Question

Железнодорожный состав двигался со скоростью v = 54 км/ч. Подъезжая к станции, он начал равномерно тормозить и, спустя время t = 3,5 мин, остановился. Определите количество вагонов в поезде, если его тормозной путь оказался в семь раз большим, чем длина всего состава. Длину электровоза и каждого вагона примите равными по l = 15 м.

Answer 1

Дано
Vo=15 м/с
t=210 c
V=0
n-?
n=L/l
l=15м
S=7L
S=Vot+0.5at²
V=Vo+at        a=(v-Vo)/t  a=-15/210=-0.07 м/с
S=15*210-0.5*0.07*210²=16065 м
L=16065/7=226.5
n=226.5/15≈15   14 вагонов и поезд

Answer 2

Problem Analysis

We are given that a train is moving with a speed of v = 54 km/h. As it approaches a station, it starts braking uniformly and comes to a stop after a time t = 3.5 minutes. The braking distance is seven times greater than the length of the entire train. The length of the locomotive and each wagon is given as l = 15 m. We need to determine the number of wagons in the train.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated motion. The equations we will use are:

1. v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. 2. s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Let's break down the problem step by step.

1. Convert the given speed from km/h to m/s: - 1 km/h = (1000 m) / (3600 s) = 5/18 m/s. - Therefore, the speed v = 54 km/h = (54 * 5/18) m/s = 15 m/s.

2. Determine the acceleration of the train during braking: - Since the train is braking uniformly, the final velocity v is 0 m/s. - The initial velocity u is 15 m/s. - The time t is given as 3.5 minutes, which is equal to (3.5 * 60) seconds = 210 seconds. - Using the equation v = u + at, we can solve for the acceleration a: 0 = 15 + a * 210. - Solving for a, we find a = -15/210 = -1/14 m/s^2. The negative sign indicates deceleration.

3. Calculate the braking distance: - The braking distance is given as seven times the length of the entire train. - The length of the locomotive and each wagon is given as l = 15 m. - Therefore, the braking distance is 7 * (length of the entire train) = 7 * (l + (number of wagons * l)). - Let's denote the number of wagons as n. The braking distance can be expressed as 7 * (15 + n * 15).

4. Use the equation s = ut + (1/2)at^2 to find the braking distance: - The initial velocity u is 15 m/s. - The time t is 210 seconds. - The acceleration a is -1/14 m/s^2. - Substituting these values into the equation, we get: 7 * (15 + n * 15) = 15 * 210 + (1/2) * (-1/14) * (210^2). - Solving this equation will give us the value of n, which represents the number of wagons in the train.

Let's calculate the number of wagons using the given information.

Calculation

1. Convert the given speed from km/h to m/s: - v = 54 km/h = 54 * (5/18) m/s = 15 m/s.

2. Determine the acceleration of the train during braking: - v = 0 m/s. - u = 15 m/s. - t = 3.5 minutes = 3.5 * 60 seconds = 210 seconds. - 0 = 15 + a * 210. - Solving for a, we find a = -1/14 m/s^2.

3. Calculate the braking distance: - Braking distance = 7 * (15 + n * 15) = 105 + 105n.

4. Use the equation s = ut + (1/2)at^2 to find the braking distance: - 105 + 105n = 15 * 210 + (1/2) * (-1/14) * (210^2). - Solving this equation will give us the value of n, which represents the number of wagons in the train.

Let's solve the equation to find the number of wagons.

Solution

Using the equation 105 + 105n = 15 * 210 + (1/2) * (-1/14) * (210^2), we can solve for n.

105 + 105n = 3150 - (1/2) * (1/14) * 210^2 105 + 105n = 3150 - (1/2) * (1/14) * 44100 105 + 105n = 3150 - 150 105 + 105n = 3000 105n = 3000 - 105 105n = 2895 n = 2895 / 105 n = 27.57

Since the number of wagons must be a whole number, we round down to the nearest whole number.

Therefore, the number of wagons in the train is 27.

Answer

The number of wagons in the train is 27.