Dispatch centre · emergency call
Golden Minutes
An emergency call has come in: a person in the park (point P) needs urgent help. An ambulance is always dispatched.
The dispatch centre also has a medical drone carrying an automated external defibrillator (AED). The drone base is at point D, and the ambulance is at point A.
System data
a – 1 block east, b – 1 block north, 1 block = 200 m
boulevard: c = a + b
DP = 6a + 8b
DA = a + 5b
Straight-line distance D – P (measured by the system): 2000 m
Is it worth sending the medical drone as well?
We will work it out step by step. Be ready: new information will arrive during the analysis, and the decision may change.
Screen 1
Where is the patient relative to the ambulance?
DP = 6a + 8b
DA = a + 5b
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Write the vector AP as a linear combination of a and b.
Screen 2
The ambulance route
The navigation system suggests two routes for the ambulance:
M1 = 2c + 3a + b
M2 = 4c + a − b
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(a) Write each route in terms of a and b.
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(b) Compare your results with AP = 5a + 3b. What can you conclude?
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(c) What distance does the ambulance travel along each route?
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(d) Which route should the ambulance take?
Screen 3
Wind and the medical drone
The drone is programmed to fly north, but there is a wind. We write the velocities using new unit vectors (a and b already describe blocks):
eE – unit vector east, eN – unit vector north
vdrone/air = 25eN m/s
vwind = 12eE − 9eN m/s
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(a) Find the drone's velocity relative to the ground: vground = vdrone/air + vwind.
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(b) Is the drone moving straight towards the patient? Compare DP = 6a + 8b and vground.
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(c) What is the drone's speed relative to the ground (the magnitude of vground)?
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(d) The distance D – P is 2000 m. Before take-off, the drone needs 30 s of preparation. The ambulance travels along M1 at an average speed of 10 m/s. Compare the times from the moment of the emergency call.
We compare only the delivery time in our teaching model. This is not a medical claim about the effect of an AED.
Screen 4
The tower
! Navigation warning
A communications tower T has been detected on the drone's flight path.
DT = 3a + 4b
DP = 6a + 8b
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(a) Find the scalar k such that DT = k · DP.
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Where is the tower T?
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(b) The flight-planning system checks the route: where will the drone be after 40 s of flight at 20 m/s? Call this point X.
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Compare the positions of X and T. What can you conclude?
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(c) After how many seconds of flight does the drone reach the tower? How many seconds are left after point X?
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(d) The drone has to fly around the tower.
Task condition: in our teaching model, the safe manoeuvre around or over the obstacle adds 15 s to the drone's total time. This is not a real aviation rule.
Screen 5
Traffic
! New traffic update
The ambulance has already left the boulevard and turned onto M1 when traffic congestion builds up. Changing route is no longer possible.
along the boulevard (2c): average speed 10 m/s
after the boulevard (3a + b): average speed 5 m/s
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(a) How long will the ambulance take for each section, and in total?
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(b) The drone's time (with the manoeuvre around the tower) is still 145 s. How many seconds earlier could the AED arrive?
Screen 6
The dispatcher's decision
| New information | AED by drone | Ambulance | Conclusion |
|---|---|---|---|
| wind, flight at 20 m/s | 130 s | 136 s | drone 6 s earlier |
| tower T, +15 s | 145 s | 136 s | ambulance 9 s earlier |
| traffic after the boulevard | 145 s | 216 s | drone 71 s earlier |
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What is the dispatcher's final decision?
What we used
- vector subtraction: AP = DP − DA;
- same displacement ≠ same distance travelled (M1 and M2);
- adding velocities: vground = vdrone/air + vwind;
- same direction ⇔ the components are in the same ratio (3 : 4);
- magnitude of a vector with perpendicular components – Pythagoras' theorem;
- collinear vectors: DT = k · DP, 0 < k < 1 ⇒ T is between D and P;
- distance, speed and time: s = v · t.
The decision changed three times – because the data changed, not the maths.
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