Question

Прогулочный катер вышел из пункта А и прошел по течению реки до пункта В 32 км где развернулся и отправился в обратный путь. Таким образом он вновь оказался в пункте А спустя 6 часов после начала движения. определите собственную скорость катера в (км/ч)если скорость течения реки составляет 4 км/ч. решение подробное)) решите пжл

Answer 1

S=36км
Vт.=4км/ч
x=скорость катера
32/(X+4) +32/X-4=6
6X^2-64x-96=0

3x^2-32x-48=0

D=1024+576=1600=40^2

x1=(32+40)/6=12

x2=(32-40)/6=-1.166666!-не возмож.

Ответ:12 км/ч

Answer 2

Problem Analysis

We are given that a pleasure boat traveled from point A to point B along the current of a river, covering a distance of 32 km. The boat then turned around and traveled back to point A, reaching it 6 hours after the start of the journey. We need to determine the speed of the boat in km/h, given that the speed of the river current is 4 km/h.

Solution

Let's assume the speed of the boat is x km/h. Since the boat is traveling downstream from point A to point B, the effective speed of the boat will be the sum of its own speed and the speed of the river current. Therefore, the effective speed of the boat while traveling downstream is (x + 4) km/h.

The time taken to travel from point A to point B is given by the formula:

Time = Distance / Speed

Substituting the values, we have:

32 / (x + 4) = t1 (Equation 1)

where t1 is the time taken to travel from point A to point B.

Next, when the boat turns around and travels upstream from point B to point A, the effective speed of the boat will be the difference between its own speed and the speed of the river current. Therefore, the effective speed of the boat while traveling upstream is (x - 4) km/h.

The time taken to travel from point B to point A is given by the formula:

Time = Distance / Speed

Substituting the values, we have:

32 / (x - 4) = t2 (Equation 2)

where t2 is the time taken to travel from point B to point A.

We are given that the total time taken for the round trip is 6 hours. Therefore, we can write:

t1 + t2 = 6 (Equation 3)

Now, we have three equations (Equations 1, 2, and 3) with three unknowns (x, t1, and t2). We can solve these equations simultaneously to find the value of x, which represents the speed of the boat.

Let's solve the equations:

From Equation 1, we have:

t1 = 32 / (x + 4)

From Equation 2, we have:

t2 = 32 / (x - 4)

Substituting these values into Equation 3, we have:

32 / (x + 4) + 32 / (x - 4) = 6

To simplify the equation, we can multiply both sides by (x + 4)(x - 4) to eliminate the denominators:

32(x - 4) + 32(x + 4) = 6(x + 4)(x - 4)

Simplifying further:

32x - 128 + 32x + 128 = 6(x^2 - 16)

64x = 6x^2 - 96

Rearranging the equation:

6x^2 - 64x - 96 = 0

Now, we can solve this quadratic equation to find the value of x.

Using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

where a = 6, b = -64, and c = -96.

Substituting the values, we have:

x = (-(-64) ± sqrt((-64)^2 - 4 * 6 * (-96))) / (2 * 6)

Simplifying further:

x = (64 ± sqrt(4096 + 2304)) / 12

x = (64 ± sqrt(6400)) / 12

x = (64 ± 80) / 12

We have two possible solutions:

x1 = (64 + 80) / 12 = 144 / 12 = 12

x2 = (64 - 80) / 12 = -16 / 12 = -4/3

Since the speed of the boat cannot be negative, we discard the negative solution.

Therefore, the speed of the boat is 12 km/h.

Answer

The speed of the pleasure boat is 12 km/h.