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Velocity and acceleration of a runner

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initially 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross?


    2. Relevant equations

    not sure...

    3. The attempt at a solution
    x(9.0)-6=x(8.0)-5
     
  2. jcsd
  3. Sep 30, 2007 #2
    Ok, the best way to have a swing at this, is to first of all, ignore the flagpole, and deal with that part of the problem afterwards, lets first find out when they cross paths, and from there we can figure out how far each has traveled and therefore find their relative distances from the flagpole.

    The runners are 6km "Left" and 5km "Right", therefore, the distance between the two is 11km, I'm not much of a fan of Km/h, so for calculations sake lets convert Km/h to M/s, this is only for me, doing the calculations in Hours and minutes is all fine, but I'm in love with metric.

    9Km/h = [tex] \frac{9*1000}{60^{2}} = 2.5ms^{-1} [/tex]
    8Km/h = [tex] \frac{8*1000}{60^{2}} = 2.2222ms^{-1}[/tex]

    Now we have some metrics, for my bennefit anyway, we could use relative velocities to find out the collision time, for runner A, runner B is approaching him at his speed + runner B's speed, this is "Relative velocity", and its the same for both runners, but we only need to do the math with runner A, so lets have runner B as a stationary object and runner A moving at Runner A's speed plus Runner B's speed, to make the maths ALOT easier.

    [tex]V_{a} + V_{b} = 2.5 + 2.222 = 4.722ms^{-1} [/tex]​

    Now, we have a distance, and a speed to work with, the rest becomes somewhat self explanitory:

    [tex]V_{speed} = \frac{Distance}{Time}[/tex]​

    So:

    [tex]Time = \frac{Distance}{V_{speed}}[/tex]​

    We find that the travel time before A meets B is 2329.5 seconds (seems like a massive number, but it is, after all, equal to ~39 minutes).

    Now, the analogy is, that if A and B where too meet, they would do so at the same time, A cannot collide with B and then B collides with A, its a mutual collision between the two runners, both must be travveling for ~39 minutes before they meet eachover in the run, it seems rather counter intuitive at first since the ~39 minutes is the time taken for A, moving at the relative velocity of A+B too meet B, who in this case, for maths, is stationary, but it does work out, since when you use the "Normal" velocities of both, rather then having one going "Super fast" and the other being "Stationary" you find that it does in fact take them 39 minutes to both collectively cover the 11km (A does, say 6 of those Kilometers, and B does 5)

    Now that you know the time that both of them have to run before they meet eachover, I hope it will be more obvious how to figure out the distance A or B has travveled before meeting the other, and from that distance, figure out how far the collision is from the pole.
     
  4. Sep 30, 2007 #3
    Thank you very much
     
  5. Sep 30, 2007 #4
    No problem, sorry I couldnt give you the full answer, but you'll probably appreciate working through the problem confidently from start to finish so you can impress your teacher/tutor if he asked how you figured it out :)

    EDIT: Also to check if your answer is definitely correct, when you find out the distances that both A and B travel, make sure they add together to create the total 11km distance at the time I mentioned earlier, that way you know for sure that 5*(the time) + 6*(the time) = The total distance between those two healthy guys and therefore it affirms that your maths is solid since youll be using the time that you figured out much like a variable that the question paper gave you or however you look at it.
     
    Last edited: Sep 30, 2007
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