Question

будь ласка !! Ліфт телевізійної вежі перед зупинкою, рухаючись рівноприскорено зі сповільненням, проходить 49 м за 14 с. знайти прискорення ліфта під час гальмування та його швидкість на початку гальмування

Answer 1

Ответ: 0,5 м/с²       7 м/с

Объяснение: Див. додаток

Answer 2

Calculation of Elevator's Acceleration and Speed

To find the acceleration of the elevator during deceleration and its speed at the beginning of deceleration, we can use the following kinematic equations:

1. Equation for displacement with constant acceleration: \[ s = ut + \frac{1}{2}at^2 \] where: - \( s \) = displacement - \( u \) = initial velocity - \( a \) = acceleration - \( t \) = time

2. Equation for final velocity with constant acceleration: \[ v = u + at \] where: - \( v \) = final velocity - \( u \) = initial velocity - \( a \) = acceleration - \( t \) = time

Given: - Displacement, \( s = 49 \, \text{m} \) - Time, \( t = 14 \, \text{s} \)

First, let's find the acceleration of the elevator during deceleration.

Using the equation for displacement with constant acceleration, we can find the acceleration: \[ s = ut + \frac{1}{2}at^2 \]

Plugging in the given values: \[ 49 = 0 + \frac{1}{2}a(14)^2 \]

Solving for \( a \): \[ a = \frac{2 \times 49}{14^2} \]

\[ a = \frac{98}{196} \]

\[ a = 0.5 \, \text{m/s}^2 \]

The acceleration of the elevator during deceleration is 0.5 m/s^2.

Next, let's find the speed of the elevator at the beginning of deceleration.

Using the equation for final velocity with constant acceleration, we can find the speed: \[ v = u + at \]

Given that the elevator is moving with constant acceleration, the initial velocity \( u \) is the speed of the elevator at the beginning of deceleration.

Plugging in the values: \[ v = 0 + 0.5 \times 14 \]

\[ v = 7 \, \text{m/s} \]

The speed of the elevator at the beginning of deceleration is 7 m/s.