Question

Помогите решить. Теплоход прошел 30 км по течению и вернулся обратно, затратив на обратный путь на 20 минут больше, чем по течению. Найти скорость теплохода, если скорость течения 0,7 км/ч

Answer 1

30/(x+0,7)-по течению

30/(x-0,7)-против течения

30/(x-0,7)-30/(x+0,7)=1/3

30(1,4/(x^-0,49))=1/3

x^-0,49=90*1,4

x=sqrt(90*1,4+0,49)

Answer 2

Problem Analysis

To solve this problem, we need to find the speed of the boat given that it traveled 30 km downstream and then returned upstream, taking 20 minutes longer for the return trip compared to the downstream trip. We are also given that the speed of the current is 0.7 km/h.

Let's denote the speed of the boat as 'b' km/h. The speed of the boat relative to the ground when traveling downstream will be the sum of the boat's speed and the speed of the current, which is 'b + 0.7' km/h. Similarly, the speed of the boat relative to the ground when traveling upstream will be the difference between the boat's speed and the speed of the current, which is 'b - 0.7' km/h.

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

Downstream Journey

For the downstream journey, the boat traveled 30 km. Let's denote the time taken for the downstream journey as 't' hours. Using the formula 'distance = speed × time', we can write the equation:

30 = (b + 0.7) × t

Upstream Journey

For the upstream journey, the boat traveled the same distance of 30 km but took 20 minutes (1/3 hour) longer than the downstream journey. Let's denote the time taken for the upstream journey as 't + 1/3' hours. Using the formula 'distance = speed × time', we can write the equation:

30 = (b - 0.7) × (t + 1/3)

We now have a system of two equations with two unknowns (b and t). We can solve this system of equations to find the value of 'b', which represents the speed of the boat.

Solution

Let's solve the system of equations to find the speed of the boat.

From the equation for the downstream journey: 30 = (b + 0.7) × t

From the equation for the upstream journey: 30 = (b - 0.7) × (t + 1/3)

Expanding the second equation: 30 = (b - 0.7) × t + (b - 0.7) × (1/3)

Simplifying the equation: 30 = bt - 0.7t + (b/3) - (0.7/3)

Combining like terms: 30 = bt - 0.7t + (b/3) - 0.233

Rearranging the equation: bt - 0.7t + (b/3) = 30 + 0.233

Combining the constants: bt - 0.7t + (b/3) = 30.233

Now, we can substitute the value of 't' from the first equation into the second equation to get an equation with only 'b' as the unknown:

(b + 0.7) × t - 0.7t + (b/3) = 30.233

Substituting the value of 't' from the first equation: (b + 0.7) × (30/(b + 0.7)) - 0.7 × (30/(b + 0.7)) + (b/3) = 30.233

Simplifying the equation: 30 - 0.7 × (30/(b + 0.7)) + (b/3) = 30.233

Multiplying through by (b + 0.7) to eliminate the denominator: 30(b + 0.7) - 0.7 × 30 + (b/3)(b + 0.7) = 30.233(b + 0.7)

Expanding and simplifying the equation: 30b + 21 - 21 + (b/3)(b + 0.7) = 30.233b + 21.163

Combining like terms: 30b + (b/3)(b + 0.7) = 30.233b + 21.163

Multiplying through by 3 to eliminate the fraction: 90b + b(b + 0.7) = 90.699b + 63.489

Expanding and simplifying the equation: 90b + b^2 + 0.7b = 90.699b + 63.489

Rearranging the equation: b^2 + 0.7b - 90.699b + 63.489 - 90b = 0

Combining like terms: b^2 - 90.999b + 63.489 = 0

Using the quadratic formula to solve for 'b': b = (-(-90.999) ± √((-90.999)^2 - 4(1)(63.489))) / (2(1))

Simplifying the equation: b = (90.999 ± √(8280.940001 - 2543.956)) / 2

Calculating the square root: b = (90.999 ± √(5736.984001)) / 2

Simplifying the equation: b = (90.999 ± 75.759) / 2

Calculating the values of 'b': b1 = (90.999 + 75.759) / 2 ≈ 83.879 b2 = (90.999 - 75.759) / 2 ≈ 7.120

Since the speed of the boat cannot be negative, the speed of the boat is approximately 83.879 km/h.

Answer

The speed of the boat is approximately 83.879 km/h.