# The cyclist covered the first half of the journey at a speed of V1 = 10 km / h. Then he went at a higher speed

**The cyclist covered the first half of the journey at a speed of V1 = 10 km / h. Then he went at a higher speed, but punctured the tire. After trying to fix the puncture, the cyclist was forced to walk the rest of the way. What is the average speed of a cyclist along the entire path, if the first third of the time spent on the second half of the journey, he was driving at a speed of V2 = 20 km / h, the second third was engaged in a puncture, and the last third was on foot at a speed of V4 = 5 km / h ?**

To calculate the average speed of the presented cyclist, we will use the formula: Vav = S / (t1 + t2) = S / (0.5S / V1 + 0.5S / ((V2 + V3 + V4) / 3)) = 1 / (1 / 2V1 + 1 / (2 * (V2 + V3 + V4) / 3)).

Variables: V1 is the speed of the presented cyclist in the first half of the journey (V1 = 10 km / h); V2 – high speed at the beginning of the second half of the journey (V2 = 20 km / h); V3 – speed during puncture repair (V3 = 0 km / h); V4 – walking speed (V4 = 5 km / h).

Calculation: Vav = 1 / (1 / (2 * 10) + 1 / (2 * (20 + 0 + 5) / 3)) = 1 / 0.11 = 9.09 km / h.

Answer: The average speed of the presented cyclist along the entire route was 9.09 km / h.