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Homework Help: Difficult Physics Question (challenge)

  1. Sep 25, 2014 #1
    • Homework-like problem originally posted in a non-homework forum

    Before anyone thinks this is a coursework question, it is not. It is a challenge problem, which I found online, and seems worth discussing.

    (Question) In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the anchorman for team B, can cover 100 m in 10.1 s. What is the largest lead the team B runner can have when the team A runner starts the final leg of the race, in order that the team A runner not lose the race?

    The answer is 3.0m, how?

    Calculus please (wherever applicable).

    First we have team A, which we can denote by [itex]A[/itex] <--- The anchorman
    Team B can be denoted by [itex]B[/itex] <-- The anchorman

    Since [itex]v_a(t) = 10.204 m/s[/itex]
    [itex]v_b(t) = 9.9009 m/s[/itex]

    [itex]x_a(t) = 10.204t[/itex]
    [itex]x_b(t) = 9.9009t[/itex]

    But [itex]x_a(t) > x_b(t)[/itex] for all real [itex]t[/itex] so, runner B cant overtake runner A?
  2. jcsd
  3. Sep 25, 2014 #2
    Hint: If they tie the race, where would runner B have to be when runner A starts the final lap? (you know how long it will be until runner A crosses the finish line)
  4. Sep 25, 2014 #3
    [itex] {\frac{100m}{9.8s}} = {\frac{100m + x(m)}{10.1s}} [/itex], so for B to tie A, it can have about a 3.06m lead, or about a 1.133second lead. Now (assuming I did this correctly) any lead B has greater then this will mean that B will overtake A, and any less of a lead for B will mean A will overtake B. You don't really need calculus for this. Think about it like this, since A is running faster, if they both started at the same time A will be the winner, but say the runners before the Anchor for A were a bit slow, or the previous runners for B were faster the Team A's, this means that if Team B was able to give their anchor a big enough lead, his speed is enough to allow him to win, but anchor B could be in a position where anchor A's speed will allow him to overtake anchor B.
  5. Sep 25, 2014 #4
    v3nture, your equation calculates the location of runner A when runner B is 100 m away from the finish line, assuming that they are going to tie. The problem asks for the opposite. If the racers are destined to tie, it asks for the location of runner B when runner A is 100 m away from the finish line.

    Also, just as a friendly heads up, you aren't supposed to give solutions to problems, just hints.
  6. Sep 25, 2014 #5


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    Well you can calculate the time difference between the 2 runners finishing the race. How much ground will runner A cover in that time?
  7. Sep 25, 2014 #6
    @v3nture and @BiGyElLoWhAt thanks to both.

    What does x(m) mean in @v3nture example?

    For @BiGyElLoWhAt, runner A can cover 100m in 9.8s, while runner b can cover 100m in 10.1s

    10.1 - 9.8 = 0.3s

    So what can we do now?
  8. Sep 25, 2014 #7


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    Well since these dudes are perfect runners, with no variance in their final leg run-time, there will always be .3 s between their run times, which means if runner B leaves the start line of the last leg (or whatever runners call it, I don't move, I'm a potatoe), if runner A want's to tie with runner B, he can't leave later than .3s later, right?

    I misstated in my previous post, (doh!) what's the distance between runner A and runner B if runner B leaves .3s before runner A?
  9. Sep 25, 2014 #8
    This is an incorrect approach, and can only give an approximation.

    This problem is simple. We know that runner A can travel 100m in 9.8 seconds. If runner A is 100 m away from the finish line, all we need to know is how far runner B can travel in 9.8 seconds. That distance tells us where runner B would have to be in order for them to tie the race.
  10. Sep 25, 2014 #9
    Hello @jz92wjaz, okay. Runner B's speed is,

    [itex]r = 100m/10.1s = 9.9009m/s[/itex]

    So Runner B can run [itex](9.9009)(9.8)m = 97.02 m[/itex] in 9..8 seconds exactly.

    So Runner A can run 100 meters in 9.8 seconds.

    So if they started running at the same time then Runner A will be [itex]100-97.02 = 2.98 m ahead[/itex] Always.

    So if Runner A starts running when Runner B is at 2.98m then they will be tied.

    But 2.98 =/= 3. The correct answer is 3 seconds.
  11. Sep 25, 2014 #10
    That is the correct approach :)

    I presume that you intended to say that the answer is 3.0 meters, not 3 seconds.

    Edit: FYI, there is a small rounding error at the 100ths place. It doesn't matter in this problem because we end up rounding to the tenth, but be sure to watch out for it in the future.
  12. Sep 25, 2014 #11


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    You won't get exactly 3 @Amad27
  13. Sep 25, 2014 #12


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    You mean the time difference between the two runners starting the final leg, no?
  14. Sep 25, 2014 #13


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    Ha, yea I did. Sorry about that.
  15. Sep 25, 2014 #14
    Well, I thought this was an easy problem not needing calculus and I took the same approach as you did Amad27. Runner A runs the final distance to finish line 0.3s faster than runner B, therefore, 100 m / 9.8s = 10.204081632653061224489795918367 m/s ; then, 0.3 s (difference to finish) * 10.204081632653061224489795918367 m/s = 3.0612244897959183673469387755102 m ; which is equal to the largest lead Team B can have and still have Team A win the race. Well, actually, that lead distance will result in a tie, therefore, Team B must have some lead that is < 3.0612244897959183673469387755102 m to allow Team A to win.
  16. Sep 25, 2014 #15


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    We set the problem up to be a tie. Actually (very technically speaking) the largest lead team B can have is what we found for the tie ##-\Delta x## where delta x is very small, and approaching (not reaching) 0. Or dx.
  17. Sep 25, 2014 #16
    bearnyc61, using your answer of 3.06 (rounded down, to reduce the lead of runner B), I calculate that Runner A loses.

    Distance for Runner A: 100m
    Distance for Runner B: 100m-3.06m = 96.94m
    Runner A speed: 100m/9.8s
    Runner B speed: 100m/10.1s
    Time to finish line for Runner A: 9.8 s
    Time to finish line for Runner B: Distance/Time = 96.94/(100/10.1) = 9.79094 s

    Edit: I take back what I said earlier about not being able to use the .3 second difference between the finish times for the racers. You can use that .3 second difference, as long as you multiply it by the correct runner's speed. It's probably the best way to solve it, though my suggestion also works.
    Last edited: Sep 25, 2014
  18. Sep 27, 2014 #17
    So why did I get =~2.98 meters then? With the other method?
  19. Sep 27, 2014 #18


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    Honestly, probably some rounding differences. Try keeping all of your calculations in fraction form for both methods and punching them out if you wish. Your calculator can only hold so many decimal places, and at some point it has to round.
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