Question

Отъезжая от остановки, автобус за 10 с развил скорость 10 м/с. Определить ускорение автобуса. Каким будет ускорение автобуса в системе отсчёта, связаной с равномерно движущимся автомобилем, проезжающим мимо остановки автобуса со скоростью 15 м/с?Пожалуйста, полное решение с ПОДРОБНЫМ объяснением. Особенно это касается последнего вопроса.

Answer 1

-относительно остановки ускорение  a = (v-vo) /t = (10-0) /10 = 1м/с2

-относительно равномерно движущегося автомобиля (скорость мобыть любая, главное , что постоянная)  ускорение такое же   a=1 м/с2

пояснение

система отсчета, связанная с автомобилем, движется в системе отсчета, связанной с остановкой БЕЗ УСКОРЕНИЯ

Answer 2

Calculation of the Bus's Acceleration

To determine the acceleration of the bus, we can use the formula:

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

Given that the bus starts from rest and reaches a speed of 10 m/s in 10 seconds, we can calculate the acceleration as follows:

Δv = 10 m/s - 0 m/s = 10 m/s Δt = 10 s

Substituting these values into the formula, we get:

a = Δv / Δt = 10 m/s / 10 s = 1 m/s²

Therefore, the acceleration of the bus is 1 m/s².

Acceleration of the Bus in the Reference Frame of the Moving Car

To determine the acceleration of the bus in the reference frame of the moving car, we need to consider the relative motion between the bus and the car.

Given that the car is moving past the bus with a speed of 15 m/s, we can consider the car as the reference frame. In this reference frame, the bus appears to be moving with a velocity of 10 m/s (since the bus is moving away from the bus stop with a speed of 10 m/s).

To calculate the acceleration of the bus in the reference frame of the car, we can use the formula:

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

Since the bus is moving with a constant velocity of 10 m/s in the reference frame of the car, the change in velocity (Δv') is 0 m/s. The time taken (Δt') is the same as the time taken in the original reference frame, which is 10 seconds.

Substituting these values into the formula, we get:

a' = Δv' / Δt' = 0 m/s / 10 s = 0 m/s²

Therefore, the acceleration of the bus in the reference frame of the moving car is 0 m/s².

Summary

In summary, the acceleration of the bus is 1 m/s² when measured in its own reference frame (moving away from the bus stop), and the acceleration of the bus is 0 m/s² when measured in the reference frame of the moving car passing by the bus stop.

Please let me know if you need any further clarification or if there's anything else I can help you with!