Question

Уклон длиной 340 м, поезд прошел за 20с. В конце развил скорость 19 м/с. Найти начальную скорость и ускорение.

Answer 1

Дано
l = 340 м
t = 20 с
Vк = 19 м/с
_______________
Найти - v₀, a - ?
Решение
1) Vк = v₀+at
a = v-v₀ / t
v₀ = 2*l / t - Vк
v₀ = 2 * 340 м / 20 с -19 м/c = 15 м/с
2) a = Vк / t - v₀ / t => a = 2*Vк / 2 - 2*l / t (Vк - l / t)
a = 2 / 20 с ( 19- 340м/ 20с) = 0,2 м/с²
Ответ: V₀ = 15 м/с ; a = 0,2 м/с²

Answer 2

Problem Analysis

We are given the length of a slope, the time it takes for a train to pass through the slope, and the final velocity at the end of the slope. We need to find the initial velocity and acceleration of the train.

Solution

To solve this problem, we can use the equations of motion. Let's denote the initial velocity as v0, the final velocity as vf, the acceleration as a, the distance traveled as s, and the time taken as t.

The equation that relates distance, initial velocity, final velocity, acceleration, and time is:

s = v0 * t + (1/2) * a * t^2 We are given the following values: - Distance (s) = 340 m - Time (t) = 20 s - Final velocity (vf) = 19 m/s

We need to find the initial velocity (v0) and acceleration (a).

Finding the Initial Velocity (v0)

To find the initial velocity, we can rearrange equation as follows:

v0 = (s - (1/2) * a * t^2) / t Substituting the given values into equation we can find the initial velocity.

Finding the Acceleration (a)

To find the acceleration, we can use the equation:

vf = v0 + a * t Substituting the given values into equation we can find the acceleration.

Calculation

Let's calculate the initial velocity and acceleration using the given values.

Using equation v0 = (340 - (1/2) * a * (20^2)) / 20

Using equation 19 = v0 + a * 20

Now we have a system of two equations with two unknowns (v0 and a). We can solve this system of equations to find the values of v0 and a.

Solving the System of Equations

To solve the system of equations, we can use substitution or elimination. Let's use substitution.

From equation we can express v0 in terms of a: v0 = 19 - a * 20

Substituting this expression for v0 into equation we get: (340 - (1/2) * a * (20^2)) / 20 = 19 - a * 20

Now we can solve this equation for a.

Calculation

Let's calculate the value of acceleration (a) using the equation derived above.

(340 - (1/2) * a * (20^2)) / 20 = 19 - a * 20

Simplifying the equation, we get: 340 - 200a = 380 - 20a

Solving for a, we find: 180a = 40 a = 40 / 180 a = 2/9 m/s^2

Now that we have the value of acceleration (a), we can substitute it back into equation to find the initial velocity (v0).

Calculation

Let's calculate the value of the initial velocity (v0) using the derived value of acceleration (a).

19 = v0 + (2/9) * 20

Simplifying the equation, we get: 19 = v0 + 40/9

Solving for v0, we find: v0 = 19 - 40/9 v0 = (171 - 40) / 9 v0 = 131 / 9 v0 ≈ 14.56 m/s

Answer

The initial velocity of the train is approximately 14.56 m/s and the acceleration is 2/9 m/s^2.

Please let me know if I can help you with anything else.