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Displacement, Time, and Average Velocity

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Two runners start simultaneously from the same point on a circular 200-m track and run in the opposite direction. One runs at a constant speed of 6.20 m/s, and the other runs at a constant speed of 5.50 m/s. when they meet, (a) for how long a time will they have been running, and (b) how far will each one have run along the track?

    2. Relevant equations

    3. The attempt at a solution

    I could picture exactly what they our describing but I can't get the solution.
    (a)When they first meet how long a time will each one have run along the track
    (b)How far will each one have run along the track

    *I graphed this and found the answer but I want to know if anyone could find a faster way to solving this?
    (a) 17 seconds
    (b) 106m , 96m

    1. The problem statement, all variables and given/known data
    -2 runners
    -start = 0, finish = 200m
    *track is circular.
    *One runner is going in a negative direction
    runner 1 average velocity = 6.20m/s
    runner 2 average velocity = 5.50m/s

    2. Relevant equations

    Average Velocity = x2 - x1 / t2 - t1
    where x1 is the initial position, x2 is the ending position and t1 is initial time, t2 is ending time.

    3. The attempt at a solution
  2. jcsd
  3. Oct 12, 2008 #2
    Hmm it seems to me that you've done something wrong. I assume they both start off at the same place. You can see, just by thinking about it that for the faster runner to be equal with the slower runner, the faster runner is going to have to complete at least 1 full lap. This is because for the entirety of his first lap he will be in the lead, as soon as he starts his second lap, he will be behind the slower runner (however the slower runner will be on his first lap).

    Therefore your answer of 106m and 96m is implausible.

    What I would do is calculate the relative velocity of the faster runner, then see how long it takes him to run a full lap.
  4. Oct 12, 2008 #3
    Hi. I've done a question like this before and I think the method I used would be more convenient for you. Here it is;

    Both of them can cover (6.2 + 5.5)m in 1s. Then divide the total distance along the track (200m) by this distance (6.2 +5.5)m and you should be able to find the time taken before they meet. Using the time, you should be able to find the distance travelled by each runner.
  5. Oct 12, 2008 #4
    Okay, completely ignore my first post. I clearly didn't read that they were going in opposite directions haha! Yeah your first method was correct, using the graphs.
    DarylMBCP's method is also correct, and a shorter way of doing it.

    It is the exact same way that I posted mine, except because they are travelling in opposite directions he added their velocities instead of dividing.

    It works because you are working out the relative velocity of one to the other, such that this velocity is the speed one is travelling towards the other, in the frame of reference such that the other appears to be standing still.

    like two cars hitting head on at 50km/h each is equivelant to one at 100km/h hitting a stationary car.
  6. Oct 12, 2008 #5


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    Homework Helper

    Rake-MC you simply misread the question. Both runners start running in opposite directions :tongue:

    A way of calculating without actually graphing (which is much quicker of course) is to use the formula:
    [tex]s=vt[/tex] where s=displacement, v=velocity, t=time.

    For runner A (5.5ms-1 runner) ~ [tex]s=5.5(t)[/tex]

    For runner B (6.2ms-1 runner) ~ [tex]200-s=6.2(t)[/tex] (notice how the displacement is 200 minus runner A's displacement, since they will meet somewhere on the track)

    From here, simply by solving the simultaneous equations to find t (by making s the subject in each, or otherwise) and then finding the displacement (s) by substituting the time (t) back into the formula.
  7. Oct 12, 2008 #6
    Yes, you are right. That way is also correct so there are many ways of solving it. It just depends on which method you prefer using.
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