Question

С одной станции одновременно в противоположных направлениях вышли два поезда Скорость одного из них 54 км в час а скорость другого на 18 км час больше Через сколько часов расстояние между ними будет равно 504 км

Answer 1

Если поймёте)
504=S1+S2

Answer 2

Problem Analysis

We are given that two trains start simultaneously from the same station in opposite directions. The speed of one train is 54 km/h, and the speed of the other train is 18 km/h faster. We need to determine the time it takes for the distance between the two trains to be 504 km.

Solution

Let's assume that the slower train is traveling at a speed of x km/h. Since the faster train is traveling at a speed 18 km/h faster, its speed would be (x + 18) km/h.

We can use the formula `distance = speed * time` to find the time it takes for each train to travel a certain distance.

For the slower train: - Distance = x km/h * t hours

For the faster train: - Distance = (x + 18) km/h * t hours

Since the two trains are moving in opposite directions, the total distance between them is the sum of the distances covered by each train.

Total distance = Distance covered by slower train + Distance covered by faster train

504 km = (x km/h * t hours) + ((x + 18) km/h * t hours)

Now, we can solve this equation to find the value of t.

Calculation

Let's solve the equation:

504 km = (x km/h * t hours) + ((x + 18) km/h * t hours)

504 km = (xt + (x + 18)t) km

504 km = (2x + 18) t km

Divide both sides of the equation by (2x + 18) to isolate t:

t = 504 km / (2x + 18) km

Now, we can substitute the given speed of one train (54 km/h) into the equation to find the value of t.

t = 504 km / (2 * 54 km/h + 18 km/h)

Calculation

Let's calculate the value of t:

t = 504 km / (2 * 54 km/h + 18 km/h) t = 504 km / (108 km/h + 18 km/h) t = 504 km / 126 km/h t ≈ 4 hours

Answer

It will take approximately 4 hours for the distance between the two trains to be 504 km.

Please note that this solution assumes constant speeds for the trains and does not consider factors such as acceleration or deceleration.