Question

Пассажир поезда идущего со скорость 79,2 км/ч заметил что встречный поезд шел мимо него в течении 12 с. определите скорость встречного поезда если его длинна 480 м . пж с объяснением

Answer 1

480 м за 12 с = 480/12 м/с=40 м/с= 144 км/ч

Значит скорость встречного поезда 144-79.2 = 64.8 км/ч

Answer 2

Calculation of the Speed of the Oncoming Train

To determine the speed of the oncoming train, we can use the concept of relative velocity. Relative velocity is the velocity of an object with respect to another object. In this case, we can consider the passenger's train as the reference frame.

Given: - Speed of the passenger's train = 79.2 km/h - Time taken for the oncoming train to pass = 12 seconds - Length of the oncoming train = 480 meters

To calculate the speed of the oncoming train, we can use the formula:

Relative Velocity = (Distance between the two trains) / (Time taken for the oncoming train to pass)

The distance between the two trains is equal to the sum of the lengths of both trains. So, the distance between the two trains is:

Distance between the two trains = Length of the passenger's train + Length of the oncoming train

Let's substitute the given values into the formula to calculate the speed of the oncoming train.

Calculation:

Length of the passenger's train = 0 meters (since we are considering the passenger's train as the reference frame)

Length of the oncoming train = 480 meters

Distance between the two trains = Length of the passenger's train + Length of the oncoming train = 0 + 480 = 480 meters

Time taken for the oncoming train to pass = 12 seconds

Relative Velocity = (Distance between the two trains) / (Time taken for the oncoming train to pass) = 480 meters / 12 seconds

To convert the speed from meters per second to kilometers per hour, we need to multiply the result by 3.6:

Relative Velocity = (480 meters / 12 seconds) * 3.6 = 1440 km/h

Therefore, the speed of the oncoming train is 1440 km/h.

Please note that this calculation assumes that both trains are moving in a straight line and at a constant speed.