Two trains one of length 100m and another of length 125 m, are moving in mutually opposite directions along parallel lines, meet each other. Each with speed 10 m/s. If their acceleration are `0.3(m)/(s^2)` and `0.2(m)/(s^2)` respectively, then the time they take to pass each other will be

Options

A. 5s
B. 10s
C. 15s
D. 20s

Explanation

To solve the problem, we need to determine the time it takes for two trains to pass each other while considering their lengths, initial speeds, and accelerations. Let's break it down step by step.

### Step 1: Identify the parameters
- Length of Train A (L1) = 100 m
- Length of Train B (L2) = 125 m
- Speed of Train A (V1) = 10 m/s
- Speed of Train B (V2) = 10 m/s
- Acceleration of Train A (a1) = 0.3 m/s²
- Acceleration of Train B (a2) = 0.2 m/s²

### Step 2: Calculate the total distance to be covered
The total distance (S) that the two trains need to cover to completely pass each other is the sum of their lengths:
\[ S = L1 + L2 = 100 \, \text{m} + 125 \, \text{m} = 225 \, \text{m} \]

### Step 3: Calculate the relative velocity
Since the trains are moving in opposite directions, the relative velocity (V) is the sum of their speeds:
\[ V = V1 + V2 = 10 \, \text{m/s} + 10 \, \text{m/s} = 20 \, \text{m/s} \]

### Step 4: Calculate the total acceleration
The total acceleration (a) is the sum of their accelerations:
\[ a = a1 + a2 = 0.3 \, \text{m/s}^2 + 0.2 \, \text{m/s}^2 = 0.5 \, \text{m/s}^2 \]

### Step 5: Use the equation of motion
We will use the equation of motion to find the time (t) it takes for the trains to pass each other:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S \) = 225 m (total distance)
- \( u \) = 20 m/s (initial relative velocity)
- \( a \) = 0.5 m/s² (total acceleration)

Substituting the values into the equation:
\[ 225 = 20t + \frac{1}{2}(0.5)t^2 \]
This simplifies to:
\[ 225 = 20t + 0.25t^2 \]

### Step 6: Rearranging the equation
Rearranging the equation gives us:
\[ 0.25t^2 + 20t - 225 = 0 \]

### Step 7: Solve the quadratic equation
To solve the quadratic equation, we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 0.25 \)
- \( b = 20 \)
- \( c = -225 \)

Calculating the discriminant:
\[ b^2 - 4ac = 20^2 - 4(0.25)(-225) \]
\[ = 400 + 225 = 625 \]

Now substituting into the quadratic formula:
\[ t = \frac{-20 \pm \sqrt{625}}{2 \times 0.25} \]
\[ = \frac{-20 \pm 25}{0.5} \]

Calculating the two possible values for \( t \):
1. \( t = \frac{5}{0.5} = 10 \) seconds (valid solution)
2. \( t = \frac{-45}{0.5} = -90 \) seconds (not valid, as time cannot be negative)

### Final Answer
The time taken for the two trains to pass each other is:
\[ \boxed{10 \, \text{seconds}} \]