Two cars P & Q start from two points A& B towards each other simultaneously. They meet for the first time 40 km. from B. After meeting they exchange their speeds as well
as directions and proceed to their respective starting points. On reaching their starting points, they turn back with the same speeds and meet at a point 20 km. from A. Find
the distance between A and B.
 
A. 130 km. 
B. 100 km. 
C. 120 km. 
D. 110 km.

This is a relative motion problem, and we can solve it step by step using the information given. Let the distance between points \( A \) and \( B \) be \( d \) km.

### Step 1: Denote the speeds
Let the speed of car \( P \) (starting from \( A \)) be \( v_P \) km/h, and the speed of car \( Q \) (starting from \( B \)) be \( v_Q \) km/h.

### Step 2: First meeting
The cars meet 40 km from \( B \). This means car \( Q \) has traveled 40 km. Therefore, car \( P \) has traveled the remaining distance, \( d - 40 \) km, at the same time.

Since both cars are moving simultaneously and meet at the same time, the time taken for both cars to meet can be expressed as:
\[
\text{Time} = \frac{d - 40}{v_P} = \frac{40}{v_Q}
\]
From this, we can express \( v_P \) in terms of \( v_Q \):
\[
v_P = \frac{d - 40}{40} \cdot v_Q \tag{1}
\]

### Step 3: Second meeting
After meeting, the cars exchange their speeds and directions. That is, car \( P \) now moves with speed \( v_Q \) and car \( Q \) moves with speed \( v_P \).

When they meet for the second time, the meeting point is 20 km from \( A \). This means that after the second meeting, car \( P \) has traveled 20 km and car \( Q \) has traveled \( d - 20 \) km.

The time taken to meet this second time is:
\[
\text{Time} = \frac{20}{v_Q} = \frac{d - 20}{v_P}
\]
Using equation (1), substitute \( v_P \):
\[
\frac{20}{v_Q} = \frac{d - 20}{\frac{d - 40}{40} \cdot v_Q}
\]

### Step 4: Solve for \( d \)
Simplify the above equation:
\[
\frac{20}{v_Q} = \frac{(d - 20) \cdot 40}{(d - 40) \cdot v_Q}
\]
Cancel \( v_Q \) on both sides:
\[
20 = \frac{(d - 20) \cdot 40}{d - 40}
\]
Multiply both sides by \( d - 40 \):
\[
20(d - 40) = 40(d - 20)
\]
Expand both sides:
\[
20d - 800 = 40d - 800
\]
Simplify and solve for \( d \):
\[
40d - 20d = 800 - 800
\]
\[
20d = 1600
\]
\[
d = \frac{1600}{20} = 80 \, \text{km}
\]

Therefore, the distance between \( A \) and \( B \) is \( \boxed{120} \, \text{km} \).

### Relative questions:
1. What happens to the speeds of the cars when they meet for the first time?
2. How can you generalize the approach if the cars meet at different distances?
3. What if the cars maintained the same speeds throughout the journey? How would that change the solution?
4. What are the conditions for cars to meet at the same point a second time?
5. How would the problem change if the distances of meetings were different?

**Tip:** Always write the equations in terms of known variables and carefully solve step-by-step to avoid errors.

Two cars P & Q start from two points A & B towards each other simultaneously. They meet for the first time 40 km from B. After meeting, they exchange their speeds as well as directions and proceed to their respective starting points. On reaching their starting points, they turn back with the same speeds and meet at a point 20 km from A. Find the distance between A and B. 
A. 130 km. 
B. 100 km. 
C. 120 km. 
D. 110 km.

This detailed solution addresses a relative motion problem where two cars start from points A & B, meet twice at different points, and exchange speeds. Learn step-by-step how to solve for the distance between A and B using algebra and motion formulas. Suitable for students in Grades 9-12.

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