Two trains `A` and `B` `, 100km` apart are travelling towards each other on different tracks with same starting speed of `50km//h`. The train `A` accelerates at `20km//h^(2)` and the train `B` retards at the rate `20km//h^(2)`. The distance covered by the train `A` when they cross each other is

Options

A. `70km`
B. `55km`
C. `65km`
D. `60km`

Answer

To solve the problem, we need to analyze the motion of both trains A and B as they travel towards each other.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Distance between the two trains, \( D = 100 \) km.
- Initial speed of both trains, \( u = 50 \) km/h.
- Acceleration of train A, \( a_A = 20 \) km/h².
- Deceleration of train B, \( a_B = -20 \) km/h².

2. **Set Up the Equations of Motion:**
- For train A, the distance covered \( H \) can be expressed using the equation of motion:
\[ H = ut + \frac{1}{2} a_A t^2 \]
- For train B, the distance covered is \( 100 - H \):
\[ 100 - H = ut + \frac{1}{2} a_B t^2 \]
- Substitute \( a_B \) with \(-20\) km/h²:
\[ 100 - H = ut - \frac{1}{2} (20) t^2 \]

3. **Combine the Equations:**
- Adding both equations:
\[ H + (100 - H) = ut + \frac{1}{2} a_A t^2 + ut - \frac{1}{2} (20) t^2 \]
\[ 100 = 2ut + \frac{1}{2} (20) t^2 - \frac{1}{2} (20) t^2 \]
- This simplifies to:
\[ 100 = 2ut \]

4. **Solve for Time \( t \):**
- Rearranging gives:
\[ t = \frac{100}{2u} \]
- Substituting \( u = 50 \) km/h:
\[ t = \frac{100}{2 \times 50} = 1 \text{ hour} \]

5. **Calculate the Distance Covered by Train A:**
- Now substitute \( t = 1 \) hour back into the equation for \( H \):
\[ H = ut + \frac{1}{2} a_A t^2 \]
- Substituting the values:
\[ H = 50 \times 1 + \frac{1}{2} \times 20 \times (1)^2 \]
\[ H = 50 + 10 = 60 \text{ km} \]

### Final Answer:
The distance covered by train A when they cross each other is **60 km**.

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