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.

mdavies23

Homework Statement


A teacher rides in an elevator that moves upwards at a constant velocity and holds a set keys 1.00m above the

floor of the elevator. After 2.00 seconds of moving upwards from the ground level of the science building, theteacher releases the keys from their hand. Relative to the ground level of the building, he keys strike the floor of the

elevator at the exact height as where they were released.

(a) Calculate the velocity of the elevator.

(b) Now consider the elevator as traveling downwards at the same constant speed found in part (c). The

keys are again released 1.00m above the floor of the elevator. What is the difference in height (relative toground level) between where the keys are released and where they hit the floor of the elevator?

Homework Equations


Vy=V0 + ay*delta t

The Attempt at a Solution


Key height = 1m
time = 2s
g=-9.8m/s^2
Free fall motion
 
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mdavies23 said:
The attempt at a solution
I would not say that qualifies as an attempt.
mdavies23 said:
time = 2s
Time for what? How does this relate to the trajectory of the keys?

Start by assigning a variable to represent the unknown velocity, then see what equations you can write involving it.
 
haruspex said:
I would not say that qualifies as an attempt.

Time for what? How does this relate to the trajectory of the keys?

Start by assigning a variable to represent the unknown velocity, then see what equations you can write involving it.
The unknown velocity would be assigned to Vy. I Honestly have no idea where to go with the problem that's why there is not a good try
 
mdavies23 said:
The unknown velocity would be assigned to Vy.
But you had
mdavies23 said:
Vy=V0 + ay*delta t
Is the elevator accelerating?
 
haruspex said:
But you had

Is the elevator accelerating?
No because it is constant velocity.
 
mdavies23 said:
No because it is constant velocity.
Right. And all motion is vertical, so we can dispense with the y suffix.
How far does the lift travel while the keys are falling?
 
haruspex said:
Right. And all motion is vertical, so we can dispense with the y suffix.
How far does the lift travel while the keys are falling?
1m
 
mdavies23 said:
1m
Right. So for how long were the keys falling?
 
haruspex said:
Right. So for how long were the keys falling?
2 seconds so V would be .5m/s
 
  • #10
haruspex said:
Right. So for how long were the keys falling?
So for part b, if V=.5m/s and the key is released at 1m again would you just plug those in along with -g into V^2=V^2 +2a(delta y)?
 
  • #11
mdavies23 said:
2 seconds
I don't see where that is coming from. There is a period of 2 seconds mentioned, but how is that defined in the question?
 
  • #12
haruspex said:
I don't see where that is coming from. There is a period of 2 seconds mentioned, but how is that defined in the question?
That is how long the elevator is moving upward
 
  • #13
mdavies23 said:
That is how long the elevator is moving upward
It is how long the elevator moved upwards until what event?
 
  • #14
haruspex said:
It is how long the elevator moved upwards until what event?
Until the key is dropped
 
  • #15
mdavies23 said:
Until the key is dropped
Right. So what has it do with the time for which the key fell?
 
  • #16
The key strikes at the same height as it fell
 
  • #17
mdavies23 said:
The key strikes at the same height as it fell
That does not answer my question. For two seconds, the lift moved up and the keys were in the teacher's hand. Then the key was let go (at some height h from the ground) and time t later it hit the floor of the lift, again at height h from the ground.
Have you any basis for claiming from that that t=2 seconds?
Would it make any difference to the answer if it were three seconds before the teacher let go of the keys?
 
  • #18
haruspex said:
That does not answer my question. For two seconds, the lift moved up and the keys were in the teacher's hand. Then the key was let go (at some height h from the ground) and time t later it hit the floor of the lift, again at height h from the ground.
Have you any basis for claiming from that that t=2 seconds?
Would it make any difference to the answer if it were three seconds before the teacher let go of the keys?
No I guess it wouldn’t matter
 
  • #19
mdavies23 said:
No I guess it wouldn’t matter
Right. So please have another go at my question in post #8. If the lift is moving up at speed v and moves 1m in the time the keys are falling, how long is that time?
 
  • #20
haruspex said:
Right. So please have another go at my question in post #8. If the lift is moving up at speed v and moves 1m in the time the keys are falling, how long is that time?
So this would have to factor -g in right?
 
  • #21
mdavies23 said:
So this would have to factor -g in right?

No. Just concentrate on these facts. You can ignore the rest of the problem for the moment:
haruspex said:
the lift is moving up at speed v and moves 1m in the time the keys are falling, how long is that time?
Hint: i can change that to "the lift is moving up at speed v and moves 1m in the time that (blah, blah...), how long is that time?"
 
  • #22
Oh 1s
 
  • #23
mdavies23 said:
Oh 1s
No.
If you move at speed v, how long does it take you to go distance x?
 
  • #24
haruspex said:
No.
If you move at speed v, how long does it take you to go distance x?
x/v=t
 
  • #25
mdavies23 said:
x/v=t
So apply that to the elevator's motion while the keys are falling.
 
  • #26
haruspex said:
So apply that to the elevator's motion while the keys are falling.
t=1/v
 
  • #27
mdavies23 said:
t=1/v
Yes, or more correctly with units, 1/v m.

Now we need to consider the trajectory of the keys.
In the "SUVAT" equations for constant acceleration there are five variables: initial velocity u, final velocity v (but we already are using v for the elevator, so call it w instead), acceleration a, time t, and displacement s. Which of these do we know for the keys?
 
  • #28
haruspex said:
Yes, or more correctly with units, 1/v m.

Now we need to consider the trajectory of the keys.
In the "SUVAT" equations for constant acceleration there are five variables: initial velocity u, final velocity v (but we already are using v for the elevator, so call it w instead), acceleration a, time t, and displacement s. Which of these do we know for the keys?
We know t, s, and a
 
  • #29
mdavies23 said:
We know t, s, and a
One more.
 
  • #30
haruspex said:
One more.
v
 

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