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 ? (in m/s)
(2)What is the acceleration a ? (in m/s2)
To solve this problem, we can break it down into three main parts:
1. The elevator's motion during the acceleration phase
2. The elevator's motion during the constant speed phase
3. The elevator's motion during the deceleration phase
Let's start with the first part:
1. The elevator's motion during the acceleration phase:
Given: Initial velocity (u) = 0 m/s, time (t) = 5 seconds
We know that the formula for calculating the distance covered during uniformly accelerated motion is:
s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time.
Since the elevator is starting from rest, the initial velocity is 0 m/s. We want to find the distance covered during this phase, which we'll call s1.
s1 = 0 + (1/2)at^2
s1 = (1/2)at^2
Now, let's move on to the second part:
2. The elevator's motion during the constant speed phase:
Given: Time (t) = 35 seconds
During this phase, the elevator is moving at a constant speed, so there is no acceleration. We want to find the distance covered during this phase, which we'll call s2.
s2 = velocity × time = v × t
Finally, let's move on to the third part:
3. The elevator's motion during the deceleration phase:
Given: Final velocity (v) = 0 m/s, time (t) = unknown
In this phase, the elevator is slowing down with an acceleration of -a. We want to find the distance covered during this phase, which we'll call s3.
The formula for the final velocity in uniformly accelerated motion is:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the final velocity is 0, we can solve for time:
0 = v + at
0 = 0 + at
t = -v/a
Now, let's calculate the distance covered during this phase:
s3 = vt + (1/2)at^2
s3 = 0 x (-v/a) + (1/2)(-a)((-v/a)^2)
s3 = 0 + (1/2)(-a)((v^2/a^2))
s3 = -v^2 / (2a)
Now, let's add up the distances covered during each phase to find the total distance traveled by the elevator:
Total distance (s_tot) = s1 + s2 + s3
s_tot = (1/2)at^2 + v × t - v^2 / (2a)
Given: Total distance (s_tot) = 320 meters
Now, we have an equation containing unknowns a and v. We can solve this equation to find our answers. Let's simplify the equation:
320 = (1/2)at^2 + vt - v^2 / (2a)
Multiply through by 2a to get rid of the denominators:
640a = at^2 + 2avt - v^2
Divide through by t to isolate a:
640a / t = at + 2av - v^2 / t
Rearrange terms:
at + v^2 / t = 640a / t - 2av
Factor out a:
a(t + v^2 / t - 640 / t + 2v) = 0
Since we know a > 0, the equation becomes:
t + v^2 / t - 640 / t + 2v = 0
Now, we have an equation involving only v and t, which we can solve to find the values of v and t.