to use two different equations, in most physics classes one of the kinematics equations is
    Δx = vo t + (1/2) a t2
    and I am quite surprised if that is not among the equations your class is working with.

    Be that as it may, the equation you wrote is correct and useful for proceeding with solving this...

    True. The tricky part of this problem is that we should now consider the motion after the runner has finished accelerating. After accelerating, he runs at a constant speed for the remainder of the race.

    So we need another equation that relates time and the distance traveled at constant speed, or a=0. As a hint, I will mention that this distance will be
    1100 m - d
    and similar reasoning applies to the time it takes to run that distance.
     
  10. Sep 13, 2010 #9
    Yes, I had that one but I kept ending up with a t term and a t squared term, so I tried the other ones but ended up with the same thing anyway.

    I plugged the equation I got into the equation above, but I end up getting a quadratic no matter what I do. I used the quadratic formula to get 63.8. I'm not too good with quadratics so I'm not sure if that is equal to the T or not.
    The problem with quadratics is there's so much room for error. I tried it again, doing it a different way in the same equation and got 200 instead of 63.8. I also got some negative answers, which I assume I can discard.
    But I don't know if I'm supposed to even be doing a quadratic. So does this require me to do a quadratic and if so, does the answer I get equal the time? And if so, is that time equal to how long he accelerates?
     
  11. Sep 14, 2010 #10

    Redbelly98

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    Yes, you will need to solve a quadratic equation to find t. That means you can get two solutions, but only one of them would be the actual answer.

    Not sure why you're getting different answers though.
     
  12. Sep 15, 2010 #11
    I solved the quadratic and got 63.48 seconds, but I can't figure out what that time tells me. It can't be how long he accelerates.
     
  13. Sep 15, 2010 #12

    Redbelly98

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    Yes, the t you are solving for in the quadratic is how long he accelerates. That is exactly how we defined t for this problem (in Post #4, "...he accelerates for t seconds...")

    Why do you think it means something else?
     
  14. Sep 15, 2010 #13
    I guess because it's such a high number. After I did it, I looked in the back of the book to find the right answer and it's 3.1 seconds. I got 63.48 seconds. I thought because it was such a huge difference, that I didn't actually find the duration of his acceleration. So I did get the wrong answer, right? If he really accelerated for that long, he'd be running like 40 miles per hour.
     
  15. Sep 15, 2010 #14
    Hello , sorry to interrupt but i am searching for kinematics problems and came across this thread and i am interested also in solving and i want to share what i have done.

    The runner has only 180 seconds left to complete the run with a distance of 1100 m. I used two formula here, 1 with acceleration and 1 with constant velocity. So for the first part of the solution, the runner must run say a distance d with an acceleration of 0.2 m/sec/sec for a time t. Then for the second part. The runner must run with constant velocity (his current rate which is 5.49 m/sec) for a distance of (1100 - d) meters and for a time of (180 - t) seconds.

    1st part i used the formula : d = Vo (t) + .5at^2
    Substituting the values : d = 5.49t + (.5)(0.2)t^2

    2nd part i used the constant velocity formula d = v(t)
    substituting the values : 1100 - d = 5.49 (180 - t)
    I solved for d in this 2 nd part

    and then equate the two equations d = d

    i got t = 33.44 sec , this is the time needed for hte runner to accelerate at 0.2 m/sec/sec and then the remaining time of 180 - 33.44 sec = 146.56 sec he can then run at the constant rate of 5.49 sec.

    I check my answers by getting the distance travelled and the two distances sum up to 1100 m .

    I wasn't able to use the quadratic formula since 5.49t cancels out when i equate d = d.

    Was my solution correct?
     
  16. Sep 15, 2010 #15
    what level is this type of question o or A level
     
  17. Sep 16, 2010 #16

    Redbelly98

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    Good point. Can you show your work? Otherwise I can't see where things went wrong.
    5.49 m/s was the runner's initial speed at the beginning of the acceleration.
     
  18. Sep 16, 2010 #17
    Hi, thanks for your input. I don't think you got the correct answer, either. If he accelerated for 33.44 seconds, he would be running about 27 miles per hour. That seems a little too fast for a human.
    I got 63.48 seconds, which is way too fast.
    After I got that value, I got all excited that I finally answered this question I've been working on for a couple of weeks. I looked at the correct answer in the back of the book and it's 3.1 seconds.
    Sure.
    I got the value of 5.49t + .1t^2 as the initial distance, which I assume is the distance he runs during his acceleration period. Then I put that back into the same equation, but this time with a constant acceleration and a value of 1100 for the final distance.
    That gave me a quadratic of .1t^2 + 10.98t - 1100 = 0. I put that into the quadratic formula and that gave me 63.48 seconds as the value of t.
     
  19. Sep 16, 2010 #18

    Redbelly98

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    Yes, good.
    I don't understand this statement. Are you now saying he runs with a constant acceleration for the full 1100 m? But you should be thinking in terms of a constant speed, for the remaining distance after he has finished accelerating.
    It sounds like you haven't written a proper equation that describes the constant speed at the final portion of the run.
     
  20. Sep 16, 2010 #19
    Sorry, I meant constant velocity. I think I was debating in my mind whether to type "constant velocity" or "zero acceleration" and combined them subconsciously.

    Does that change anything, or is my equation still messed up?
     
  21. Sep 16, 2010 #20

    Redbelly98

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    Okay.
    Since I had also gotten the book's 3.1s answer some days ago, it looks like your equation is incorrect. Perhaps you could show the steps of your work in between
    d = 5.49t + .1t^2​
    and
    .1t^2 + 10.98t - 1100 = 0​
    Of particular interest is the equation you wrote to express the constant speed part of the run.
     
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