Average speed is the whole distance over the whole time, never the mean of the speeds
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.
Speed · A Van's Trip in Two Legs at Two Different Speeds
QuestionA delivery van drove 80 km from Town A to Town B at an average speed of 40 km/h. It then drove 180 km from Town B to Town C at an average speed of 60 km/h. (a) How long did the whole trip from Town A to Town C take? (b) What was the van's average speed for the whole trip?
1.The first leg is 80 km at 40 km/h. Each hour covers 40 km, so the first leg takes 80 ÷ 40 = 2 hours.
A to B: 80 km at 40 km/h. Each hour covers 40 km, so it takes 2 hours. 2.The second leg is 180 km at 60 km/h. Each hour covers 60 km, so the second leg takes 180 ÷ 60 = 3 hours.
B to C: 180 km at 60 km/h. Each hour covers 60 km, so it takes 3 hours. 3.(a) The whole trip takes 2 + 3 = 5 hours.
(a) The whole trip takes 2 + 3 = 5 hours. 4.The whole distance is 80 + 180 = 260 km.
The whole road is 80 + 180 = 260 km. 5.(b) Share the 260 km equally over the 5 hours: 260 ÷ 5 = 52 km in each hour. The average speed is 52 km/h. Check: 52 km/h is between 40 and 60 km/h, and nearer 60, because the van spent more hours at 60 km/h.
(b) The 260 km shared equally over the 5 hours: 260 ÷ 5 = 52 km in each hour, so 52 km/h.
Answer: (a) 5 hours; (b) 52 km/h
Watch out for
- Taking the mean of the two speeds: (40 + 60) ÷ 2 = 50 km/h. The van drove for 2 hours at 40 km/h but for 3 hours at 60 km/h, so the faster speed lasted longer and pulls the average above 50 km/h.
- Adding the two speeds: 40 + 60 = 100 km/h. The van never went faster than 60 km/h, so its average speed cannot be more than 60 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.
Speed · A Journey Slowed by Traffic After the First Third
QuestionMr. Tan drove 360 km from City A to City B. For the first 13 of the distance, he drove at an average speed of 80 km/h. Heavy traffic then slowed him down for the rest of the way. The whole journey took 512 hours. (a) How long did the first 13 of the journey take? (b) What was his average speed for the rest of the journey?
1.Cut the 360 km into 3 equal units. Each unit is 360 ÷ 3 = 120 km.
Three equal units of 360 ÷ 3 = 120 km. 2.(a) He drove the first 120 km at 80 km/h, which took 120 ÷ 80 = 112 hours.
(a) The first unit at 80 km/h: 120 ÷ 80 = 112 hours. 3.The rest of the journey is the other 2 units: 2 × 120 = 240 km.
The other two units are 2 × 120 = 240 km. 4.The whole journey took 512 hours, so the rest took 512 − 112 = 4 hours.
Of the 512 hours, 512 − 112 = 4 hours are left for those 240 km. 5.(b) His average speed for the rest was 240 ÷ 4 = 60 km/h. Check: 112 hours at 80 km/h and 4 hours at 60 km/h cover 120 + 240 = 360 km.
(b) Speed for the rest: 240 ÷ 4 = 60 km/h.
Answer: (a) 1.5 hours; (b) 60 km/h
Watch out for
- Taking 13 of the time instead of 13 of the distance: 512 ÷ 3 hours at 80 km/h. The question says the first 13 of the distance, so it is 120 km, and its time is 120 ÷ 80 = 112 hours.
- Taking the 112 hours away from 5 hours instead of 512 hours, and getting 312 hours for the rest. The whole journey took 512 hours, so the rest took 4 hours.
Your turn
1.120 km at 60 km/h, then 120 km at 40 km/h. Average speed?