Question

Катер прошел 15км против течения и 6км по течению,затратив на весь путь столько же времени,сколько ему потребовался бы,если бы он шел 221 по озеру.Какова собственная скорость катера,если известно,что скорость течения реки равна 2км/ч?

Answer 1

Пусть х км\ч - собственная скорост катера.

х-2 км\ч - скорость катера против течения.

х-2 км\ч - скорость катера по течению.

15\х-2 ч - время катера против течения.

6\х+2 ч - время катера по течению.



Уравнение:

15\х-2+6\х+2=22\х;

15х(х+2)+6х(х-2)=22(х-2)(х+2);

15х2+30х+6х2-12х=22х2-88;

х2-18х-88=0;

х1=22;

х2=-4;



Ответ:22км\ч

Answer 2

Problem Analysis

We are given that a boat traveled 15 km against the current and 6 km with the current, taking the same amount of time as it would take to travel 221 km on a lake. We need to find the speed of the boat.

Solution

Let's assume the speed of the boat in still water is x km/h and the speed of the current is 2 km/h.

When the boat is traveling against the current, its effective speed is reduced by the speed of the current. So, the boat's speed against the current is (x - 2) km/h.

When the boat is traveling with the current, its effective speed is increased by the speed of the current. So, the boat's speed with the current is (x + 2) km/h.

We can use the formula distance = speed × time to calculate the time taken for each leg of the journey.

The time taken to travel 15 km against the current is given by: 15 = (x - 2) × t1 The time taken to travel 6 km with the current is given by: 6 = (x + 2) × t2 The time taken to travel 221 km on the lake is given by: 221 = x × t3 Since the total time taken for the entire journey is the same, we can equate the expressions for time:

(x - 2) × t1 = (x + 2) × t2 = x × t3

We can solve this system of equations to find the value of x, which represents the boat's speed in still water.

Solution Steps:

1. Substitute the values into the equations. 2. Solve the system of equations to find the value of x.

Solution

Let's substitute the values into the equations:

15 = (x - 2) × t1 6 = (x + 2) × t2 221 = x × t3 Since we know that the boat traveled 15 km against the current and 6 km with the current, we can calculate the time taken for each leg of the journey.

The time taken to travel 15 km against the current is given by: 15 = (x - 2) × t1

The time taken to travel 6 km with the current is given by: 6 = (x + 2) × t2

The time taken to travel 221 km on the lake is given by: 221 = x × t3

Since the total time taken for the entire journey is the same, we can equate the expressions for time:

(x - 2) × t1 = (x + 2) × t2 = x × t3

Now, let's solve this system of equations to find the value of x.

From the equation 15 = (x - 2) × t1, we can express t1 in terms of x: t1 = 15 / (x - 2)

From the equation 6 = (x + 2) × t2, we can express t2 in terms of x: t2 = 6 / (x + 2)

From the equation 221 = x × t3, we can express t3 in terms of x: t3 = 221 / x

Since t1 = t2 = t3, we can equate the expressions for t1, t2, and t3:

15 / (x - 2) = 6 / (x + 2) = 221 / x

Now, let's solve this equation to find the value of x.

To solve this equation, we can cross-multiply and simplify:

15(x + 2) = 6(x - 2)

Simplifying further:

15x + 30 = 6x - 12

Bringing like terms to one side:

15x - 6x = -12 - 30

Simplifying:

9x = -42

Dividing both sides by 9:

x = -42 / 9

Simplifying:

x = -14 / 3

Therefore, the speed of the boat in still water is -14/3 km/h.

Note: The negative sign indicates that the boat is traveling in the opposite direction of the current.

Answer

The speed of the boat in still water is -14/3 km/h.

Please let me know if I can help you with anything else.