Math Problem Statement

Two friends P and Q simultaneously start running from same point around a circular track. They

run in the same direction. P runs at 6 m/sec and Q runs at b m/sec. If they cross each other at exactly

two points on the circular track and b is a natural number less than 6, then how many values can b

take?

Solution

To solve this problem, we need to understand the conditions under which two runners, P and Q, cross each other at exactly two points on a circular track.

Given:

  • P runs at a speed of 66 m/s.
  • Q runs at a speed of bb m/s, where bb is a natural number less than 6.
  • They run in the same direction.
  • They cross each other exactly at two points on the circular track.

Step 1: Relative speed and crossing points

Since both runners are moving in the same direction, the relative speed between them is: Relative Speed=6b m/s.\text{Relative Speed} = |6 - b| \text{ m/s}. In this case, since b<6b < 6, the relative speed is 6b6 - b m/s.

The number of times they meet while running around the track is determined by the ratio of their speeds and the total distance of the track. They meet at distinct points if the relative speed divides evenly into the total speed difference.

Step 2: Condition for exactly two crossings

For the two runners to meet exactly twice on the track, the ratio of their speeds must allow only two points of intersection. This implies that the relative speed should create two crossings in one complete cycle around the track.

Thus, the speed difference 6b6 - b must be such that the runners meet exactly twice per full lap of the track.

Step 3: Possible values for bb

We are told that bb is a natural number less than 6, so bb can be one of these values: 1,2,3,4,51, 2, 3, 4, 5.

Let's check for each case:

  1. b=5b = 5, relative speed = 65=16 - 5 = 1 m/s.
  2. b=4b = 4, relative speed = 64=26 - 4 = 2 m/s.
  3. b=3b = 3, relative speed = 63=36 - 3 = 3 m/s.
  4. b=2b = 2, relative speed = 62=46 - 2 = 4 m/s.
  5. b=1b = 1, relative speed = 61=56 - 1 = 5 m/s.

We want the condition that ensures exactly two crossings. In this case, the number of crossings is determined by the relative speed as a divisor of the total number of potential crossings.

Since the problem specifies that they meet exactly two times, the valid relative speed would allow this outcome.

Therefore, the correct number of values that bb can take is 2, corresponding to b=4b = 4 and b=2b = 2.

Final Answer:

bb can take 2 values.


Would you like further clarification or details on any step? Here are some additional related questions:

  1. How does the relative speed of two objects affect their meeting points?
  2. What happens if bb is greater than 6 in a similar scenario?
  3. How would the problem change if they were running in opposite directions?
  4. Can we generalize this for any other speeds besides 6 m/s for P?
  5. What if there was a third person running on the same track?

Tip: The number of crossings between two moving objects on a circular track can often be tied to the greatest common divisor (GCD) of their speeds.

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Math Problem Analysis

Mathematical Concepts

Relative Speed
Circular Motion
Speed Ratios

Formulas

Relative speed = |Speed of P - Speed of Q|

Theorems

Meeting points on a circular track depend on relative speed and track length

Suitable Grade Level

Grades 9-11