Why Isn't Average Speed Simply the Arithmetic Mean of Two Speeds?

[Natasha1]

Homework Statement

If you walk to school at 4 mph and jog home at 6mph, what was your average speed?

Homework Equations

S = D/T

The Attempt at a Solution

Is the average speed not just (4 + 6)/2 = 5mph ?

If not, why not?

 

[mjc123]

Work it out. Suppose the school is 2 miles away. How far have you traveled there and back? how long did it take? What was your average speed?

 

[Natasha1]

Work it out. Suppose the school is 2 miles away. How far have you traveled there and back? how long did it take? What was your average speed?

4 miles in total
way there = 2/4 = 30 mins
way back = 2/6 = 20 mins

Speed = 4/(50/60) = 4.8 mph

Is this correct?

 

[mjc123]

Yes. Technically, the average speed is the harmonic mean (not the arithmetic mean) of the two speeds, i.e. 1/S =1/2*(1/S₁ + 1/S₂). Or more generally, when the distances are not equal,
S = (D₁ + D₂)/(T₁ + T₂)
(D₁ + D₂)/S = T₁ + T₂ = D₁/S₁ + D₂/S₂
Hence S is the harmonic mean of S₁ and S₂ when weighted by distance, as is commonly the case in questions like "go there at one speed and back at another". However, we can also write
(T₁ + T₂)*S = D₁ + D₂ = T₁S₁ + T₂S₂
S = (T₁S₁ + T₂S₂)/(T₁ + T₂)
So S is the arithmetic mean speed when weighted by time. The common error is to assume it is the arithmetic mean when weighted by distance, which is not true.