4.4 Time, Speed & Distance

4.4 Time, Speed & Distance

Quick Answer: Distance = speed × time. For equal-distance legs, average speed is the harmonic mean 2ab/(a+b), not the arithmetic mean. Convert km/h to m/s by multiplying by 5/18.

Time-speed-distance (TSD) problems are pilot-relevant: a GD pilot constantly computes ground speed, time to waypoint, and relative motion. The PAF verbal battery tests the same arithmetic under time pressure.

1. The Core Formula

distance = speed × time — and its rearrangements speed = distance / time, time = distance / speed.

Keep units consistent. If speed is in km/h and time in minutes, convert minutes to hours (divide by 60) before multiplying.

2. Unit Conversion

Example: 72 km/h in m/s = 72 × (5/18) = 4 × 5 = 20 m/s.

Example: 25 m/s in km/h = 25 × (18/5) = 5 × 18 = 90 km/h.

3. Average Speed — the Harmonic-Mean Trap

Average speed = total distance / total time. When two equal distances are travelled at speeds a and b:

average speed = 2ab / (a + b) (the harmonic mean).

Worked Example

A pilot flies to a base at 600 km/h and returns along the same route at 400 km/h. Average speed?

average = 2 × 600 × 400 / (600 + 400) = 480 000 / 1000 = 480 km/h.

The arithmetic mean (600 + 400)/2 = 500 km/h is wrong — the aircraft spends more time at the slower speed, dragging the average down.

If distances differ, use average = total distance / total time directly: compute each leg's time, sum, and divide.

4. Relative Speed

Worked Example — Meeting Time

Two towns are 180 km apart. A car leaves A at 60 km/h while a bus leaves B at 40 km/h toward A. When do they meet?

Relative speed (opposite directions) = 60 + 40 = 100 km/h. Time to close 180 km = 180 / 100 = 1.8 hours = 1 h 48 min.

Worked Example — Overtaking

A truck at 50 km/h is 30 km ahead of a car at 80 km/h going the same way. When does the car catch up?

Relative speed (same direction) = 80 − 50 = 30 km/h. Time to close 30 km = 30 / 30 = 1 hour.

5. Boats and Streams

Let the boat's speed in still water = b and the stream's speed = c.

• Downstream speed (with the current) = b + c
• Upstream speed (against the current) = b − c
• Still-water speed b = (downstream + upstream) / 2
• Stream speed c = (downstream − upstream) / 2

Worked Example

A boat covers 60 km downstream in 2 hours and the same 60 km upstream in 3 hours. Find the boat's speed in still water and the stream's speed.

• Downstream speed = 60 / 2 = 30 km/h.
• Upstream speed = 60 / 3 = 20 km/h.
• Still-water speed b = (30 + 20) / 2 = 25 km/h.
• Stream speed c = (30 − 20) / 2 = 5 km/h.

6. Common Traps

1. Arithmetic mean for average speed. Use 2ab/(a+b) for equal distances, never (a+b)/2.
2. Mixing units. A speed in km/h with a time in minutes must be reconciled before applying d = s × t.
3. Stops included in total time. If a train stops for 10 minutes, that 10 minutes counts in the total time for average-speed calculations.
4. Same-direction subtraction order. Relative speed is the absolute difference |s₁ − s₂|; the faster object must be behind to catch up.
5. Crew/boat confusion. "Speed of the boat in still water" is the boat's own capability; "speed of the stream" is the current — never swap them.

Procedure for Any TSD Item

1. Write down what is given (s, d, or t) with units.
2. Convert all units to a single system (SI or km/h-hours).
3. Identify the sub-type (single leg, two legs, relative, boat).
4. Apply the right formula.
5. Sanity-check the answer's magnitude.

7. Circular Tracks and Meeting Points

When two runners share a circular track, the relative-speed logic still holds, but the 'distance to close' is the track length, not a straight-line gap. Two cadets start together on a 400 m track, one jogging at 4 m/s and the other at 6 m/s in the same direction. Relative speed = 6 − 4 = 2 m/s. The faster cadet laps the slower one when the relative distance equals one full lap: time = 400 / 2 = 200 seconds. If they run in opposite directions, the relative speed is 4 + 6 = 10 m/s and they meet every 400 / 10 = 40 seconds. The trap is treating 'meet again' like 'meet for the first time from a standing start' — on a circle, every lap-length of relative separation produces another meeting, so divide the track length by the relative speed each time.

8. Worked Chain with a Stop

A bus covers 120 km at 60 km/h, halts for 15 minutes, then returns 120 km at 40 km/h. Find the average speed for the whole journey. Leg 1 time = 120 / 60 = 2 h. Stop = 15 min = 0.25 h. Leg 2 time = 120 / 40 = 3 h. Total time = 2 + 0.25 + 3 = 5.25 h. Total distance = 240 km. Average speed = 240 / 5.25 ≈ 45.7 km/h. The stop pulls the average below the 48 km/h you would get by ignoring it — and below the harmonic-mean shortcut, which assumes no stop. Whenever a rest or halt is mentioned, abandon the 2ab/(a+b) shortcut and compute total distance / total time directly.

9. Pilot-Relevant Relative Motion

A formation leader flies north at 500 km/h while a wingman trails 50 km behind at 560 km/h on the same heading. Relative speed (same direction) = 560 − 500 = 60 km/h. Time to close 50 km = 50 / 60 ≈ 0.83 h ≈ 50 minutes. If the wingman were closing head-on (opposite direction), the relative speed would be 500 + 560 = 1060 km/h and the closure time would be 50 / 1060 ≈ 2.8 minutes — a stark reminder that direction flips the formula. GD pilot aptitude items reuse exactly this relative-motion reasoning, so practising it as arithmetic now pays in the later flying-aptitude stage.