Question

Поезд отправляется с вокзала. После разгона скорость поезда 108км/ч. Поезд все время увеличивал скорость с постоянным ускорением 0,5 м / с2. 1.За сколько времени поезд разогнался до скорости 108 км / ч? 2.Какое расстояние проехал поезд при разгоне до 108 км / ч? Максимально подробное решение пожалуйста

Answer 1

Ответ: t=60 с, s=900 м.

Объяснение:

Пусть a=0,5 м/с² - ускорение поезда. Скорость поезда v=108/3,6=30 м/с.

1) время разгона t=v/a=30/0,5=60 с.

2) расстояние s=a*t²/2=0,5*3600/2=900 м.

Answer 2

Problem Statement

We have a train that starts from a station and accelerates with a constant acceleration of 0.5 m/s^2. After reaching its maximum speed, the train travels at a constant speed of 108 km/h. We need to find the following:

1. The time it takes for the train to accelerate to a speed of 108 km/h. 2. The distance traveled by the train during the acceleration phase until it reaches a speed of 108 km/h.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated linear motion. The equations we will use are:

1. v = u + at: Final velocity (v) is equal to initial velocity (u) plus acceleration (a) multiplied by time (t). 2. s = ut + (1/2)at^2: Distance (s) is equal to initial velocity (u) multiplied by time (t) plus half the acceleration (a) multiplied by time squared (t^2).

Let's solve each part of the problem step by step.

Part 1: Time to Accelerate to 108 km/h

We are given that the acceleration of the train is 0.5 m/s^2 and the final speed is 108 km/h. We need to find the time it takes for the train to reach this speed.

First, let's convert the final speed from km/h to m/s: - 108 km/h = 108 * (1000 m / 1 km) / (3600 s / 1 h) = 30 m/s.

Using the equation v = u + at, where the initial velocity (u) is 0 m/s, the final velocity (v) is 30 m/s, and the acceleration (a) is 0.5 m/s^2, we can solve for time (t): - 30 = 0 + 0.5t - 0.5t = 30 - t = 30 / 0.5 - t = 60 seconds.

Therefore, it takes the train 60 seconds to accelerate to a speed of 108 km/h.

Part 2: Distance Traveled During Acceleration

We need to find the distance traveled by the train during the acceleration phase until it reaches a speed of 108 km/h.

Using the equation s = ut + (1/2)at^2, where the initial velocity (u) is 0 m/s, the acceleration (a) is 0.5 m/s^2, and the time (t) is 60 seconds, we can solve for distance (s): - s = 0 * 60 + (1/2) * 0.5 * (60^2) - s = 0 + 0.5 * 0.5 * 3600 - s = 0 + 0.25 * 3600 - s = 0 + 900 - s = 900 meters.

Therefore, the train travels a distance of 900 meters during the acceleration phase until it reaches a speed of 108 km/h.

Summary

To summarize: 1. The train takes 60 seconds to accelerate to a speed of 108 km/h. 2. The train travels a distance of 900 meters during the acceleration phase until it reaches a speed of 108 km/h.

Please let me know if you need any further clarification or assistance.