# Average speed question: forwards then backwards

## Homework Statement

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?

## Homework 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.

## 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.

HallsofIvy
Science Advisor
Homework Helper

## Homework Statement

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?

## Homework 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.

## 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.
Yes, it takes 2 seconds to go 20 m at 10m/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.
I would have written this as 20/6= 10/3 s.

Overall, the journey from A to B and back again takes (2 + 3.33) s = 5.33 s.
I would have written it as 2+ 10/3= (6+ 10)/3= 16/3 s (or 5 and 1/3 seconds)

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.).
40/(16/3)= 40(3/16)= 120/16= 16/2= 7.5 m/s (NOT "approximate).

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.
You don't say what value was in the textbook or what the numbers in the problem actually were but you don't seem to have done anything wrong here.

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.

nasu
Gold Member
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.

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