Calculating the Velocity of an Elevator

Click For Summary
The discussion focuses on calculating the velocity of an elevator and the subsequent motion of keys released inside it. The elevator moves upward at a constant velocity, and after 2 seconds, the teacher releases the keys from a height of 1 meter. The group explores the relationship between the elevator's velocity and the time it takes for the keys to fall, emphasizing that the keys strike the elevator floor at the same height from which they were released. The conversation highlights the importance of understanding the motion of both the elevator and the keys, particularly in applying the correct equations of motion. Ultimately, the key takeaway is that the initial and final velocities of the keys are equal to the elevator's velocity at the moment of release.
  • #31
mdavies23 said:
v
Can you be a bit clearer? We are now discussing the trajectory of the keys. Of the five variables I listed, acceleration, time, displacement, initial velocity and final velocity you correctly posted that we know the first three: a, t, s. Which velocity do we know?
 
Physics news on Phys.org
  • #32
haruspex said:
Can you be a bit clearer? We are now discussing the trajectory of the keys. Of the five variables I listed, acceleration, time, displacement, initial velocity and final velocity you correctly posted that we know the first three: a, t, s. Which velocity do we know?
We know both initial=final
 
  • #33
mdavies23 said:
We know both initial=final
Explain more. What do you think the initial speed is? How do you deduce it is the same as the final?
 
  • #34
haruspex said:
Explain more. What do you think the initial speed is? How do you deduce it is the same as the final?
The initial speed is the same as the elevator and the final speed is also the speed of the elevator
 
  • #35
If I understand where @haruspex is leading you, you can get an equation for key position above the starting point as a function of time, elevator velocity and gravity. You can get a simpler equation for elevator floor position above the starting point as a function of time, elevator velocity and gravity. You can then equate the two and solve for time.

It would seem that a simpler (but arguably equivalent) approach would be to switch for a moment to a reference frame in which the elevator is at rest. Figure out how long the keys take to drop one meter to the floor using this frame. Then switch back to the ground frame.
 
  • #36
mdavies23 said:
The initial speed is the same as the elevator
Yes.
mdavies23 said:
the final speed is also the speed of the elevator
No. This is a common blunder. The equations are for constant acceleration. As soon as the keys touch the ground the acceleration is no longer constant, so the the equations don't apply. We can only apply them up to the instant before they hit the ground, and at that time they will have quite a different velocity from the elevator's.

So we have these four: time, acceleration, initial velocity and displacement. Which SUVAT equation has those four?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
5K
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
7K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
4K
  • · Replies 18 ·
Replies
18
Views
8K
  • · Replies 18 ·
Replies
18
Views
12K