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.

Question

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.

(a) What is the maximal speed v of the elevator ? (in m/s)
(b) What is the acceleration a ? (in m/s2)

Answer 1

divide the distance 320 by time 40seconds to get speed. for acceleration, divide this speed by 5.

Answer 2

To solve this problem, we can use the equations of motion for uniformly accelerated motion:

1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as

where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
- s is the displacement

Let's go step by step to find the answers:

(a) What is the maximal speed v of the elevator?

First, we need to find the time it takes for the elevator to reach its maximum speed. We know that it accelerates for 5 seconds and then moves at a constant speed for 35 seconds. Therefore, the time taken to reach the maximum speed is 5 seconds.

Using equation (1), we can find the initial velocity u:
u = v - at

Since the elevator starts from rest, its initial velocity u is 0. So we have:
0 = v - a * 5

Simplifying the equation, we get:
v = 5a

Next, we need to find the displacement during the acceleration phase. Using equation (2):
s = ut + (1/2)at^2

Given that the acceleration is a > 0 and the time is 5 seconds, we have:
s = 0 * 5 + (1/2) * a * (5^2)

Simplifying the equation, we get:
s = (25/2) * a

Now, let's find the displacement during the deceleration phase. We know that the elevator comes to a halt at the top floor, which is 320 meters above the ground floor. As the elevator is moving upwards, the displacement is positive. Using equation (2):
s = ut + (1/2)at^2

Here, the acceleration is -a (opposite direction) and the time is 35 seconds. So we have:
320 = 0 * 35 + (1/2) * (-a) * (35^2)

Simplifying the equation, we get:
320 = (1/2) * (-a) * 1225

Dividing both sides by 612.5, we get:
0.52 = a

Substituting this value of a into the equation we found earlier for v, we get:
v = 5 * 0.52
v ≈ 2.6 m/s

Therefore, the maximal speed of the elevator is approximately 2.6 m/s.

(b) What is the acceleration a?

From our calculations above, we found that the acceleration a is approximately 0.52 m/s^2.