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?
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.
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.