Question

Поезд подходит к станции со скоростью 36км/ч и останавливается через минуту после начала торможения.С каким ускорением двигался поезд?

Answer 1

Vo=36 км/ч=10м/с t=60c  a=(V-V0)/t   V=0   Дерзай

Answer 2

Calculation of Acceleration

To determine the acceleration of the train, we need to use the formula:

Acceleration (a) = Change in velocity (Δv) / Time taken (t)

Given that the train approaches the station with a speed of 36 km/h and comes to a stop after 1 minute of braking, we need to convert the speed to m/s and the time to seconds for consistent units.

Converting the speed from km/h to m/s: - 1 km/h = 1000 m / 3600 s = 5/18 m/s

Therefore, the initial velocity (v1) of the train is: - v1 = 36 km/h * (5/18) m/s = 10 m/s

The time taken (t) for the train to stop is 1 minute, which is equal to 60 seconds.

Now, we can calculate the acceleration (a) using the formula mentioned earlier:

Acceleration (a) = Change in velocity (Δv) / Time taken (t)

Since the train comes to a stop, the change in velocity (Δv) is equal to the initial velocity (v1) because the final velocity (v2) is 0.

Acceleration (a) = v1 / t

Substituting the values: - v1 = 10 m/s - t = 60 s

Acceleration (a) = 10 m/s / 60 s = 1/6 m/s²

Therefore, the train was moving with an acceleration of 1/6 m/s².