Question

Вертолет пролетел по ветру расстояние 120 км и обратно вернулся, потратив на весь путь 6 час. Найдите скорость ветра если скорость в штиль составляет 45 км/час.

Answer 1

X ( км/час ) скорость ветра ; ( 120 / ( 45 + X )) + ( 120 / ( 45 - X )) = 6 ; X не равен 45 ; X > 0 ; 120 * ( 45 - X ) + 120 * ( 45 + X ) = 6 * ( 2025 - X ^ 2 ) ; 5400 - 120X + 5400 + 120X = 12150 - 6X^2 ; 10800 = 12150 - 6X^2 ; 6X^2 = 1350 ; X^2 = 225 X1 = 15 ; X2 = - 15 ( не подходит ) ; ОТВЕТ 15 км/час

Answer 2

Calculation of Wind Speed

To find the wind speed, we need to determine the time it took for the helicopter to fly against the wind and with the wind. Let's assume the speed of the helicopter in still air is 45 km/h.

Let's denote the speed of the wind as w km/h.

The time taken to fly against the wind for a distance of 120 km can be calculated using the formula:

Time against wind = Distance / (Speed of helicopter - Speed of wind)

The time taken to fly with the wind for a distance of 120 km can be calculated using the formula:

Time with wind = Distance / (Speed of helicopter + Speed of wind)

Given that the total time for the round trip is 6 hours, we can write the equation:

Time against wind + Time with wind = 6 hours

Now, let's substitute the values and solve for the wind speed.

Time against wind = 120 / (45 - w)

Time with wind = 120 / (45 + w)

120 / (45 - w) + 120 / (45 + w) = 6

To solve this equation, we can multiply both sides by (45 - w)(45 + w) to eliminate the denominators:

120(45 + w) + 120(45 - w) = 6(45 - w)(45 + w)

Simplifying the equation:

120(45 + w) + 120(45 - w) = 6(45^2 - w^2)

Now, let's solve for w.

120(45 + w) + 120(45 - w) = 6(45^2 - w^2)

Simplifying further:

5400 + 120w + 5400 - 120w = 6(2025 - w^2)

Combining like terms:

10800 = 12150 - 6w^2

Rearranging the equation:

6w^2 = 12150 - 10800

6w^2 = 1350

Dividing both sides by 6:

w^2 = 225

Taking the square root of both sides:

w = ±15

Since wind speed cannot be negative, the wind speed is 15 km/h.

Therefore, the wind speed is 15 km/h.