In physics, two cyclists start out on the same trail together that is overall 20km long. At what point along the path should the second cyclist...

Cyclist A completes the entire 40km course -- she travels the first 20 km at 20km/hr, and the final leg at 15 km/hr.

Cyclist B will not complete the entire course and travels at a constant 10km/hr.

We are asked to find the point at which cyclist B should turn around in order to meet cyclist A at the starting point.

I. One way to approach this is to find the time spent on the...

Cyclist A completes the entire 40km course -- she travels the first 20 km at 20km/hr, and the final leg at 15 km/hr.

Cyclist B will not complete the entire course and travels at a constant 10km/hr.

We are asked to find the point at which cyclist B should turn around in order to meet cyclist A at the starting point.

I. One way to approach this is to find the time spent on the course for the cyclists. The time will be the same for each cyclist as they end at the starting point at the same time.

Cyclist A travels the first 20km at 20 km/hr which takes 1hr. The return trip at 15km/hr requires 4/3 hour (80 minutes). So the entire trip takes 7/3 hr (140 minutes.)

Cyclist B will also need to travel for the same time: the distance travelled at 10km/hr for 7/3 hours is 70/3km. This is the total distance -- to ascertain where Cyclist B should stop we divide by 2 to get 35/3 km (or 11 2/3 km).

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Cyclist B should stop at the 11 2/3 km mark and return to the starting point to meet cyclist A.

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II. An alternative is to find the "average" speed of cyclist A to determine the timne spent on the course. You must realize that you cannot take the arithmetic mean of the rates, you must use the harmonic mean.

Here the average speed for cyclist A is (2*20*15)/(20+15)=120/7 km/hr.

At this rate, cyclist A will take (40km)/(120/7 km/hr)=7/3 hour and you can proceed as above. (Note that the arithmetic mean is (20+15)/2=17.5 while the harmonic mean is 120/7 or approximately 17.14285714...)

The important point is that you must use the harmonic mean when averaging rates.