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Homework Help: Runner in circular track

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    A runner travels 1,5 laps around a circular trackin a time of 50s.
    The diameter of the track is 40m and its circumference is 126m.

    2. Relevant equations
    a) Find the average speed of the runner
    b) Find the magnitude of the runner's average velocity.

    3. The attempt at a solution
    d = r*t
    r = d/t = (126*1.5)/50 = 189/50 = 3.78 m/s
  2. jcsd
  3. Sep 25, 2011 #2


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    Yes. Part b?
  4. Sep 25, 2011 #3


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    Fine, you found average speed correctly using (total distance) / (time)

    For part b - average velocity, you want displacement over time

    SO what is that?
  5. Sep 25, 2011 #4
    im still thinking about it, i think i need some law
  6. Sep 25, 2011 #5


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    Not a law - some thought.

    If the runner had completed only 1 lap, the displacement, and average velocity, would have been zero, since after one lap, you are back where you started.

    This runner at least do 1.5 laps, so is on the other side of the circle. Displacement represents the straight-line distance from start point to finish point. [there is a name for that distance in this particular example - starts with d]
  7. Sep 26, 2011 #6
    well, that displacement is 189m ?
  8. Sep 26, 2011 #7


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    Imagine this.
    Before the runner starts, he drives a peg into the ground at his start point, and attaches a piece of rope - 200m long to allow for every eventuality.
    He ties the other end to his waist, then runs 1.5 laps of the circular track - with no obstacle in the in-field to snag the rope.
    He then stops and starts dragging in the rope until it goes tight.

    How long will the tight piece of rope from where he finished, to where he started, be?
  9. Sep 27, 2011 #8
    63m ?
  10. Sep 27, 2011 #9


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    if you run 1.5 laps, you are on the other side of the circle when you finish.

    The diameter of this circle is 40 m .

    The piece of tight rope will be less than 63m.
  11. Oct 5, 2011 #10
    I don't get it :(
  12. Oct 5, 2011 #11


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    Displacement is shortest distance between starting point and finishing point.

    Displacement pays no attention to the path followed; the path only affects distance covered.

    My car has covered 126 000 km since it was new, 15 years ago. The displacement in that time is about 5km. I live about 5km [as the crow flies] from the dealer.

    If your computer is at home right now, there is probably a wall behind it. Imagine you were to get up and walk to the other side of that wall.
    Your displacement would be about 1 or 2 metres [straight through the wall] The distance you covered to get there will be quite a few metres.

    If you have someone stand at the starting position, and someone stand at the finish position, the shortest distance between them is a diameter of the circle - the finish is straight across the circle from where the start is.

    Does one of these examples help you out?
  13. Oct 5, 2011 #12
    wow, so the displacement is 40m ?
    and therefore the average velocity= 40/50 = 0.8 m/s ?
  14. Oct 5, 2011 #13


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    AT LAST!! YES !!!!!!!

    If you were given details of where the start was - for example on the East side of the track - you would even know the displacement, and thus average velocity, had the direction West.

    That is why the question only asked for the magnitude of the velocity - it takes the direction out of it, so they don't have to specify where the runner started.
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