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?

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 \( 6 \) m/s.
- Q runs at a speed of \( b \) m/s, where \( b \) 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:
\[
\text{Relative Speed} = |6 - b| \text{ m/s}.
\]
In this case, since \( b < 6 \), the relative speed is \( 6 - b \) 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 \( 6 - b \) must be such that the runners meet exactly twice per full lap of the track.

### Step 3: Possible values for \( b \)

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

Let's check for each case:

1. \( b = 5 \), relative speed = \( 6 - 5 = 1 \) m/s.
2. \( b = 4 \), relative speed = \( 6 - 4 = 2 \) m/s.
3. \( b = 3 \), relative speed = \( 6 - 3 = 3 \) m/s.
4. \( b = 2 \), relative speed = \( 6 - 2 = 4 \) m/s.
5. \( b = 1 \), relative speed = \( 6 - 1 = 5 \) 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 \( b \) can take is **2**, corresponding to \( b = 4 \) and \( b = 2 \).

### Final Answer: 
\( b \) 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 \( b \) 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.

Two friends P and Q simultaneously start running from the 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, how many values can b take?

Discover how to determine the values of speed b for two runners crossing each other exactly twice on a circular track. This problem involves concepts of relative speed, circular motion, and speed ratios. Ideal for students in Grades 9-11.

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