Вопрос задан 01.05.2019 в 00:37. Предмет Алгебра. Спрашивает Логунова Анастасия.

автомобиль, пройдя путь от А до В равный 300 км ,повернул назад и через 1 час 12 мин после выхода

из В увеличил скорость на 16 км\ч .В результате на обратный путь он затратил на 48 мин меньше,чем на путь от А до В. Найдите первоночальную скорость автомобиля.
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Отвечает Демков Костя.

Решение в тестовом документе. (Просто так формулы удобнее писать)

 

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Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

Problem Analysis

We are given that a car travels from point A to point B, a distance of 300 km, and then turns back. After 1 hour and 12 minutes from leaving point B, the car increases its speed by 16 km/h. As a result, the car takes 48 minutes less for the return trip than it took for the trip from A to B. We need to find the initial speed of the car.

Solution

Let's assume the initial speed of the car is x km/h. We can break down the problem into two parts: the time taken for the trip from A to B and the time taken for the return trip.

# Time taken for the trip from A to B

The time taken for the trip from A to B can be calculated using the formula time = distance / speed. Substituting the values, we get: time_AB = 300 / x.

# Time taken for the return trip

After 1 hour and 12 minutes from leaving point B, the car increases its speed by 16 km/h. Let's denote the increased speed as x + 16 km/h. The time taken for the return trip can be calculated using the formula time = distance / speed. Substituting the values, we get: time_return = 300 / (x + 16).

# Time difference

We are given that the car takes 48 minutes less for the return trip than it took for the trip from A to B. We can convert 48 minutes to hours by dividing it by 60: 48 / 60 = 0.8 hours. Therefore, the time difference can be expressed as: time_AB - time_return = 0.8.

# Solving the equation

We can now solve the equation time_AB - time_return = 0.8 to find the initial speed of the car.

Let's substitute the values of time_AB and time_return into the equation: (300 / x) - (300 / (x + 16)) = 0.8.

To simplify the equation, we can multiply both sides by x(x + 16) to eliminate the denominators: 300(x + 16) - 300x = 0.8x(x + 16).

Simplifying further, we get: 4800 = 0.8x^2 + 12.8x.

Rearranging the equation, we have: 0.8x^2 + 12.8x - 4800 = 0.

Now we can solve this quadratic equation to find the value of x.

Calculation

Using the quadratic formula x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = 0.8, b = 12.8, and c = -4800, we can calculate the value of x.

Substituting the values into the quadratic formula, we get: x = (-12.8 ± sqrt(12.8^2 - 4 * 0.8 * -4800)) / (2 * 0.8).

Simplifying further, we have: x = (-12.8 ± sqrt(163.84 + 15360)) / 1.6.

x = (-12.8 ± sqrt(15523.84)) / 1.6.

Taking the square root, we get: x = (-12.8 ± 124.6) / 1.6.

Simplifying further, we have two possible solutions: 1. x = (-12.8 + 124.6) / 1.6. 2. x = (-12.8 - 124.6) / 1.6.

Calculating each solution, we get: 1. x = 111.8 / 1.6 = 69.875. 2. x = -137.4 / 1.6 = -85.875.

Since the speed of a car cannot be negative, we can discard the second solution.

Answer

Therefore, the initial speed of the car is 69.875 km/h.

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