Question

Потяг, рухаючись рівносповільнено, протягом часу t=1хв зменшує свою швидкість від v0=72 км/год до v=54 км/год. Знайти прискорення а потяга і визначити відстань s, яку він пройшов під час гальмування.

Answer 1

Ответ:

дано:

t= 1 хв = 60 секунд

v1=72 км/год

v2=54 км/год

знайти:

а(з вектором) - ?

s - ?

розв‘язок :

а= v1-v2/t

a=54-72/60= -18/60= -0,3 м/с

Объяснение:

Answer 2

Calculation of Acceleration and Distance Traveled by the Train

To find the acceleration and distance traveled by the train during braking, we can use the following equations:

1. The equation for acceleration is given by:

a = (v - v0) / t

where: - a is the acceleration - v is the final velocity - v0 is the initial velocity - t is the time taken

2. The equation for distance traveled is given by:

s = v0 * t + (1/2) * a * t^2

where: - s is the distance traveled

Let's calculate the acceleration and distance traveled by the train.

Given: - Initial velocity (v0) = 72 km/h - Final velocity (v) = 54 km/h - Time (t) = 1 min = 1/60 hour

Calculation:

Converting the velocities from km/h to m/s:

v0 = 72 km/h = (72 * 1000) / 3600 m/s = 20 m/s

v = 54 km/h = (54 * 1000) / 3600 m/s = 15 m/s

Converting the time from minutes to hours:

t = 1 min = 1/60 hour

Now, let's calculate the acceleration using the first equation:

a = (v - v0) / t = (15 - 20) / (1/60) = -300 m/s^2

The negative sign indicates that the train is decelerating or slowing down.

Next, let's calculate the distance traveled using the second equation:

s = v0 * t + (1/2) * a * t^2 = 20 * (1/60) + (1/2) * (-300) * (1/60)^2 = 1/3 - 1/720 = 239/720 ≈ 0.3319 km

Therefore, the acceleration of the train during braking is approximately -300 m/s^2, and the distance it travels during braking is approximately 0.3319 km.

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