At 11:00 the boat left point A to point B, located 15 km from A. Having passed the puncture at 1 hour 20 minutes

At 11:00 the boat left point A to point B, located 15 km from A. Having passed the puncture at 1 hour 20 minutes, the boat set off back and returned back to point A at 15:00. Determine (in km / h) the boat’s own speed if it is known that the speed of the river flow is 2 km / h.

Let’s designate the speed of the boat through “k”, the speed downstream and against it will be: (k + 3) and (k – 3), respectively.

The time for passing along the river and back is (15 – 11 – 1 1/3) = 4 – 4/3 = 8/3 (hours).

Let’s make an equation, knowing the time for moving along the river and back is equal to 8/3 hours.

15 / (k + 3) + 15 / (k – 3) = 8/3; 15 * (k + 3 + k – 3) * 3 = 8 * (k ^ 2- 9);

15 * 2 * k * 3 = 8 * (k ^ 2 – 9); 8 * k ^ 2 – 90 * k – 72 = 0; (divide by 8)

k ^ 2 – 45/4 * k – 9 = 0; k1,2 = 45/8 + – √ [(45/8) ^ 2 + 9] =

45/8 + – √ (2025 + 64 * 9) / 9 = 45/8 + – √2601 / 64 = 45/8 + – 51/8.

Positive root k1 = 96/8 = 12 (km / h).