Lesson Problems on bodies moving on a circle

Problems on bodies moving on a circle

Problem 1

When Nina will complete her first loop,  Jenna will be in  2  seconds ahead and will complete   =   of her second loop.

 
When Nina will complete her second loop,  Jenna will win additional    of the loop and will be ahead just in    of the loop.


Every time when Nina will complete her next  20-seconds loop,  Janna will win additional    of the loop.


Finally,  when Nina will complete her  9-th loop  (180 seconds),  Jenna will complete her  10-th loop, 
Thus,  in  180  seconds after start,  they both will be at the front door at the same time.


Answer.  In  180  seconds after start it will happen at the first time that both girls be at the front door simultaneously.

Problem 2

1.  The readings difference between clocks Q and P is increasing at the rate of 4 min per hour.

    It means that their readings will coincide every 15 hours, when their minute hands are at the same positions  "verically up".


    It happens every 15 hours, and there is no other time moments, when their positions coincide.



2.  The readings difference between clocks R and P is increasing at the rate of 10 min per hour.

    It means that their readings will coincide every 6 hours, when their minute hands are at the same positions  "verically up".


    It happens every 6 hours, and there is no other time moments, when their positions coincide.



3.  The readings difference between clocks Q and R is increasing at the rate of 6 min per hour.

    It means that their readings will coincide every 10 hours, when their minute hands are at the same positions  "verically up".


    It happens every 10 hours, and there is no other time moments, when their positions coincide.



4.  It implies that the next and closest time moment when the readings of all 3 clock coincide comes in  LCM (15,6,10) which is 30 hours.

    LCM is "Leatt Common Multiple).


Answer.  In  30  hours.

Problem 3

Intuitively, it is clear that the cars A and B will be simultaneously at the point P next time in N seconds,
where N is least common multiple of the numbers of 5 and 7, i.e. after 35 seconds. 

The car A will complete 7 loops at that time, while car B will make 5 loops.


Mathematically, the condition that the cars A and B will be simultaneously at the point P after making their loops, 
is this equality


5n = 7m seconds,


where n and m are natural numbers.


Thus the number 5n, which is the same as 7m, is a common multiple of numbers 5 and 7. 

Problem 4

When Tracy and Kelly are running on the track in opposite directions, the distance they cover together from one meeting point 
to the next meeting point is exactly the length of the track D.

     It is the key moment to understand solving this problem.


Therefore, if "t" and "k" are the rates of each girl/women, then their summary (combined) rate is

t + k =  per second in terms of the length D,    (1)

where D is the length of the track.


From the other side, we know that the Tracy's rate is t =  per second in terms of the length D.


Therefore, from (1) we have 

k =  =  =  =  = .


Thus, Kelly's rate is k =  per second in terms of D.


It means that Kelly will cover the entire distance D (one lap) in  = 36 seconds.


Answer.  Kelly's lap will take 36 seconds.

         In other words, Kelly's rate is , while Tracy's rate is . 

Check.   Their combined rate when they are running in opposite directions is

                 =  =  = .   Check !

Problem 5

Let their rates are "u" and "v" ft per second, where "u" stands for the faster track-man.

Then the first equation is 

 = u + v.          (1)

This equation is for the case then they are running in opposite directions.
     (Paradoxically, moving in opposite directions means moving towards each other along the circumference in this case !)

The left side says that they together cover 300 ft in 10 seconds.

The right side is the rate of decreasing the distance between them measured along the circumference. 
   (Thinking on this equation, keep in mind that they meet for every circle counting after the first meeting).

So the first equation is 

u + v = 30.             (1')


The second equation is 

 = 300.    (2)

This equation is for the case they are running in one direction, and the equation says that for the faster track-man 
the path from one meeting point to the next meeting point is in one full circumference longer than for the slower track-man.
 
The equation (2) is equivalent to

u - v = 6.         (2')

The equations (1') and (2') are the governing equations, and we can easily solve them by adding. Doing so, you get

2u = 30 + 6  --->  2u = 36   --->  u =  = 18.

So the faster track-man speed is 18 ft/s.

Then the slower track-man speed is 30-18 = 12 ft/s.

Check.   =  = 10 seconds.

         =  = 50 seconds.

Answer.  18 ft/s  and  12 ft/s.