Calculating the Velocity of an Elevator

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SUMMARY

The discussion focuses on calculating the velocity of an elevator and the subsequent motion of keys released within it. The elevator moves upwards at a constant velocity, and after 2 seconds, the keys are released from a height of 1.00m. The calculated velocity of the elevator is determined to be 0.5 m/s. When the elevator moves downwards at the same speed, the difference in height between the release point and where the keys hit the floor is analyzed using the SUVAT equations for constant acceleration.

PREREQUISITES
  • Understanding of constant velocity motion
  • Familiarity with the SUVAT equations for constant acceleration
  • Basic knowledge of free fall motion and gravitational acceleration
  • Ability to manipulate equations involving initial and final velocities
NEXT STEPS
  • Study the SUVAT equations in detail, focusing on their applications in various motion scenarios
  • Learn about gravitational effects on objects in free fall
  • Explore the concept of reference frames in physics
  • Practice problems involving constant velocity and acceleration to reinforce understanding
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding motion dynamics in elevators and similar systems.

  • #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?
 
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  • #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?
 

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