# Task 2.

While a lion cub, who was 6 minutes away from the water hole, set off to drink, a second one, having already quenched his thirst, started heading back along the same path at 1.5 times the speed of the first. At the same time, a turtle, who was 32 minutes away from the water hole, set off along the same path to drink. After some time, the first lion cub stepped on the turtle, and 2 minutes and 24 seconds later, the second lion cub stepped on her. How many minutes after the second incident did the turtle reach the water hole, given that all three moved at constant speeds?
Answer: 28.8

Solution. Let $x$ be the speed of the turtle, and $y$ be the speed of the first lion cub. Then the speed of the second lion cub is $1.5 y$. The entire path to the watering hole for the first lion cub is $6 y$, and for the turtle, it is $32 x$. Therefore, the initial distance between them was $6 y - 32 x$, and the first incident occurred after $(6 y - 32 x) / (y - x)$ minutes. The second incident occurred after $32 x / (x + 1.5 y)$ minutes. Therefore,

$$
\frac{32 x}{x + 1.5 y} - \frac{6 y - 32 x}{y - x} = 2.4 \text{, from which } 2.4 x^2 + 75.2 x y - 12.6 y^2 = 0
$$

or $63(y / x)^2 - 376(y / x) - 12 = 0$. This quadratic equation in terms of $y / x$ has roots of different signs. The positive root is 6. Thus, $y = 6 x$ and the time for the turtle to reach the watering hole after the second incident is

$$
32 - \frac{32 x}{x + 1.5 y} = 32 - \frac{32 x}{x + 9 x} = 32 - 3.2 = 28.8 \text{ min. }
$$

## B-2

At the time when a lion cub, located 5 minutes from the watering hole, set off for it, the second, having quenched his thirst, set off back along the same path at 1.5 times the speed of the first. At the same time, a turtle, located 33 minutes from the watering hole, set off along the same path. After some time, the first lion cub stepped on the turtle, and 3 minutes and 54 seconds later, the second lion cub did the same. How many minutes after the second incident did the turtle reach the watering hole, given that all three moved at constant speeds?

## Answer: 29.7

## B-3

At the time when a lion cub, located 6 minutes from the watering hole, set off for it, the second, having quenched his thirst, set off back along the same path at 2 times the speed of the first. At the same time, a turtle, located 33 minutes from the watering hole, set off along the same path. After some time, the first lion cub stepped on the turtle, and 42 seconds later, the second lion cub did the same. How many minutes after the second incident did the turtle reach the watering hole, given that all three moved at constant speeds?

Answer: 30.8

## B-4

At the time when a lion cub, located 5 minutes from the watering hole, set off for it, the second, having quenched his thirst, set off back along the same path at 2 times the speed of the first. At the same time, a turtle, located 36 minutes from the watering hole, set off along the same path. After some time, the first lion cub stepped on the turtle, and 2 minutes and 34 seconds later, the second lion cub did the same. How many minutes after the second incident did the turtle reach the watering hole, given that all three moved at constant speeds?

Answer: 33.6

## B-5

A crocodile swam across a river 30 meters wide in a straight line in 6 seconds. During this time, it was carried downstream by 36 meters. After resting and gathering strength, it returned to the same point along the same straight line, but this time fighting the current, in 10 seconds. How many seconds would it take for the crocodile to swim the same distance in still water? Round the answer to the nearest hundredth.

## Answer: 7.34

Solution. We will provide a general solution for the following problem statement.

A crocodile swam across a river $a$ meters wide in a straight line in $t$ seconds. During this time, it was carried downstream by $b$ meters. After resting and gathering strength, it returned to the same point along the same straight line, but this time fighting the current, in $T$ seconds. How many seconds would it take for the crocodile to swim the same distance in still water? Round the answer to the nearest hundredth.

In this formulation, the required time is $\sqrt{a^2 + b^2} / v$, where $v$ is the speed of the crocodile in still water. The longitudinal and transverse components of the speed are $u_1 = b / t$, $w_1 = a / t$ in the first case and $u_2 = b / T$, $w_2 = a / T$ in the second case. The above longitudinal components of the speed were obtained as the sum or difference of the longitudinal component of the crocodile's own speed and the speed of the current. The transverse and longitudinal components of the crocodile's own speed can differ in the first and second cases, but in each case, the sum of their squares is equal to $v^2$. Let $u$ be the speed of the current. Then we have the system

$$
\left\{
\begin{array}{l}
\left(u_1 - u\right)^2 + w_1^2 = v^2 \\
\left(u_2 + u\right)^2 + w_2^2 = v^2
\end{array}
\right.
$$

Subtracting these equations, we get the equation from which we can find the speed of the current:

$$
u = \frac{w_1^2 - w_2^2 + u_1^2 - u_2^2}{2(u_1 + u_2)} = \frac{a^2\left(\frac{1}{t^2} - \frac{1}{T^2}\right) + b^2\left(\frac{1}{t^2} - \frac{1}{T^2}\right)}{2 b\left(\frac{1}{t} + \frac{1}{T}\right)} = \frac{(a^2 + b^2)(T - t)}{2 b T t}
$$

After this, we can find the square of the crocodile's own speed:

$$
\begin{array}{r}
v^2 = \left(\frac{b}{T} + \frac{(a^2 + b^2)(T - t)}{2 b T t}\right)^2 + \left(\frac{a}{T}\right)^2 = \frac{(2 b^2 t + (a^2 + b^2)(T - t))^2 + 4 a^2 b^2 t^2}{(2 b T t)^2} = \\
= \frac{4 a^2 b^2 t^2 + 4 b^4 t^2 + (a^2 + b^2)^2(T - t)^2 + 4 b^2 t(a^2 + b^2)(T - t)}{(2 b T t)^2} = \\
= \frac{(a^2 + b^2)}{(2 b T t)^2}(4 b^2 t^2 + 4 b^2 t(T - t) + (a^2 + b^2)(T - t)^2)
\end{array}
$$

and then the required time:

$$
\sqrt{\frac{a^2 + b^2}{v^2}} = \frac{2 b T t}{\sqrt{4 b^2 t^2 + 4 b^2 t(T - t) + (a^2 + b^2)(T - t)^2}}
$$

## B-6

A crocodile swam across a river 30 meters wide in a straight line in 6 seconds. During this time, it was carried downstream by 24 meters. After resting and gathering strength, it returned to the same point along the same straight line, but this time fighting the current, in 15 seconds. How many seconds would it take for the crocodile to swim the same distance in still water? Round the answer to the nearest hundredth.

Answer: 7.56

## B-7

A crocodile swam across a river 36 meters wide in a straight line in 6 seconds. During this time, it was carried downstream by 24 meters. After resting and gathering strength, it returned to the same point along the same straight line, but this time fighting the current, in 9 seconds. How many seconds would it take for the crocodile to swim the same distance in still water? Round the answer to the nearest hundredth.

Answer: 6.90

## B-8

A crocodile swam across a river 56 meters wide in a straight line in 8 seconds. During this time, it was carried downstream by 42 meters. After resting and gathering strength, it returned to the same point along the same straight line, but this time fighting the current, in 14 seconds. How many seconds would it take for the crocodile to swim the same distance in still water? Round the answer to the nearest hundredth.

Answer: 9.57

## B-9

A crocodile swam across a river 42 meters wide in a straight line in 6 seconds. During this time, it was carried downstream by 28 meters. After resting and gathering strength, it returned to the same point along the same straight line, but this time fighting the current, in 14 seconds. How many seconds would it take for the crocodile to swim the same distance in still water? Round the answer to the nearest hundredth.

Answer: 7.20

## B-10

A crocodile swam across a river 36 meters wide in a straight line in 6 seconds. During this time, it was