Lesson Two runners on a circular track

Two runners on a circular track

Problem 1

        Andrew passes Bob every time, when he covers one circumference more distance than Bob.



Andrew runs     of the circular track length in one minute.

Bob    runs   of the circular track length in one minute.

Every minute, the distance between Andrew and Bob increases by   -   of the track length.

          //  assuming the distance is measured along the circumference in the same direction as they run.


     -  =  -  =  -  =  =   of the track length.


Hence, every 5 minutes Andrew covers 1 track's circumference length more than Bob.


Making one circumference length more is exactly the condition for Andrew to pass Bob.


Hence, the answer to the problem is 5 minutes.


Again, the answer to the problem is x= 5 minutes.

Problem 2

When they run in one direction, they will meet for the first time when the faster runner will cover 
the distance in one circumference longer that the slower runner.


So, let assume that "u" is the rate running of Albert and "v" is the rate of Bob,
and let assume that Albert is faster than Bob.


Then the criterion :first time meet after start" is this equation


    u*t - v*t = 600  meters,  or

    (u - v)*12 = 600,   which implies

    u - v = 600/12 = 50 meter per minute.



When they run in opposite direction, the criterion to meet is


    u*t + v*t = 600 meters,  or

    (u + v)*4 = 600,  which implies

     u + v = 600/4 = 150  meters per second.


Thus you have these two equations to find unknowns "u" and "v"

    u + v = 150     (1)

    u - v =  50     (2)


To solve the system, add the equations. You will get  

    2u = 150 + 50 = 200,  u = 200/2 = 100 m/sec.


Then from equation (1) you find  v = 150 - u = 150 - 100 = 50 m/sec.


Thus the rates "u" and "v" are just found.


Then you make the last step to get the answer:


    time for Albert = 600/100 = 6 minuter per lap,  and

    time for Bob    = 600/50 = 12 minuter per lap.

Problem 3

Let "u" be the Albert's rate running and "v" be the Bob's rate.


We are given that Albert catches with Bob second time 20 minutes after the first catch.


It means that during these 20 minutes Albert covered the distance which is one circumference longer than the distance covered by Bob during this time


    20u - 20v = 400,  or

      u - v = 20.


So, Albert's rate is 20 meters per second greater than Bob's rate.


It means that Bob's headstart was 6*20 = 120 meters ahead Albert.


ANSWER.  Head start distance x is 120 meters.

Problem 4

Let  L  be the circumference of the circle.

Then the faster cyclist will catch the slower cyclist first time  when the faster cyclist will cover 
the distance which exactly 1 circumference longer than the distance covered by the slower cyclist 

      9t - 5t = L.


It gives the time to get first meeting point

       t =  =   hours


and the distance which the faster cyclist covered during this time is

        = 9t =   miles.


The distance which the slower cyclist covered during this time is 

        = 5t = .


The meeting point is geometrically the same point on the circle for both cyclists, , 
and its angle measure on the circle is


         =  =  =   of the full angle of  radians, or 90 degrees.


So, they started simultaneously, and their first meeting point is at the 90 degrees angle.


Next, they start from this point  SIMULTANEOUSLY  and . . . and everything is repeated.


Hence, their next meeting point is the point on the circle with the angle of 180 degrees.



    It can be proved using strict mathematical arguments, but it should be ABSOLUTELY CLEAR to you without any arguments.



So, there are 4 remarkable points on the circle: first point is the starting point, and 3 other points 

(the points where whey meet/catch each other) are the images of the starting point, rotated 90°, 180°, and 270° along the circle.


ANSWER.  There are 4 points on the circle, where the faster cyclist catches the slower cyclist.

Problem 5

In this problem, the key to the solution is to find the way/(the method) to analyze, the way to organize thoughts and 
the way to present the results.


Each of the two runners runs cyclically.


The cycle by Jerry is to run 100 m in 20 seconds; then to rest 10 seconds.  So, his cycle is 30 seconds long.

The cycle by Kenny is to run 100 m in 25 seconds; then to rest 10 seconds.  So, his cycle is 35 seconds long.



    I will analyze their positions (as the time and the coordinate along the circumference, beginning from Kerry's starting point) 
    at the end of the each runner cycle.


        Notice and memorize that Jerry is faster.


For Jerry,  the time   = 30*(i-1)  and  the position   = -100 + 100*(i-1),  "i" is the number of a cycle.

For Kenny,  the time   = 35*(i-1)  and  the position   =        100*(i-1),  "i" is the number of a cycle.


Jerry first time will catch up with Kenny, when their cycles will have common ending time  AND  the positions will be the same
or will differ by a multiple of the length of the circumference.


Since 30 = 5*6  and 35 = 5*7, the first candidate to check is the 7-th cycle for Jerry and the 6-th cycle for Kenny.


And if you will calculate their time-space coordinates at the end of their corresponding circles,

    you will see that they are at the same time-space point at the time t= 210 seconds and the coordinate on the circle 600 meters,

    which is the same as 200 m away along the circle clockwise from the Kenny starting point.


Since t = 210 seconds is the end of the Jerry's 10-second resting period and the end of the Kenny's 10-second resting period,

    it means, that ACTUALLY Jerry will catch up Kenny first time at the time moment t= 210-10 = 200 second after start running.


ANSWER.  Jerry will catch Kenny in 200 seconds.