Two trains are approaching each other on the same straight track. One is moving at a constant speed, while the other is decelerating. Which of the graphs shows the correct change in their separation as a function of time?

Correct option is D

Solution:
At first, when both trains are moving, the separation decreases quickly.
As the decelerating train slows down, separation decreases more slowly (curve flattening).
When the decelerating train stops, the constant-speed train alone moves toward the stationary one, causing separation to reduce rapidly (sharp vertical drop).
Thus, the graph:
First → Curved decrease (both moving).
Then → Straight sharp fall (only one moving at constant speed).

​$$\text{Distance at any time:} \\d = \text{initial separation} - (v_1 \times t) - \left( v_{2,\text{initial}} \times t - \frac{1}{2} a t^2 \right)\\[10pt]\text{After Train 2 stops:} \\d = d_0 - v_1 \times t$$​​

The correct graph is Option 4, showing:
Smooth decrease first,
Sharp drop later (after second train stops).
Final Answer: Option (d)