Question

Условие к задаче: с железнодорожной станцией в 12 часов вышел скорый поезд со скоростью 70 км ч . на 3 часа раньше с этой же станцией был отправлен в том же направлении товарный поезд. в 16 ч скорый поезд догнал товарный найдите скорость товарного поезда

Answer 1

S = v * t - формула пути
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Пусть х (км/ч) - скорость товарного поезда; 16 - 9 = 7 (ч) - время в пути.
70 км/ч - скорость скорого поезда; 16 - 12 = 4 (ч) - время в пути. Уравнение: 
х * 7 = 70 * 4
7х = 280
х = 280 : 7
х = 40
Ответ: 40 км/ч - скорость товарного поезда.

Answer 2

Problem Analysis

We are given that a fast train left a railway station at 12:00 with a speed of 70 km/h. Three hours earlier, a freight train was sent in the same direction from the same station. At 16:00, the fast train caught up with the freight train. We need to find the speed of the freight train.

Solution

Let's assume the speed of the freight train is x km/h.

To find the speed of the freight train, we can use the formula: speed = distance / time.

We know that the fast train traveled for 4 hours (from 12:00 to 16:00) and the freight train traveled for 7 hours (from 9:00 to 16:00).

The distance traveled by the fast train is 70 km/h * 4 h = 280 km.

The distance traveled by the freight train is x km/h * 7 h = 7x km.

Since the fast train caught up with the freight train, the distance traveled by both trains must be the same. Therefore, we can set up the equation:

280 km = 7x km

Now we can solve for x to find the speed of the freight train.

Calculation

Let's solve the equation:

280 km = 7x km

Divide both sides of the equation by 7:

40 km = x km

Therefore, the speed of the freight train is 40 km/h.

Answer

The speed of the freight train is 40 km/h.