The speed of the river flow is 1 km / h. The distance between points A and B

The speed of the river flow is 1 km / h. The distance between points A and B along the river is 12 km. The motor boat travels from A and B and back in 2 hours 12 minutes. Find the speed of the boat in still water.

1. Let x km / h be the boat speed in still water. Since there is no river of its own in stagnant water, the boat moves only at its own speed. Therefore, the boat’s speed in still water and the boat’s own speed are the same.

2. By the condition of the problem, the speed of the river flow is 1 km / h. Then the speed downstream of the river can be written as (x + 1) km / h, and the speed upstream (x – 1) km / h.

3. Let the flow of the river have a direction from A to B. The time during which the boat passes this path can be written as 12 / (x + 1). Then the time it takes for the boat to travel back (from B to A) can be written as 12 / (x – 1).

4. It is known that the boat travels there and back in 2 hours 12 minutes. Or in 2.2 hours (12 minutes / 60 minutes = 0.2). Then you can write the following equation:

12 / (x – 1) + 12 / (x + 1) = 2.2,

(12 * (x + 1) + 12 * (x – 1)) / (x – 1) / (x + 1) = 2.2,

12x + 12 + 12x – 12 = 2.2 (x + 1) (x – 1),

24x = 2.2 (x * x – 1),

12x = 1.1 (x * x – 1), multiply both sides of the equation by 10 to arrive at integer coefficients,

120x = 11x * x -11,

11x * x – 120x – 11 = 0,

a = 11, b = -120, k = b / 2 = -60, c = -11,

D / 4 = D1 = k * k – ac = 60 * 60 – (-11) * 11 = 3721 = 61 * 61.

x1 = (-k + √ (D1)) / a = (60 + 61) / 11 = 11,

x2 = (-k – √ (D1)) / a = (60 – 61) / 11 = -1/11.

The equation has two roots. Since the boat speed is positive, we leave the positive root x = 11 km / h.

Answer: boat speed in still water is 11 km / h.