Question

Поезд проехал прямолинейный участок между двумя железнодорожными станциями A и B со средней скоростью V. На разгон и торможение машинист потратил суммарно минуту τ=1 минуту. Остальное время поезд двигался равномерно со скоростью 1,2V. За какое время машинист проехал участок AB, если разгон и торможение происходили с постоянными, но не равными между собой по модулю, ускорениями? Ответ выразить в минутах, округлив до целых

Answer 1

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Answer 2

To solve this problem, we need to break down the motion of the train into different phases: acceleration, constant speed, and deceleration. Let's denote the time taken for acceleration and deceleration as t1 and t2, respectively.

Acceleration Phase:

During the acceleration phase, the train starts from rest and reaches the speed of V. The time taken for acceleration is t1.

Constant Speed Phase:

After the acceleration phase, the train maintains a constant speed of 1.2V. The duration of this phase can be calculated by subtracting the time taken for acceleration and deceleration from the total time of the journey, which is 1 minute.

Deceleration Phase:

During the deceleration phase, the train slows down from the speed of 1.2V to rest. The time taken for deceleration is t2.

To find the total time taken for the journey, we need to calculate t1 and t2.

Let's assume the distance between stations A and B is D.

During the acceleration phase, the train covers a distance of D/2, as it starts from rest and reaches the midpoint between A and B. The equation for distance covered during constant acceleration is given by:

D/2 = (1/2) * a * t1^2, where a is the acceleration.

Similarly, during the deceleration phase, the train covers another distance of D/2, as it slows down from the midpoint to the station B. The equation for distance covered during constant deceleration is given by:

D/2 = (1/2) * a * t2^2

Since the accelerations during the acceleration and deceleration phases are not equal in magnitude, we can denote them as a1 and a2, respectively.

Now, let's solve the equations to find t1 and t2.

From the first equation, we can express a in terms of t1:

a = (2 * D) / (t1^2)

From the second equation, we can express a in terms of t2:

a = (2 * D) / (t2^2)

Since the magnitudes of the accelerations are not equal, we can equate the two expressions for a:

(2 * D) / (t1^2) = (2 * D) / (t2^2)

Simplifying the equation, we get:

t1^2 = t2^2

Taking the square root of both sides, we have:

t1 = t2

Therefore, the time taken for acceleration is equal to the time taken for deceleration.

Now, let's calculate the time taken for the constant speed phase.

The total time for the journey is given as 1 minute. Subtracting the time taken for acceleration and deceleration, we have:

1 - 2 * t1 = t_const

Now, we can calculate the total time taken for the journey:

Total time = t1 + t_const + t2

Since t1 = t2, we can rewrite the equation as:

Total time = 2 * t1 + t_const

Substituting the value of t_const, we have:

Total time = 2 * t1 + (1 - 2 * t1) = 1 - t1

Therefore, the total time taken for the journey is 1 - t1 minutes.

Now, let's calculate t1.

From the equation D/2 = (1/2) * a * t1^2, we can express a in terms of t1:

a = (2 * D) / (t1^2)

Since the average speed during the acceleration phase is V, we can express a in terms of V:

a = V / t1

Equating the two expressions for a, we have:

(2 * D) / (t1^2) = V / t1

Simplifying the equation, we get:

t1 = sqrt((2 * D) / V)

Now, substituting the value of t1 in the equation for the total time, we have:

Total time = 1 - sqrt((2 * D) / V)

Therefore, the time taken for the train to travel from station A to station B is 1 - sqrt((2 * D) / V) minutes.

Please note that we need the value of D (the distance between stations A and B) and V (the average speed of the train) to calculate the exact time.