Two trains ` A and B` of length `400 m` each are moving on two parallel tracks with a uniform speed of ` 72 km h^(-1)`
in the same direction with ` A` ahead of `B` .The driver of B decides to overtake A
and accelerates by 1m/s².if after 50s ,the guard of B just passes the driver of A , what was the original distance between them ?

Options

A. 750 m
B. 1000 m
C. 1250 m
D. 2250 m

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the speed of the trains from km/h to m/s
Given that the speed of the trains is 72 km/h, we can convert this to meters per second using the conversion factor (1 km/h = 5/18 m/s).

\[ \text{Speed in m/s} = 72 \times \frac{5}{18} = 20 \, \text{m/s} \]

### Step 2: Identify the initial conditions
Both trains A and B are initially moving at the same speed of 20 m/s. The length of each train is 400 m.

### Step 3: Determine the acceleration of train B
Train B accelerates at a rate of 1 m/s².

### Step 4: Set up the relative motion
Since both trains are moving in the same direction, we need to find the relative velocity and relative acceleration of train B with respect to train A.

- Initial relative velocity (V_BA) = V_B - V_A = 20 m/s - 20 m/s = 0 m/s
- Relative acceleration (A_BA) = A_B - A_A = 1 m/s² - 0 m/s² = 1 m/s²

### Step 5: Use the equation of motion to find the distance
We will use the second equation of motion to find the distance (x) that train B covers relative to train A in 50 seconds.

The equation is:
\[ x = V_{BA} \cdot t + \frac{1}{2} A_{BA} \cdot t^2 \]

Substituting the values:
- \(V_{BA} = 0 \, \text{m/s}\)
- \(A_{BA} = 1 \, \text{m/s}^2\)
- \(t = 50 \, \text{s}\)

Calculating:
\[ x = 0 \cdot 50 + \frac{1}{2} \cdot 1 \cdot (50)^2 \]
\[ x = 0 + \frac{1}{2} \cdot 1 \cdot 2500 \]
\[ x = \frac{2500}{2} = 1250 \, \text{m} \]

### Step 6: Conclusion
The original distance between the two trains A and B is 1250 meters.

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