Question

Теплоход прошёл 28 км против течения реки и вернулся обратно потратив на обратный путь на 4 мин меньше. Найдите скорость теплохода в стоячей воде, если скорость течения реки равна 1км/ч

Answer 1

Дано:                                                  Решение:
S = 28 км
t₂ = t₁ - 1/15 ч                Скорость теплохода против течения:
v₀ = 1 км/ч                                            v₁ = v - v₀ = S/t₁
------------------                Скорость теплохода по течению:
Найти: v - ?                                          v₂ = v + v₀ = S/t₂ = S/(t₁-1/15)
                            Тогда:
                                  v₂ - v₁ = 2v₀ = S/(t₁-1/15) - S/t₁
                                              (St₁-S(t₁-1/15))/(t₁²-t₁/15) = 2
                                               St₁-S(t₁-1/15) = 2(t₁²-t₁/15)
                                               28t₁-28t₁+28/15 = 2t₁²-2t₁/15
                                               2t₁²-2t₁/15-28/15 = 0
                                               15t₁² - t₁ - 14 = 0      D=b²-4ac=1+840=841=29²
                                   t₁₁ = (1-29)/30 = -28/30 - противоречит условию
                                   t₁₂ = (1+29)/30 = 1 (ч)      
                  
v₁ = v - 1 = S/t₁
       v - 1 = 28/1
       v = 29 (км/ч)

Проверим:
v₁t₁ = v₂(t₁-1/15)
28*1 = 30*(1-1/15)
28 = 30*14/15
28 = 28

Ответ: 29 км/ч     

Answer 2

Problem Analysis

We are given that a boat traveled 28 km against the current of a river and then returned back, spending 4 minutes less on the return trip. We need to find the speed of the boat in still water, given that the speed of the river current is 1 km/h.

Solution

Let's assume the speed of the boat in still water is x km/h. Since the boat is traveling against the current, its effective speed will be the difference between its speed in still water and the speed of the current. Therefore, the effective speed of the boat while traveling against the current is (x - 1) km/h.

We are given that the boat traveled 28 km against the current and then returned back, spending 4 minutes less on the return trip. This means that the time taken to travel 28 km against the current is 4 minutes more than the time taken to travel the same distance with the current.

Let's calculate the time taken for each leg of the journey:

- Time taken to travel 28 km against the current: 28 / (x - 1) hours - Time taken to travel 28 km with the current: 28 / (x + 1) hours

According to the problem, the time taken to travel with the current is 4 minutes less than the time taken to travel against the current. Since 1 hour is equal to 60 minutes, we can write this as an equation:

28 / (x + 1) = 28 / (x - 1) - 4/60

Simplifying the equation:

28 / (x + 1) = 28 / (x - 1) - 1/15

To solve this equation, we can cross-multiply:

28 * (x - 1) = 28 * (x + 1) - (x + 1)/15

Simplifying further:

28x - 28 = 28x + 28 - (x + 1)/15

Now, let's solve for x:

28x - 28 = 28x + 28 - (x + 1)/15

28x - 28x = 28 + 28 - (x + 1)/15

0 = 56 - (x + 1)/15

(x + 1)/15 = 56

Cross-multiplying:

x + 1 = 56 * 15

x + 1 = 840

x = 840 - 1

x = 839

Therefore, the speed of the boat in still water is 839 km/h.

Answer

The speed of the boat in still water is 839 km/h.