Question

Электропоезд, двигающийся со скоростью, модуль которой V0=30 м/с, начал тормозить с постоянным ускорением. Определите время торможения электропоезда до полной остановки и модуль его ускорения, если тормозной путь электропоезда s=180м

Answer 1

Вот решение для данной задачи

Answer 2

Calculation of Time and Acceleration for the Braking of an Electric Train

To determine the time it takes for the electric train to come to a complete stop and the magnitude of its acceleration, we can use the given information:

- Initial velocity (V0) of the train: 30 m/s - Braking distance (s) of the train: 180 m

We can use the following kinematic equation to solve for the time (t) and acceleration (a) of the train:

s = V0t + (1/2)at^2

Since the train is braking, the final velocity (V) will be 0 m/s. Therefore, the equation becomes:

s = V0t + (1/2)at^2 = 0

Substituting the given values:

180 = 30t + (1/2)at^2

To solve for t and a, we need another equation. We can use the equation that relates the final velocity (V), initial velocity (V0), acceleration (a), and time (t):

V = V0 + at

Since the final velocity is 0 m/s, the equation becomes:

0 = 30 + at

Solving this equation for t, we get:

t = -30/a Now we can substitute this value of t into the first equation:

180 = 30(-30/a) + (1/2)a(-30/a)^2

Simplifying the equation:

180 = -900/a + (1/2)(900/a^2)

Multiplying both sides by a^2 to eliminate the denominators:

180a^2 = -900a + 450

This equation can be rearranged to form a quadratic equation:

180a^2 + 900a - 450 = 0

Simplifying further:

2a^2 + 10a - 5 = 0

Using the quadratic formula, we can solve for the value of a:

a = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values:

a = (-10 ± √(10^2 - 4(2)(-5))) / (2(2))

Simplifying:

a = (-10 ± √(100 + 40)) / 4

a = (-10 ± √140) / 4

The two possible values for acceleration are:

a1 = (-10 + √140) / 4

a2 = (-10 - √140) / 4

To determine which value is correct, we need to consider the physical context of the problem. Since the train is braking, the acceleration should be negative. Therefore, the correct value for acceleration is:

a = (-10 - √140) / 4 Now we can substitute this value of a into equation to solve for t:

t = -30 / ((-10 - √140) / 4)

Simplifying:

t = 12 / (5 + √35)

Therefore, the time it takes for the electric train to come to a complete stop is approximately t = 12 / (5 + √35) seconds, and the magnitude of its acceleration is approximately a = (-10 - √140) / 4 m/s^2.

Please note that the above calculations are based on the given information and assumptions.