P and Q started from points X and Y at the same time and running towards Y and X at speed of 25 km/h and 15 km/h respectively. After reaching their respective destinations, they turn back and speed of Q increased by 3 km/h in every hour and of P decreased by 1 km/h. Find the total time taken for their second meeting, if distance between X and Y is 300 km?

To solve this problem, we need to consider both the first and second meetings of P and Q as they move towards each other and then turn back.

### Step 1: Time for the first meeting
Let the distance between points X and Y be \( d = 300 \) km. 

P's speed = \( v_P = 25 \) km/h.  
Q's speed = \( v_Q = 15 \) km/h.

When P and Q are running towards each other, their relative speed is \( v_P + v_Q = 25 + 15 = 40 \) km/h.

The time taken to meet the first time, \( t_1 \), is given by:
\[
t_1 = \frac{d}{v_P + v_Q} = \frac{300}{40} = 7.5 \text{ hours}
\]

### Step 2: Distance covered by each during the first meeting
P covers \( 25 \times 7.5 = 187.5 \) km.  
Q covers \( 15 \times 7.5 = 112.5 \) km.

### Step 3: Time for the second meeting
After reaching their destinations, they turn back. Now, Q's speed increases by 3 km/h every hour, and P's speed decreases by 1 km/h every hour.

- Initially, Q's speed is 15 km/h, and P's speed is 25 km/h when they turn back.
- Let \( t \) be the time taken for their second meeting after they turn back.

During the second meeting:
- Q’s speed after \( t \) hours = \( 15 + 3t \) km/h.
- P’s speed after \( t \) hours = \( 25 - t \) km/h.

The distance covered by P and Q during \( t \) hours can be expressed as:
\[
\text{Distance covered by Q} = \int_0^t (15 + 3t) \, dt = 15t + \frac{3t^2}{2}
\]
\[
\text{Distance covered by P} = \int_0^t (25 - t) \, dt = 25t - \frac{t^2}{2}
\]

They will meet when the sum of these distances equals the total distance between X and Y:
\[
15t + \frac{3t^2}{2} + 25t - \frac{t^2}{2} = 300
\]

Simplifying the equation:
\[
40t + \frac{2t^2}{2} = 300
\]
\[
40t + t^2 = 300
\]
\[
t^2 + 40t - 300 = 0
\]

### Step 4: Solving the quadratic equation
The quadratic equation is:
\[
t^2 + 40t - 300 = 0
\]
We can solve it using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 40 \), and \( c = -300 \):
\[
t = \frac{-40 \pm \sqrt{(40)^2 + 4 \times 300}}{2 \times 1}
\]
\[
t = \frac{-40 \pm \sqrt{1600 + 1200}}{2}
\]
\[
t = \frac{-40 \pm \sqrt{2800}}{2}
\]
\[
t = \frac{-40 \pm 52.92}{2}
\]

So the two possible solutions are:
\[
t = \frac{-40 + 52.92}{2} = 6.46 \text{ hours (approximately)}
\]
\[
t = \frac{-40 - 52.92}{2} = -46.46 \text{ hours (not physically meaningful)}
\]

### Conclusion
The total time taken for their second meeting after they turn back is approximately **6.46 hours**.

Would you like further details or explanations on any part of this solution?

---

**Here are five related questions to expand your understanding:**

1. What would be the total distance covered by P before the second meeting?
2. How would the problem change if P's speed increased instead of decreasing?
3. Can you find the exact time of the second meeting using a different method?
4. What would happen if Q's speed also decreased?
5. How does the quadratic nature of the equation affect the problem's solution?

**Tip:** Always verify your quadratic solution by checking the physical meaning of both roots. Only positive time values make sense in real-world problems.

Learn how to calculate the total time for the second meeting of two runners in a speed and distance problem. Includes step-by-step solution involving initial speeds, turning back, and changing speeds over time. Suitable for high school students.

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