30

Step 1: Determine the positions of both persons at any time after noon.
The first person (Person 1) walks along track AOB at a speed of 5 km/h. By noon (12:00 PM), they are at point O. When t = 0, their position is at O.
Position of Person 1 at time t (in hours):
Distance covered = speed × time = 5t km.
So their position at time t can be represented as (5t, 0) in the coordinate system with O as (0, 0).
Step 2: The second person (Person 2) walks along track COD and reaches O at 1:30 PM, which is 1.5 hours after noon. This means that at noon, they are still 1.5 hours away from O. Therefore, at 12:00 PM (t = 0), they are still 7.5 km away from O moving toward it since they cover the distance in 1.5 hours at 5 km/h. Hence, their speed is also 5 km/h, and they will take 1.5 hours to reach O.
Position of Person 2 at time t:
When they start at t = -1.5 hours (1:30 PM), they would be at position (0, 7.5) at t = 0.
For t = 0 (12:00 PM) to t hours later, their position is (0, 7.5 - 5t).
Step 3: Distance D between the two persons at time t (after noon):
Using distance formula: D = √[(5t - 0)² + (0 - (7.5 - 5t))²]
= √[(5t)² + (5t - 7.5)²]
= √[25t² + (5t - 7.5)²]
= √[25t² + 25t² - 75t + 56.25]
= √[50t² - 75t + 56.25].
Step 4: To find the minimum distance, we find the derivative of D² wrt t and set it to 0.
D² = 50t² - 75t + 56.25. The minimum occurs when the derivative d(D²)/dt = 0:
d(D²)/dt = 100t - 75 = 0 → t = 0.75 hours (which is 45 minutes after noon).
Step 5: Thus, the time at which they are closest is 12:45 PM or 12:T PM where T = 45. However, T represents the minutes after 12:00 PM; therefore second's person reach O at 1:30 PM hence T=30.