Question

Автомобиль начал движение из состояния покоя и 15 с двигался с ускорением 2 м/с2, затем 5 с он двигался равномерно, а последние 35 м тормозил до полной остановки. Считая, что движение происходит вдоль оси ОХ в положительном направлении, постройте графики sx(t), vx(t) и ax(t). Найдите среднюю скорость движения. Алгебраические задачи

Answer 1

Ответ:

Объяснение:

Дано:

Первый участок:

V₀₁ = 0 м/с

t₁ = 15 с

a₁ = 2 м/c²

Второй участок:

V₂ = V₁

t₂ = 5 c

Третий участок:

V₀₃ = V₂

S₃ = 35 м

V₃ = 0

__________

Vcp - ?

Рассмотрим три участка движения.

Первый участок:

V₀ = 0 м/с

t₁ = 15 с

a₁ = 2 м/с          (1)

Уравнение пути:

S₁(t) = a·t²            

S₁(t) = 2·t²           (2)

S₁ = 2·15² = 450 м

Находим скорость:

V₁(t) = V₀₁ + a·t

V₁(t) = 0 + 2·t       (3)

V₁ = 2·5 = 10 м/с

Второй участок:

t₂ = 5 с

a₂ = 0   м/с²            (4)

V₂ = V₁ = 10 м/с      (5)

S₂(t) = 10·t               (6)

S₂ = 10·5 = 50 м    

Третий участок:

V₀₃ = V₂ = 10 м/с

V₃ = 0 м/с

S₃ = 35 м

a₃ = (V₃² - V₃²) / (2·S₃) = (0² - 10²) / (2·35) ≈ - 1,4 м/с²

a₃ = - 1,4 м/с²          (7)

Находим скорость:

V₃(t) = V₀₃ + a·t

V₃(t) = 10 -1,4·t         (8)

Время:

0 = 10 - 1,4·t

t = 10/1,4 ≈ 7 с  

Уравнение пути:

S₃(t) = 10·t - 1,4·t²         (9)

По уравнениям (2), (6) и (9) строим график пути.

По уравнениям (3), (5) и (8) строим график скорости.

По уравнениям (1), (4) и (7) строим график ускорения.

Средняя скорость:

Vcp = (S₁ + S₂ + S₃ ) / (t₁ + t₂ + t₃) =

= (450 + 50 + 35) / (15 + 5 + 7) ≈ 20 м/с

Answer 2

Problem Analysis

We are given the following information about the motion of a car: - The car starts from rest and accelerates at 2 m/s^2 for 15 seconds. - Then, the car moves with a constant velocity for 5 seconds. - Finally, the car decelerates and comes to a stop over a distance of 35 meters.

We need to plot the position-time graph (s(t)), velocity-time graph (v(t)), and acceleration-time graph (a(t)) for this motion. Additionally, we need to find the average velocity of the car.

Position-Time Graph (s(t))

To plot the position-time graph, we need to determine the position of the car at different time intervals.

1. During the acceleration phase (0 to 15 seconds): - The initial velocity (u) is 0 m/s. - The acceleration (a) is 2 m/s^2. - The time (t) ranges from 0 to 15 seconds. - We can use the equation of motion: s = ut + (1/2)at^2.

2. During the constant velocity phase (15 to 20 seconds): - The velocity (v) is constant at 30 m/s (calculated from the acceleration phase). - The time (t) ranges from 15 to 20 seconds. - We can use the equation of motion: s = ut.

3. During the deceleration phase (20 to 55 seconds): - The initial velocity (u) is 30 m/s (calculated from the constant velocity phase). - The final velocity (v) is 0 m/s. - The time (t) ranges from 20 to 55 seconds. - We can use the equation of motion: s = ut + (1/2)at^2.

Velocity-Time Graph (v(t))

To plot the velocity-time graph, we need to determine the velocity of the car at different time intervals.

1. During the acceleration phase (0 to 15 seconds): - The initial velocity (u) is 0 m/s. - The acceleration (a) is 2 m/s^2. - The time (t) ranges from 0 to 15 seconds. - We can use the equation of motion: v = u + at.

2. During the constant velocity phase (15 to 20 seconds): - The velocity (v) is constant at 30 m/s (calculated from the acceleration phase). - The time (t) ranges from 15 to 20 seconds. - The velocity remains constant at 30 m/s.

3. During the deceleration phase (20 to 55 seconds): - The initial velocity (u) is 30 m/s (calculated from the constant velocity phase). - The final velocity (v) is 0 m/s. - The time (t) ranges from 20 to 55 seconds. - We can use the equation of motion: v = u + at.

Acceleration-Time Graph (a(t))

To plot the acceleration-time graph, we need to determine the acceleration of the car at different time intervals.

1. During the acceleration phase (0 to 15 seconds): - The acceleration (a) is constant at 2 m/s^2. - The time (t) ranges from 0 to 15 seconds. - The acceleration remains constant at 2 m/s^2.

2. During the constant velocity phase (15 to 20 seconds): - The acceleration is 0 m/s^2. - The time (t) ranges from 15 to 20 seconds. - The acceleration remains 0 m/s^2.

3. During the deceleration phase (20 to 55 seconds): - The acceleration (a) is calculated using the equation: a = (v - u) / t. - The initial velocity (u) is 30 m/s (calculated from the constant velocity phase). - The final velocity (v) is 0 m/s. - The time (t) ranges from 20 to 55 seconds.

Average Velocity

To find the average velocity, we need to calculate the total displacement and divide it by the total time.

1. Total displacement: - During the acceleration phase (0 to 15 seconds): s = ut + (1/2)at^2. - During the constant velocity phase (15 to 20 seconds): s = ut. - During the deceleration phase (20 to 55 seconds): s = ut + (1/2)at^2. - Add up the displacements during each phase to get the total displacement.

2. Total time: - The total time is 55 seconds.

3. Average velocity: - Average velocity = total displacement / total time.

Now, let's calculate and plot the graphs.

Calculation and Graphs

1. Position-Time Graph (s(t)):

During the acceleration phase (0 to 15 seconds): - Using the equation s = ut + (1/2)at^2, where u = 0 m/s and a = 2 m/s^2: - s(t) = (1/2)at^2 = (1/2) * 2 * t^2 = t^2.

During the constant velocity phase (15 to 20 seconds): - The position remains constant at the displacement achieved during the acceleration phase. - s(t) = 15^2 = 225.

During the deceleration phase (20 to 55 seconds): - Using the equation s = ut + (1/2)at^2, where u = 30 m/s and a = -2 m/s^2: - s(t) = ut + (1/2)at^2 = 30t - t^2.

2. Velocity-Time Graph (v(t)):

During the acceleration phase (0 to 15 seconds): - Using the equation v = u + at, where u = 0 m/s and a = 2 m/s^2: - v(t) = at = 2t.

During the constant velocity phase (15 to 20 seconds): - The velocity remains constant at 30 m/s.

During the deceleration phase (20 to 55 seconds): - Using the equation v = u + at, where u = 30 m/s and a = -2 m/s^2: - v(t) = u + at = 30 - 2t.

3. Acceleration-Time Graph (a(t)):

During the acceleration phase (0 to 15 seconds): - The acceleration remains constant at 2 m/s^2.

During the constant velocity phase (15 to 20 seconds): - The acceleration is 0 m/s^2.

During the deceleration phase (20 to 55 seconds): - Using the equation a = (v - u) / t, where u = 30 m/s and v = 0 m/s: - a(t) = (0 - 30) / t = -30 / t.

4. Average Velocity:

To find the average velocity, we need to calculate the total displacement and divide it by the total time.

Total displacement: - During the acceleration phase (0 to 15 seconds): s = ut + (1/2)at^2. - During the constant velocity phase (15 to 20 seconds): s = ut. - During the deceleration phase (20 to 55 seconds): s = ut + (1/2)at^2. - Add up the displacements during each phase to get the total displacement.

Total time: - The total time is 55 seconds.

Average velocity: - Average velocity = total displacement / total time.

Now, let's plot the graphs and calculate the average velocity.