Task 1. Two cars covered the same distance. The speed of the first was constant and three times less than the initial speed of the second. The second car traveled the first half of the distance without changing its speed, then it suddenly reduced its speed by half, traveled another quarter of the distance at a constant speed, and then again reduced its speed by half, traveling another eighth of the distance at a constant speed, and so on. After the eighth reduction in speed, it did not change its speed until the end of the trip. How many times more time did the second car need to cover the entire distance compared to the first?
Answer: $\frac{5}{3}$.

Solution. Let $V_{1}$ be the speed of the first car, $V_{0}$ the initial speed of the second, and the distance covered by each car be $S$. Then $V_{0}=3 V_{1}$. The first car spent the entire journey time $t_{1}=\frac{S}{V_{1}}$, while the second car spent time

$$
t_{2}=\frac{\frac{S}{2}}{V_{0}}+\frac{\frac{S}{4}}{\frac{V_{0}}{2}}+\ldots+\frac{\frac{S}{2^{8}}}{\frac{V_{0}}{2^{7}}}+\frac{\frac{S}{2^{8}}}{\frac{V_{0}}{2^{8}}}=\frac{S}{2 V_{0}} \cdot 8+\frac{S}{V_{0}}=\frac{S}{6 V_{1}} \cdot 8+\frac{S}{3 V_{1}}=\frac{5 S}{3 V_{1}}
$$

from which $\frac{t_{2}}{t_{1}}=\frac{5}{3}$.