A trip in two legs
A driver goes 60 km to a town at 60 km/h, then drives the same 60 km home at 30 km/h because of heavy traffic. Each part of a trip like this is called a leg.
First find how long each leg took: time = distance ÷ speed. The first leg takes 60 ÷ 60 = 1 hour. The second leg is just as long but half as fast, so it takes 60 ÷ 30 = 2 hours.
Each block is one hour of driving, marked with the distance covered in that hour. The first leg fills 1 hour and the second leg fills 2.
Whole distance over whole time
The average speed of a trip is the whole distance divided by the whole time. The whole trip is 60 + 60 = 120 km, and it took 1 + 2 = 3 hours, so the average speed is 120 ÷ 3 = 40 km/h.
This is the steady speed that makes the same trip in the same time. A car driving at a steady 40 km/h for 3 hours also covers 3 × 40 = 120 km.
The distance covered in each of the 3 hours: 60 km, then 30 km, then 30 km. The dashed line is their mean, 120 ÷ 3 = 40 km in each hour.
Why the answer is not 45
It is tempting to find the mean of the two speeds: (60 + 30) ÷ 2 = 45 km/h. That would be right only if the driver spent the same time at each speed. Here the driver spent 1 hour at 60 km/h and 2 hours at 30 km/h, so the slow speed counts twice.
Check 45 against the trip: at a steady 45 km/h for 3 hours, the car would cover 3 × 45 = 135 km, but the trip was only 120 km. The average speed is 40 km/h, closer to the slow speed, because more of the time was spent going slowly.
The two speeds, one bar for each leg, and their mean, 45. This picture gives each leg the same weight, but the second leg lasted twice as long as the first.
The usual mistakes
Finding the mean of the speeds. The mean of 60 km/h and 30 km/h is 45 km/h, but the trip took 3 hours to go 120 km, which is 40 km/h.
Adding the speeds. 60 + 30 = 90 km/h is faster than either leg, and no average can be faster than the fastest part of the trip.
Giving the slow leg's speed. The first hour at 60 km/h pulls the average above 30 km/h. The average speed always lies between the slowest and the fastest speeds.
Worked example: Average Speed for a Multi-Leg Single Journey
Question A delivery van traveled from Point A to Point B, a distance of 80 km, at an average speed of 40 km/h. It then continued from Point B to Point C, a distance of 180 km, at an average speed of 60 km/h. Find the average speed of the delivery van for the entire journey from Point A to Point C.
1.Draw Segment 1: Bar of 80 km. Since rate is 40 km per 1 hour, box count for time = 80 ÷ 40 = 2 hours.
A to B: 80 km at 40 km/h. Each hour-box holds 40 km, so 2 boxes. 2.Draw Segment 2: Bar of 180 km. Since rate is 60 km per 1 hour, box count for time = 180 ÷ 60 = 3 hours.
B to C: 180 km at 60 km/h. Each hour-box holds 60 km, so 3 boxes. 3.Combine segments into one master bar: Total distance = 80 + 180 = 260 km.
One trip: 80 + 180 = 260 km. 4.Combine time boxes: Total time = 2 + 3 = 5 hours.
One time: 2 + 3 = 5 hours. 5.Average speed = 260 ÷ 5 = 52 km/h.
Average speed is the whole distance over the whole time: 260 ÷ 5 = 52 km/h, not the midpoint of 40 and 60.
Answer: 52 km/h
Common mistakes
- Averaging the two speeds directly: 40 + 602 = 50 km/h.
- Adding the speeds together: 40 + 60 = 100 km/h.
Working back from the whole journey
The same rule works backward. If a question gives the whole distance and the whole time, subtract what is known about the first leg to find the distance and the time left for the second leg. Then the second leg's speed is its own distance divided by its own time.
Worked example: Two-Stage Journey with Mid-Way Speed Reduction
Question Mr. Tan drove from City A to City B, covering a total distance of 360 km. For the first 13 of the distance, he drove at an average speed of 80 km/h. Heavy traffic caused him to reduce his speed for the remaining journey. If the entire journey took 512 hours, what was his average speed for the remaining journey?
1.Total bar of 360 km split into 3 units of 120 km each.
Thirds of 360 km: three units of 120 km. 2.First unit = 120 km. At 80 km/h, time taken = 120 ÷ 80 = 1.5 hours.
The first unit at 80 km/h: 120 ÷ 80 = 1.5 hours. 3.Remaining 2 units = 2 × 120 = 240 km.
The remaining two units are 240 km. 4.Total time is 5.5 hours, so time left for the 2 units = 5.5 − 1.5 = 4 hours.
Of the 5.5 hours, 5.5 − 1.5 = 4 remain for those 240 km. 5.Speed for remaining 2 units = 240 ÷ 4 = 60 km/h.
Speed for the rest: 240 ÷ 4 = 60 km/h.
Answer: 60 km/h
Common mistakes
- Applying 80 km/h across the first 13 of the time (5.5 ÷ 3) instead of the distance.
- Subtracting 1.5 hours from 5 hours instead of 5.5 hours.