1. Jul 17, 2017

Natasha1

1. The problem statement, all variables and given/known data
If you walk to school at 4 mph and jog home at 6mph, what was your average speed?

2. Relevant equations
S = D/T

3. The attempt at a solution
Is the average speed not just (4 + 6)/2 = 5mph ?

If not, why not?

2. Jul 17, 2017

mjc123

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

3. Jul 17, 2017

Natasha1

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?

4. Jul 17, 2017

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.