Question

Answer!
A car travels from A to B at a speed of 30 mph then returns, using the same road, from B to A at a speed of 65 mph. What is the average speed for the round trip?

Answer 1

Here is an example for your question



Average with respect to what? If we take the time-weighted average, the two legs of the trip will be weighted differently (the leg from A to B taking 60/40 = 3/2 times as long as the leg from B to A), and we will get an average of 48 MPH.

But the time-weighted average is not the only average we can consider. We could also consider the distance-weighted average (which might be useful if we were considering, say, the wear on the road caused by such travel); then the two legs would be weighted equally and the average would be 50 MPH.

Often, when people speak about average speed, it's the former of these notions which is implicitly meant (since speed itself is distance over time, we have the nice property that the time-weighted average of speeds will always be the total distance over the total time), but the latter is a perfectly legitimate notion as well. Averages are always averages with respect to some weighting, and different weightings will give different averages.

Answer 2

Answer:

         41 mph

Step-by-step explanation:

s₁ = s₂ = x       {the distance from A to B}

s = s₁ + s₂ = x + x = 2x      {the distance of the round trip}

t = t₁ + t₂               {the time of the round trip}

v₁ = 30 mph      

v₂ = 65 mph

t₁ = s₁/v₁ = x/30 h       {the time of trip from A to B}

t₂ = s₂/v₂ = x/65 h      {the time of trip from B to A}

The average speed:

$v=\dfrac st=\dfrac{2x}{\frac x{30}+\frac x {65}}=\dfrac{2x}{x(\frac 1{30}+\frac1 {65})}=\dfrac{2}{\frac{13}{390}+\frac6{390}}=2\cdot\frac{390}{19}=41.05263...\approx41$