Average speed question: forwards then backwards

1. Nov 9, 2014

nobahar

1. The problem statement, all variables and given/known data
An object starts at the origin, position A; then travels 20 m to position B; then returns to position A. From A to B the average speed is 10 m/s; on its way back the average speed is 6 m/s. What is the average speed for the entire trip?

2. Relevant equations
average speed = distance/time
distance is the "length" covered. This is not - according to my understanding - the same as velocity, which is displacement/time. In this example, the velocity would by 0 m/s, because the numerator, the displacement, is zero, as the object has returned to the origin.

3. The attempt at a solution
From A to B the distance is 20 m and the average speed is 10 m/s. Therefore:
20 m/time = 10 m/s and time = 20 m/ 10 m/s = 2 s.
From B to A the distance is 20 m and the average speed is 6 m/s. Therefore:
20 m/time = 6 m/s and time = 20 m/ 6 m/s = 3.33 s.
Overall, the journey from A to B and back again takes (2 + 3.33) s = 5.33 s.
The total distance covered (not displacement) is 40 m (20 m from A to B and 20 m from B to A).
Therefore, the average speed = distance/time = 40 m/ 5.33 s = 7.50 m/s (approx.).

I have changed the numbers from the textbook to see if I was making some type of numerical error, but I have applied the same reasoning. The textbook has a different value from what I obtain (it is lower). Where am I going wrong?

Any help appreciated, many thanks.

2. Nov 9, 2014

HallsofIvy

Staff Emeritus

3. Nov 10, 2014

nobahar

In the textbook the value from A to B is 8.8 m/s and from B to A it is 6.6 m/s. They give the answer as something like 2.6 m/s. I got 7.54 (to 2 d.p.):
40 m / [(20/8.8) + (20/6.6)] s = 7.542857143 m/s.

4. Nov 10, 2014

nasu

2.6 m/s does not make sense. It cannot be less than the 6.6 m/s, the lowest of the two average speeds.

5. Nov 16, 2014

nobahar

Thanks for the reply. I guess it's just an error in the textbook.