Question

расстояние в 400 км скорый поезд прошел на час быстрее товарного. какова скорость каждого поезда, если скорость товарного поезда на 20 км\ч меньше, чем скорого?

Answer 1

пусть скорость скорого поезда х км.ч, тогда скорость товарного (х-20) км.ч Время за которое прошел товарный поезд составляет 400/x-20, а скорого 400/х Составляем и решаем ур-е:

400/x-400/x-20=1 

400(x-20)-400x/x(x-20) -1=0

-8000/x(x-20)-1=0

-8000-x2+20x/x(x-20)=0

ОДЗ:х неравно 0; (х-20) неравно 0

-8000-х2+20х=0

-х2+20х-8000=0 *(-1)

х2-20х+8000=0

D=(b/2)2-ac=100+8000=8100

x1= 20-90=-70 не верно х2= 20+90=110 км.ч, а скорость товарного 90 км.ч

Answer 2

Problem Analysis

We are given that a fast train traveled a distance of 400 km in 1 hour less time than a freight train. We are also told that the speed of the freight train is 20 km/h less than the speed of the fast train. We need to find the speed of each train.

Solution

Let's assume the speed of the fast train is x km/h. Since the speed of the freight train is 20 km/h less than the speed of the fast train, the speed of the freight train is x - 20 km/h.

We can use the formula distance = speed × time to find the time taken by each train to travel 400 km.

For the fast train: - Distance = 400 km - Speed = x km/h - Time = 400 / x hours

For the freight train: - Distance = 400 km - Speed = x - 20 km/h - Time = 400 / (x - 20) hours

We are given that the fast train took 1 hour less time than the freight train. So, we can set up the equation:

400 / x = 400 / (x - 20) + 1

Let's solve this equation to find the value of x.

Calculation

To solve the equation 400 / x = 400 / (x - 20) + 1, we can cross-multiply and simplify:

400(x - 20) = 400x + x(x - 20)

Expanding and simplifying:

400x - 8000 = 400x + x^2 - 20x

Rearranging and simplifying:

x^2 - 20x - 8000 = 0

We can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -20, and c = -8000. Plugging in these values:

x = (-(-20) ± √((-20)^2 - 4(1)(-8000))) / (2(1))

Simplifying further:

x = (20 ± √(400 + 32000)) / 2

x = (20 ± √32400) / 2

x = (20 ± 180) / 2

We have two possible solutions for x:

1. x = (20 + 180) / 2 = 200 / 2 = 100 2. x = (20 - 180) / 2 = -160 / 2 = -80

Since speed cannot be negative, we discard the second solution.

Answer

The speed of the fast train is 100 km/h and the speed of the freight train is 100 - 20 = 80 km/h.

Please note that the above solution assumes a linear relationship between speed and time. In reality, factors such as acceleration and deceleration may affect the actual time taken by the trains.