# The motor ship goes along the river to its destination 780 km and after stopping returns to the point of departure.

**The motor ship goes along the river to its destination 780 km and after stopping returns to the point of departure. Find the speed of the current if the speed of the ship in still water is 28 km / h, the stay lasts 4 hours and the ship returns to the point of departure 60 hours after leaving it.**

Let us designate the V course of the river as Vflow. – x km / h, then V of the ship downstream V downstream. = (28 + x) km / h, and V of the ship against the current Vpr. tech. = (28 – x) km / h.

The distance between the points is 780 km, let us express tpo. and tpr. tech .:

t by current. = S: Von flow = 780: (28 + x);

tpr. tech. = S: Vpr. tech = 780: (28 – x).

The entire voyage lasted 60 hours, of which the stay was 4 hours, which means that the ship was in motion for 56 hours.

Let us express the time spent by the motor ship on the way by the equation:

tpo current + tpr. tech. = 56;

780 / (28 + x) + 780 / (28 – x) = 56;

We bring to a common denominator:

[780 · (28 – x) + 780 · (28 + x) – 56 (28 – x) · (28 + x)] / [(28 – x) · (28 + x)] = 0;

The fraction is 0 when its numerator is 0:

780 (28 – x) + 780 (28 + x) – 56 (28 – x) (28 + x) = 0;

780 56 – 56 (282 – x ^ 2) = 0;

56 * (780- 784 + x ^ 2) = 0;

x ^ 2 – 4 = 0;

x ^ 2 = 4;

x1 = – 2 – does not satisfy the condition of the problem;

x2 = – 2 (km / h) – Vcur.

Answer: the speed of the river is 2 km / h.