Question

Поезд начал разгон и за 2 минуты достиг скорости 108 км в час,после чего 3 минуты двигалсяравномерно,а потом полностью затормозил за 5 минут.Найти пройденый поездом путь

Answer 1

Вот Решение задачи задачи в отсканированном документе

Answer 2

Problem Analysis

We are given the following information about a train's motion: - The train accelerates and reaches a speed of 108 km/h in 2 minutes. - The train then moves at a constant speed for 3 minutes. - Finally, the train comes to a complete stop in 5 minutes.

We need to find the total distance traveled by the train.

Solution

To find the total distance traveled by the train, we can break down the motion into three parts: acceleration, constant speed, and deceleration.

1. Acceleration: - The train starts from rest and reaches a speed of 108 km/h in 2 minutes. - We can use the formula `v = u + at`, where `v` is the final velocity, `u` is the initial velocity, `a` is the acceleration, and `t` is the time. - Since the train starts from rest, the initial velocity `u` is 0. - We can rearrange the formula to solve for acceleration: `a = (v - u) / t`. - Plugging in the values, we get `a = (108 km/h - 0) / 2 min`. - We need to convert the time from minutes to hours to match the units of velocity: `2 min = 2/60 hours`. - Calculating the acceleration, we get `a = (108 km/h) / (2/60) h`. - Simplifying, we find `a = 3240 km/h^2`.

2. Constant Speed: - The train moves at a constant speed for 3 minutes. - Since the speed is constant, the acceleration is 0. - The formula to calculate distance when acceleration is 0 is `s = ut`, where `s` is the distance, `u` is the initial velocity, and `t` is the time. - Plugging in the values, we get `s = (108 km/h) * (3 min)`. - Again, we need to convert the time from minutes to hours: `3 min = 3/60 hours`. - Calculating the distance, we find `s = (108 km/h) * (3/60) h`. - Simplifying, we get `s = 5.4 km`.

3. Deceleration: - The train comes to a complete stop in 5 minutes. - The final velocity is 0 km/h. - We can use the same formula as in the acceleration phase: `v = u + at`. - Rearranging the formula to solve for acceleration, we get `a = (v - u) / t`. - Plugging in the values, we have `a = (0 km/h - 108 km/h) / 5 min`. - Converting the time to hours: `5 min = 5/60 hours`. - Calculating the acceleration, we find `a = (-108 km/h) / (5/60) h`. - Simplifying, we get `a = -1296 km/h^2`.

Now, we can calculate the distance traveled during deceleration using the formula `s = ut + (1/2)at^2`. - Plugging in the values, we have `s = (108 km/h) * (5/60) h + (1/2) * (-1296 km/h^2) * (5/60 h)^2`. - Simplifying, we find `s = 9 km`.

Total Distance Traveled

To find the total distance traveled by the train, we sum up the distances from each phase: - Distance during acceleration: 0 km - Distance during constant speed: 5.4 km - Distance during deceleration: 9 km

Adding these distances together, we find that the train traveled a total distance of 14.4 km.

Answer

The train traveled a total distance of 14.4 km.