Question

Лифт, поднимаясь равноускоренно в течение первого промежутка времени t1 = 2 с, достигает скорости v1 = 4 м/с, с которой продолжает подъем в течение второго промежутка времени t2 = 4 с. Затем лифт движется равнозамедленно и к концу третьего промежутка t3 = 3 с останавливается. Определить высоту подъема лифта. Решить задачу также графически

Answer 1

Problem Analysis

We are given the following information about the elevator's motion: - The elevator accelerates uniformly for the first time interval, t1 = 2 s, and reaches a speed of v1 = 4 m/s. - The elevator continues to ascend at a constant speed of v1 for the second time interval, t2 = 4 s. - The elevator decelerates uniformly for the third time interval, t3 = 3 s, and comes to a stop.

We need to determine the height the elevator has ascended during this motion.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated motion. Let's break down the problem into three parts: the first interval of acceleration, the second interval of constant speed, and the third interval of deceleration.

# First Interval: Acceleration

During the first interval, the elevator accelerates uniformly from rest to a speed of v1 = 4 m/s in a time of t1 = 2 s. We can use the equation:

v = u + at

where: - v is the final velocity (4 m/s), - u is the initial velocity (0 m/s), - a is the acceleration, and - t is the time interval (2 s).

Since the initial velocity is 0, the equation simplifies to:

v = at

Solving for acceleration (a), we have:

a = v / t

Substituting the given values, we find:

a = 4 m/s / 2 s = 2 m/s^2

Now, we can use the equation for displacement during uniformly accelerated motion:

s = ut + (1/2)at^2

Since the initial velocity (u) is 0, the equation simplifies to:

s = (1/2)at^2

Substituting the values of acceleration (a) and time (t1), we find:

s1 = (1/2)(2 m/s^2)(2 s)^2 = 4 m

Therefore, during the first interval, the elevator ascends a height of 4 meters.

# Second Interval: Constant Speed

During the second interval, the elevator continues to ascend at a constant speed of v1 = 4 m/s for a time of t2 = 4 s. Since the speed is constant, there is no acceleration, and the displacement is given by:

s2 = v2 * t2

Substituting the values, we find:

s2 = (4 m/s)(4 s) = 16 m

Therefore, during the second interval, the elevator ascends a height of 16 meters.

# Third Interval: Deceleration

During the third interval, the elevator decelerates uniformly and comes to a stop in a time of t3 = 3 s. The final velocity is 0, and we can use the equation:

v = u + at

where: - v is the final velocity (0 m/s), - u is the initial velocity (v1 = 4 m/s), - a is the acceleration, and - t is the time interval (t3 = 3 s).

Solving for acceleration (a), we have:

a = (v - u) / t

Substituting the given values, we find:

a = (0 m/s - 4 m/s) / 3 s = -4/3 m/s^2

The negative sign indicates that the elevator is decelerating.

Using the equation for displacement during uniformly accelerated motion, we have:

s = ut + (1/2)at^2

Since the final velocity (v) is 0, the equation simplifies to:

s = ut + (1/2)at^2

Substituting the values of initial velocity (u = v1 = 4 m/s), acceleration (a), and time (t3), we find:

s3 = (4 m/s)(3 s) + (1/2)(-4/3 m/s^2)(3 s)^2 = 6 m - 6 m = 0 m

Therefore, during the third interval, the elevator does not ascend or descend.

# Total Displacement

To find the total displacement, we add the displacements from each interval:

Total displacement = s1 + s2 + s3 = 4 m + 16 m + 0 m = 20 m

Therefore, the elevator ascends a total height of 20 meters.

Graphical Solution

To represent the elevator's motion graphically, we can plot a velocity-time graph and a displacement-time graph.

# Velocity-Time Graph

The velocity-time graph will consist of three segments: - The first segment represents the acceleration phase, where the velocity increases from 0 m/s to 4 m/s in 2 seconds. - The second segment represents the constant speed phase, where the velocity remains constant at 4 m/s for 4 seconds. - The third segment represents the deceleration phase, where the velocity decreases from 4 m/s to 0 m/s in 3 seconds.

The graph will look like this:

   ^
v  |         .
   |        .
   |       .
   |      .
   |     .
   |    .
   |   .
   |  .
   | .
   |.
   +----------------------------------------------------> t
        0    2    4    6    8    10   12   14   16   18   20

# Displacement-Time Graph

The displacement-time graph will consist of three segments: - The first segment represents the acceleration phase, where the displacement increases from 0 m to 4 m in 2 seconds. - The second segment represents the constant speed phase, where the displacement increases linearly from 4 m to 20 m in 4 seconds. - The third segment represents the deceleration phase, where the displacement remains constant at 20 m for 3 seconds.

The graph will look like this:

   ^
s  |         .
   |        .
   |       .
   |      .
   |     .
   |    .
   |   .
   |  .
   | .
   |.
   +----------------------------------------------------> t
        0    2    4    6    8    10   12   14   16   18   20

The area under the displacement-time graph represents the total displacement, which is 20 m in this case.

Conclusion

The elevator ascends a total height of 20 meters during its motion. This can be determined both analytically and graphically.