Question

Катер прошёл 28 км по течению и 12 км против течения, затратив на весь путь 2,15 ч.Какова скорость в стоячей воде, если скорость течения реки равна 2км.ч??

Answer 1

Пусть скорость катера х км/ч, тогда на путь по течению он затратил 28/(х+2) часа, на путь против течения 12/(х-2) часа общее время 2,15 часа (надеюсь правильно записано в условии) тогда вот такое уравнение
28/(х+2)+12/(х-2)=2,15

Answer 2

#### Calculation of Speed in Still Water To calculate the speed of the boat in still water, we can use the concept of relative velocity. The speed of the boat in still water can be represented as the average of the speeds when going upstream (against the current) and downstream (with the current). Let's denote the speed of the boat in still water as **x** km/h, and the speed of the river current as **2** km/h. Given that the boat traveled **28** km downstream (with the current) and **12** km upstream (against the current), and the total time taken for the entire journey was **2.15** hours, we can set up the following equation: **28 / (x + 2) + 12 / (x - 2) = 2.15** To solve this equation, we can multiply both sides by the least common multiple (LCM) of the denominators to eliminate the fractions. In this case, the LCM is **(x + 2)(x - 2)**. After simplifying and rearranging the equation, we get: **28(x - 2) + 12(x + 2) = 2.15(x + 2)(x - 2)** Solving this equation will give us the value of **x**, which represents the speed of the boat in still water. Let's calculate the value of **x** using the given equation. #### Calculation: 28(x - 2) + 12(x + 2) = 2.15(x + 2)(x - 2) 28x - 56 + 12x + 24 = 2.15(x^2 - 4) 40x - 32 = 2.15x^2 - 8.6 2.15x^2 - 40x + 23.4 = 0 Using the quadratic formula, we can find the value of **x**: **x = (-b ± √(b^2 - 4ac)) / (2a)** where **a = 2.15**, **b = -40**, and **c = 23.4**. Calculating the discriminant: **D = b^2 - 4ac = (-40)^2 - 4(2.15)(23.4) = 1600 - 199.2 = 1400.8** Since the discriminant is positive, we have two real solutions for **x**. Calculating the solutions: **x = (-(-40) ± √(1400.8)) / (2 * 2.15)** **x = (40 ± √(1400.8)) / 4.3** **x ≈ 11.30** or **x ≈ -3.98** Since the speed of the boat cannot be negative, we can discard the negative solution. Therefore, the speed of the boat in still water is approximately **11.30 km/h**. #### Answer: The speed of the boat in still water is approximately **11.30 km/h**.