In a large hotel, a fast elevator takes you from the ground floor to the 21st floor. The elevator takes 17 s for this trip: 5 s at constant acceleration, 7 s at constant velocity, and 5 s at constant deceleration. Each floor in the hotel has a height of 2.5 m. Calculate the values of the acceleration and deceleration (assume they are equal). Calculate the maximum speed of the elevator.
4.4
To solve this problem, we will use the kinematic equations of motion. Let's break down the motion of the elevator into three parts:
1. First, let's find the acceleration (a) during the first 5 seconds of the elevator's motion. We can use the equation:
vf = vi + at
Here, vi is the initial velocity which we assume to be 0 because the elevator starts from rest. vf is the final velocity after 5 seconds, which is the velocity it reaches at constant acceleration. Rearranging the equation, we get:
vf = at
Since we know the time (t) is 5 seconds and the elevator moves through one floor (2.5m), we can substitute the values:
2.5 = a * 5
Solving for acceleration (a):
a = 2.5/5 = 0.5 m/s^2
2. Next, let's find the maximum speed (v_max) that the elevator reaches during constant velocity. During this time, the elevator neither accelerates nor decelerates, so the acceleration is 0. We know the elevator maintains this constant speed for 7 seconds. The equation we can use is:
vf = vi + at
Since the acceleration (a) is 0, the equation simplifies to:
vf = vi + 0
vf = vi
We know the elevator is at a constant speed during this period, which means vf is the maximum speed. Therefore, the maximum speed (v_max) is the same as the velocity reached at the end of the first phase, which is 0.5 m/s.
3. Finally, let's find the deceleration (a) during the last 5 seconds of the elevator's motion. Here again, we can use the equation:
vf = vi + at
This time, vf is 0 because the elevator stops at the final floor. vi is the final velocity it reaches after the constant velocity phase, which is also 0.5 m/s. The time (t) is 5 seconds. Substituting these values:
0 = 0.5 - 5a
Solving for acceleration (a):
0.5 = 5a
a = 0.5/5 = 0.1 m/s^2
Therefore, the deceleration is 0.1 m/s^2.
In summary:
- The acceleration (a) during the first 5 seconds of the elevator's motion is 0.5 m/s^2.
- The deceleration (a) during the last 5 seconds of the elevator's motion is 0.1 m/s^2.
- The maximum speed (v_max) reached by the elevator is 0.5 m/s.