Patrol Interception With Changing Wind
Two aircraft, A and B, are 480 km apart on the same straight line, flying towards each other. A flies at 320 km/h and B at 240 km/h in still air. A steady wind of 60 km/h blows from A towards B (i.e. a tailwind for A, headwind for B). A third aircraft C is stationed at the midpoint between A and B and must rendezvous with whichever aircraft reaches it first. All speeds are constant; no turns are made.
1. Compute the ground speeds, the time to meet, and the meeting point. Which aircraft reaches the midpoint first?
Read the worked answer
Ground speeds:
- A: 320 + 60 = 380 km/h (tailwind)
- B: 240 − 60 = 180 km/h (headwind)
Closing speed = 380 + 180 = 560 km/h
Time to meet = 480 / 560 = 6/7 h ≈ 51.4 min
Meeting point from A = 380 × 6/7 = 325.7 km from A's start (so 154.3 km from B's start).
Midpoint is at 240 km from A.
- Time for A to reach midpoint = 240 / 380 ≈ 37.9 min
- Time for B to reach midpoint = 240 / 180 ≈ 80 min
A reaches the midpoint first, about 42 minutes before B. C should rendezvous with A.
2. If the wind direction reversed (now a headwind for A), how does the answer change, and what does this reveal about the trap in such problems?
Read the worked answer
Reversed wind:
- A: 320 − 60 = 260 km/h
- B: 240 + 60 = 300 km/h
- Closing speed = 560 km/h (unchanged!)
- Time to meet = 480/560 = 6/7 h (unchanged)
But the meeting point shifts: from A = 260 × 6/7 ≈ 222.9 km (was 325.7 km).
Midpoint times:
- A: 240/260 ≈ 55.4 min
- B: 240/300 = 48 min
Now B reaches the midpoint first. The closing speed and total time are identical; only the distribution changes.
The trap: Candidates assume the answer depends on the sum of speeds, which is wind-invariant. In fact, the rendezvous geometry is entirely wind-dependent. The lesson: identify which quantity is invariant (closing speed) and which is not (position of meeting). In military terms, the intercept point - not the intercept time - is what the controller must recompute when wind shifts.