Problems on Circular Motion - Time, Speed & Distance


  • QA/DILR Mentor | Be Legend


    Authored by Nitin Gupta, Founder, Director at AlphaNumeric.

    These types of problems deal with athletes running on a circular track and questions being asked about when, where and how often would two or more athletes cross each other.

    Meeting for the First Time :

    1. When running in opposite directions:
      Consider two friends, A and B, who are separated by 1500 meters and they move towards each other at speeds of 40 m/s and 10 m/s. After how much time would they meet?
      This is a straight question on relative speed and the time taken to meet = 1500/(40+10) = 30 sec

    2. When running in the same direction:
      Consider a police traveling at 40 m/s chasing a thief traveling at 10 m/s. If the distance between the police and thief when the chase starts is 1500 meters, find the time after which the police catches up with the thief.
      Again, the solution should be known by now as 1500/(40-10) = 50sec

    Time taken to meet = Track Length /Relative Speed

    Frequency of meeting :

    Say two athletes start running from a same point on a circular track and meet after t units of time. When they meet they are again together at a point (not necessarily the same point where they started but nevertheless they are together). And if they continue running in the direction they were running and at their respective speeds, then this instance can be thought of as a new beginning, two athlete starting to run from a same point. And thus, from this instant onwards, they would again meet after t units of time. And the reasoning
    could be again argued similarly at their 2nd meeting. So considering the 2nd meeting as a fresh start, they will again meet after t duration of time. Thus, the time after which they meet for the first time is also the frequency with which they keep meet if they continue running at their respective speeds and respective directions. If the first meeting takes place after t duration after start, the nth meeting will take place after n × t duration after start.

    Where would the first meeting take place?

    Once the time is known after which two athletes, running on a circular track, meet, we can find the distance run individually by them (rather by any one of them) and find the exact position on the track where the meeting takes place. Rather than measuring this distance in absolute terms, it is more beneficial to measure this distance in terms of the track length.

    Meeting for the First Time at Start :

    The question of identifying after how much time would two or more athletes meet at the starting point is an application of LCM rather than a problem of Time Speed Distance. The key word here is meeting at the starting point. Please realize that this may not be the first time that they meet. The could have crossed each other (met) at some other point on the track but then that would not be counted as a meeting point for this question as it has not occurred at the starting point.

    Consider two athletes, A and B running on a circular track and taking x and y units of time to complete one full circle. A would reach the starting point for the first time after x units and thereafter would be at the starting point after every x units. Similarly B would reach the starting point for the first time after y units and thereafter would be at the starting point after every y units. Thus, A and B would both be at the starting time after common multiple of x and y and the first time that this would occur would be the LCM of x and y.

    For these types of problems, it does not matter in which direction the two athletes are running. Even if both are running clockwise or if one is running clockwise and other anticlockwise, the time when they would be at the starting point would remain the same.

    Relative Distances or Rounds run :

    1. When running in opposite directions : they meet whenever together they have covered one full round and hence at the nth meeting, together they would have covered n rounds. This is easy to understand. Since both start from common point and run in opposite directions, at 1st meeting, if one runs f fraction of the track, the other has to run (1 – f ) fraction of the track so that he is at same point. Thus, sum of distances run = f + (1 – f ) i.e. 1 full round. This is new beginning, and from now onwards, till 2nd meeting they would together again cover 1 full round i.e. they would together cover 2 rounds since start. And so on.

    2. When running in same directions : They meet whenever the faster one has covered 1 round more than the slower one and hence at the nth meeting, the faster one would have covered n rounds more than the slower one. Consider both start from same point and run in same direction. Focus on the gap between them. The faster one will race ahead of the slower one and a gap will start emerging between them. As time passes the gap will start increasing

    Three or more people meeting :

    To find time when three or more people meet, find the time after which pairs of athletes meet, such that at least one athlete is common to all the pairs (say, A & B, A & C, A & D, ……). The respective pairs will keep meeting after any multiple of the time found. At the LCM of these durations of time, we are sure that A & B have met, that A and C have met, that A & D have met and so on. But the only way that A could have met all these people at the same instant is when all of them are together.

    Position of the meeting points :

    These types of questions do not deal with when or how often do the athletes meet when they are running on a circular track. These questions pertain to the number of points and their placement on the circular track where the athletes can possibly meet.

    1. When running in the opposite directions : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be a + b

    2. When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be |a – b|

    Number of the meeting points ( for 3 or more people) :

    Find the hcf of meeting points of all possible pair.

    Two men , Jain & Sood, walk round a circle 1200 metres in circumference. Jain walks at the rate 150 meters/minute , and sood walks @80 minutes per minute. if both start at the same time from the same point & walk in the same direction.
    (a) when will they 1st be together again at starting point.
    (b) when will they be together again .
    (c) no. of distinct meeting points.
    (d) find the distance travelled & time required by Sood & Jain when they meet for 100th time.
    (e) when they meet for 100th time what is the distance from the starting point to the point at which they will meet (in anticlock wise direction)

    a) lcm (time taken by jain to comlete 1 round, time taken by sood to complete 1 round)
    = lcm (1200/150,1200/80) = 120 min.
    c) since ratio of speed is 15 : 8 ===> no. of distinct point = 15 - 8 = 7
    b) from a & c we can say that they are meeting 7 times in 120 min ==> 1st meeting will take after 120/7 min
    d) ratio of speed = ratio of no. of rounds ===> no. of rounds covered by jain to sood = 15:8 ==> here there is a gap of 7 but for first meeting there must be a gap of 1 round so distance covered by jain & sood when they meet for the first time = 15/7 & 8/7( in terms of rounds) so at 100th meeting distance covered is 15/7 * 100 & 8/7 * 100 ( in terms of round 1 round =1200m) & time required = 120/7 * 100.( avoid using relative speed work on ratio)
    (e) since there are 7 distinct points : 100 = 7k + 2; so 100th meeting point will coincide with 2nd meeting point & when they meet for 2nd time no. of rounds covered covered by sood = 8/7 * 2 = 16/7 = 2 + 2/7 ==> 2/7 clockwise or anticlockwise depending upon whether they started clockwise or anticlockwise respectively. So this cannot be determined.

    Consider athletes A , B & C running at speeds of 150 m/s , 70 m/s &110 m/s respectively on a circular track of 1000 meters, A & B running clockwise and C anti-clockwise. If they keep running indefinitely, at how many distinct point on the circle would they meet? what is the distance covered by each when they meet for 100th time.

    a : b = 15 : 7 so A & B will meet @ 8 points (as direction is same).
    b : c = 7 : 11 so B & C will meet at 18 points (as direction is opposite)
    so no. of distinct meeting points for all 3 = hcf(8,18) = 2

    In a circular track, there are two points P and Q which are diametrically opposite.C starts running clockwise with a speed of 4m/s from P. At the same time, from Q, A starts running clockwise with a speed of 3 m/s and B starts running anti clockwise with a speed of 5 m/s. If the length of the track is 300m, then after how much time will C be equidistant from A and B for the first time.

    B and C meets after 150/(4 + 5) = 50/3 seconds.
    At this point distance between A and C is 150 + 50 - 200/3 = 400/3
    So, suppose that from this moment on wards it take them t time to reach the situation when C is equidistant from A and B
    After time t, distance between A and C will be 400/3 + 3t - 4t = 400/3 - t and distance between B and C will be 9t
    9t = 400/3 - t
    => t = 40/3
    So, after 50/3 + 40/3 = 30 seconds C will be equidistant from A and B for the first time

    Sarah and Neha start running simultaneously from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduces to half of their current speeds. If the length of the circular track is 1500 m, how many times will Sarah and Neha meet on the track?
    (1) 6
    (2) 9
    (3) 11
    (4) 7
    (5) 8

    Total distance Sarah can cover = 10/60 * (15 + 15/2 + 15/4 + ...) = 5 km
    Total distance Neha can cover = 10/60 * (25 + 25/2 + 25/4 + ...) = 25/3 km
    First time they cover 750 m and subsequently they cover a distance of 1500 m to meet.
    The total distance they cover together is 40/3 km.
    Number of meetings possible is 1 + (40000/3 - 750)/1500 = 9.4
    => choice (B) is the right answer

    Three boys A, B and C start running at constant speeds from the same point P along the circumference of a circular track. The speeds of A, B and C are in the ratio 5:1:1. A and B run clockwise while C runs in the anticlockwise direction. Each time A meets B or C on the track he gives them a card.What is the difference in the number of cards received by B and C if A distributes 33 cards in all?

    Let speed of A , B , C = 5 , 1 , 1 km/hr
    A and B runs in clock wise direction so , their total meeting points will be 4 and C runs in anti clock wise direction so , their meeting points will be 6
    let the distance be 5km
    => after 1st 5 rounds B = 4 cards and C = 6 cards
    => after 2nd 5 rounds B = 8 cards and C = 12 cards
    => after 3rd 5 rounds B = 12 cards and C = 18 cards
    => after 4th round B = 13 cards and C = 20 cards
    so , difference = 20 - 13 = 7

    Practice Problems:

    1. Seven children A, B, C, D, E, F and G started walking from the same point at the same time, with speeds in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 respectively and they are running around a circular park. Each of them carry flags of different colours and whenever two or more children meet, they place their respective flag at that point. However nobody places more than 1 flag at a same point. They are running in anti-clockwise direction. How many flags will be there in total, when there will be no scope of putting more flags?

    2. Three athletes A, B and C are running on a circular track of length 1200 meters with speeds 30 m/s, 50 m/s and 80 m/s. A is running clockwise and B and C are running anticlockwise. Find the time after which A and B will meet for the first time and the frequency (in seconds) after which they will keep meeting. Also find the sum of the distance (in fraction of the track length) run by them till their first meeting.

    3. Which of the following cannot be the ratio of speeds of two joggers running on a circular jogging track if while running they meet at a diametrically opposite point to the point from where both of them started?
      (1) 3 : 5
      (2) 1 : 3
      (3) 1 : 5
      (4) 2 : 5



  • Can some one please explain the solution to practice problem 1 (Seven children A, B, C, D, E, F and G started...). I found its solution to be 56 (21 + 15 + 10 + 6 + 3 + 1) and it somehow seems to be related to number of meeting points (A-B: 1, B-C: 1 A-C: 2 and so on ) but still not clear. Please help


  • Being MBAtious!


    @test1234

    Funda : When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be |a – b|

    A and B will meet at |1 - 2| = 1 point.
    A and C will meet at |1 - 3| = 2 points
    A and D will meet at |1 - 4| = 3 points
    A and E will meet at |1 - 5| = 4 points
    A and F will meet at |1 - 6| = 5 points
    A and G will meet at |1 - 7| = 6 points
    So A will put 1 + 2 + 3 + 4 + 5 + 6 = 21 flags.

    similarly B and C will meet at |2 - 3| = 1 point
    B and D will meet at |2 - 4| = 2 points
    B and E will meet at |2 - 5| = 3 points
    B and F will meet at |2 - 6| = 4 points
    B and G will meet at |2 - 7| = 5 points
    So B will put 1 + 2 + 3 + 4 + 5 = 15 flags

    Similarly find for C, D, E and F.

    We will get 21 + 15 + 10 + 6 + 3 + 1 = 56 flags

    clear ?



  • Option 4


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