Question

Лодка прошла по течению реки 10 км и против течения 6 км ,затратив га весь путь 2 часа . Чему равна собственная скорость лодки ?,если скорость течения реки 2 км/ч.

Answer 1

Пусть xкм\ч-собственная скорость лодки. Тогда (х+3)км\ч- скорость лодки по течению, а (х-3)км\ч- скорость лодки против течения. По условию по течению лодка проплыла 10км, тогда время, затраченое на движение лодки по течению будет равно ч. Время , затраченое на движение соответственно будет равно ч.

По условию верное равенство: 

x=1-не подходит по условию задачи, т.к. если подставить в выражение  , то t будет меньше 0.

Ответ:7км\ч

Answer 2

Problem Analysis

We are given the following information: - The boat traveled downstream for 10 km. - The boat traveled upstream against the current for 6 km. - The total time taken for the entire journey was 2 hours. - The speed of the river current is 2 km/h.

We need to find the speed of the boat in still water.

Solution

Let's assume the speed of the boat in still water is x km/h.

When the boat is traveling downstream, its effective speed is the sum of its own speed and the speed of the current. So, the effective speed downstream is (x + 2) km/h.

When the boat is traveling upstream against the current, its effective speed is the difference between its own speed and the speed of the current. So, the effective speed upstream is (x - 2) km/h.

We can use the formula distance = speed × time to calculate the time taken for each leg of the journey.

The time taken to travel downstream for 10 km is given by: 10 = (x + 2) × t1 (Equation 1)

The time taken to travel upstream for 6 km is given by: 6 = (x - 2) × t2 (Equation 2)

We also know that the total time taken for the entire journey is 2 hours: t1 + t2 = 2 (Equation 3)

We can solve this system of equations to find the value of x, which represents the speed of the boat in still water.

Calculation

Let's solve the system of equations using substitution or elimination method.

From Equation 1, we can express t1 in terms of x: t1 = 10 / (x + 2)

Substituting this value of t1 in Equation 3, we get: (10 / (x + 2)) + t2 = 2

Simplifying, we have: 10 + t2(x + 2) = 2(x + 2) 10 + t2x + 2t2 = 2x + 4

Rearranging, we get: t2x - 2x = 4 - 10 - 2t2 x(t2 - 2) = -6 - 2t2 x = (-6 - 2t2) / (t2 - 2) (Equation 4)

Now, let's substitute the given values into Equation 4 and calculate the value of x.

Calculation

Substituting t2 = 2 - t1 into Equation 4, we get: x = (-6 - 2(2 - t1)) / (2 - t1 - 2) x = (-6 - 4 + 2t1) / (-t1) x = (2t1 - 10) / t1

Simplifying, we have: x = 2 - (10 / t1) (Equation 5)

Now, let's substitute the given values into Equation 5 and calculate the value of x.

Calculation

Substituting t1 = 10 / (x + 2) into Equation 5, we get: x = 2 - (10 / (10 / (x + 2))) x = 2 - (10 / (10 / (x + 2))) x = 2 - (10(x + 2) / 10) x = 2 - (x + 2) x = 2 - x - 2 2x = 0 x = 0

Therefore, the speed of the boat in still water is 0 km/h.

Answer

The speed of the boat in still water is 0 km/h.