Question

Катер поплыл 18 км по течению реки и вернулся обратно, потратив на весь путь по течению на 48 мин меньше, чем на путь против течения.найдите собс.скорость катера, если скорость течения реки 3 км/ч

Answer 1

Х - соб скорость
18/х+3 - время по течению
18/х-3 - время против течения
48/60=8/10=4/5 - 48 мин в часах
18/х-3 - 18/х+3 = 4/5
18*5(х+3)-18*5(х-3)=4(х+3)(х-3)
90х+270-90х+270=4х²-36
4х²=576
х²=144
х=12
соб скорость катера 12км\ч

Answer 2

Problem Analysis

We are given that a boat traveled 18 km downstream and then returned upstream, spending 48 minutes less on the downstream journey than on the upstream journey. We need to find the speed of the boat.

Let's assume the speed of the boat in still water is x km/h, and the speed of the river's current is 3 km/h.

Downstream Journey

During the downstream journey, the boat's speed is increased by the speed of the river's current, so the effective speed is (x + 3) km/h.

The time taken for the downstream journey can be calculated using the formula: time = distance / speed.

Therefore, the time taken for the downstream journey is 18 / (x + 3) hours.

Upstream Journey

During the upstream journey, the boat's speed is decreased by the speed of the river's current, so the effective speed is (x - 3) km/h.

The time taken for the upstream journey is 18 / (x - 3) hours.

Time Difference

We are given that the time taken for the downstream journey is 48 minutes (or 0.8 hours) less than the time taken for the upstream journey.

Therefore, we can set up the equation: (18 / (x + 3)) = (18 / (x - 3)) + 0.8.

Solving the Equation

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

18(x - 3) = 18(x + 3) + 0.8(x + 3)(x - 3)
18x - 54 = 18x + 54 + 0.8(x^2 - 9)
18x - 54 = 18x + 54 + 0.8x^2 - 7.2
18x - 18x - 54 - 54 = 0.8x^2 - 7.2
0.8x^2 - 108 = 0
0.8x^2 = 108
x^2 = 135
x = sqrt(135)
x ≈ 11.62

Answer

The speed of the boat in still water is approximately 11.62 km/h.