An elevator is stopped at the ground floor. It starts moving upwards at constant acceleration a>0 for 5 seconds. It then keeps a constant speed for 35 seconds. Finally, it slows down with an acceleration of the same magnitude (but opposite direction) −a , until it comes to a halt at the top floor. The top floor is 320 meters above the ground floor.
(1) What is the maximal speed v of the elevator ?
(2)What is the acceleration a
To answer these questions, we need to use the equations of motion for uniformly accelerated motion. Let's break down the motion of the elevator into three stages:
Stage 1: Constant Acceleration
During the first 5 seconds, the elevator is moving upwards with a constant acceleration, a>0. Let's call the initial velocity (when t=0) as u and the final velocity (when t=5 seconds) as v1. We need to find v1.
The equation connecting initial velocity (u), final velocity (v1), acceleration (a), and time (t) is:
v1 = u + a * t
Since we know that the initial velocity u is 0 (the elevator starts from rest), the equation simplifies to:
v1 = a * t
Given that t=5 seconds, we can calculate v1 by substituting the values:
v1 = a * 5
Stage 2: Constant Speed
During the next 35 seconds, the elevator moves with a constant speed. Let's call this speed v1 (the same as the final velocity from the previous stage). The distance covered during this stage depends on the speed and time.
Using the equation:
distance = speed * time
The distance covered during this stage is:
distance = v1 * 35
Stage 3: Deceleration
In the final stage, the elevator slows down with an acceleration of -a (negative because it's in the opposite direction of the initial acceleration). This stage continues until the elevator comes to a halt at the top floor.
We need to find the time taken during this deceleration stage. Let's call this time t2. The initial velocity at this stage is v1, and the final velocity when the elevator comes to a halt (t=40 seconds) is 0. The equation for this stage is:
0 = v1 - a * t2
Simplifying and solving for t2:
t2 = v1 / a
Finally, we can find the distance covered during this deceleration stage using the equation:
distance = v1 * t2
Given that the top floor is 320 meters above the ground floor, the total distance covered is:
320 = v1 * t2
Now we have two equations with two unknowns (v1, a) which are linked. We can solve these equations to find the answers.
(1) What is the maximal speed v of the elevator?
The maximal speed occurs at the end of the first stage, so v = v1.
(2) What is the acceleration a?
We can solve the equations above to find the value of a.