A car travels in a straight line through points A, B and C, which are spaced 12 km apart. It covers AB in 20 minutes and BC in 30 minutes. Compare the average speeds on AB, BC and AC.

Step-by-step solution

1. 1

   Convert each segment time to hours. $20\text{ min}=\tfrac13\text{ h}$ and $30\text{ min}=\tfrac12\text{ h}$, so the results will come out in $\text{km/h}$.

2. 2

   Compute the speed on AB. $v_{AB}=\frac{12}{\tfrac13}=36\text{ km/h}$

3. 3

   Compute the speed on BC. $v_{BC}=\frac{12}{\tfrac12}=24\text{ km/h}$ so $v_{AB}>v_{BC}$: the car was faster on the first segment.

4. 4

   Compute the overall speed on AC. The whole trip is $24\text{ km}$ in $50\text{ min}=\tfrac56\text{ h}$: $v_{AC}=\frac{24}{\tfrac56}=\frac{144}{5}=28.8\text{ km/h}$

5. 5

   Read off the ordering. $v_{BC}=24<v_{AC}=28.8<v_{AB}=36$. The overall average lies strictly between the two segment speeds — it is not the arithmetic mean $30\text{ km/h}$, because more time was spent on the slower segment.