Question

Товарный поезд был задержан в пути на 18 мин, а затем на расстоянии в 60 км наверстал это время, увеличив скорость на 10 км/ч. Найдите первоначальную скорость поезда.

Answer 1

Пусть х - первоначальная скорость поезда, тогда увеличенная скорость - х+10.
18 минут=0,3 часа
Составим уравнение и решим его
$\frac{60}{x} - \frac{60}{x+10}=0,3\\ 60(x+ 10)-60x=0,3( x^{2} +10x)\\ 60x+600-60x=0,3 x^{2} +3x\\ 0,3 x^{2} +3x-600=0\\ D>0$
Уравнение имеет два корня
$x_{1} =-50\\ x_{2} =40$
Смыслу задачи соответствует второй корень
Ответ: первоначальная скорость поезда 40 км/ч

Answer 2

Calculation of Initial Speed of the Train

To find the initial speed of the train, we can use the information provided: the train was delayed by 18 minutes and then made up for the lost time by increasing its speed by 10 km/h over a distance of 60 km.

Let's assume the initial speed of the train is v km/h.

The time taken to cover a distance of 60 km at the initial speed is given by:

Time taken = Distance / Speed

The time taken to cover a distance of 60 km at a speed increased by 10 km/h is given by:

Time taken = Distance / (Speed + 10)

Since the train made up for the lost time, the two time intervals are equal:

Time taken at initial speed = Time taken at increased speed

Using the above equations, we can set up the following equation:

60 / v = 60 / (v + 10)

To solve this equation, we can cross-multiply:

60(v + 10) = 60v

Simplifying the equation:

60v + 600 = 60v

We can see that the variable 'v' cancels out, leaving us with:

600 = 0

This equation is not possible to solve, as it results in a contradiction. Therefore, there is no valid initial speed of the train that satisfies the given conditions.

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