Question

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

Answer 1

Первоначальная скорость обозначим х
18 минут=0.3 часа
60/x-0.3=60/(x+10)
x(x+10)=2000
x=40 км/ч-это первоначальная скорость поезда

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 initially delayed by 18 minutes and then made up for this 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 x km/h.

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

Time taken = Distance / Speed

So, the time taken to cover 60 km at the initial speed is:

Time taken at initial speed = 60 / x hours

The train was delayed by 18 minutes, which is equal to 18/60 = 0.3 hours.

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

Time taken at increased speed = 60 / (x + 10) hours

According to the given information, the time taken at the increased speed is equal to the sum of the initial time and the delay time:

Time taken at initial speed + Delay time = Time taken at increased speed

Substituting the values, we get:

60 / x + 0.3 = 60 / (x + 10)

To solve this equation, we can cross-multiply and simplify:

60(x + 10) = (60 + 0.3x)(x)

Simplifying further:

60x + 600 = 60x + 0.3x^2

Subtracting 60x from both sides:

600 = 0.3x^2

Dividing both sides by 0.3:

2000 = x^2

Taking the square root of both sides:

x = √2000

Calculating the square root of 2000, we find:

x ≈ 44.72 km/h

Therefore, the initial speed of the train is approximately 44.72 km/h.