Calculating the Velocity of an Elevator

[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

 

[haruspex]

The attempt at a solution

I would not say that qualifies as an attempt.

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.

 

[mdavies23]

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

 

[haruspex]

The unknown velocity would be assigned to Vy.

But you had

Vy=V0 + ay*delta t

Is the elevator accelerating?

 

[mdavies23]

But you had

Is the elevator accelerating?

No because it is constant velocity.

 

[haruspex]

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?

 

[mdavies23]

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

 

[haruspex]

1m

Right. So for how long were the keys falling?

 

[mdavies23]

Right. So for how long were the keys falling?

2 seconds so V would be .5m/s

 

[mdavies23]

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)?

 

[haruspex]

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?

 

[mdavies23]

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

 

[haruspex]

That is how long the elevator is moving upward

It is how long the elevator moved upwards until what event?

 

[mdavies23]

It is how long the elevator moved upwards until what event?

Until the key is dropped

 

[haruspex]

Until the key is dropped

Right. So what has it do with the time for which the key fell?

 

[mdavies23]

The key strikes at the same height as it fell

 

[haruspex]

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?

 

[mdavies23]

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

 

[haruspex]

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?

 

[mdavies23]

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?

 

[haruspex]

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:

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?"

 

[mdavies23]

Oh 1s

 

[haruspex]

Oh 1s

No.
If you move at speed v, how long does it take you to go distance x?

 

[mdavies23]

No.
If you move at speed v, how long does it take you to go distance x?

x/v=t

 

[haruspex]

x/v=t

So apply that to the elevator's motion while the keys are falling.

 

[mdavies23]

So apply that to the elevator's motion while the keys are falling.

t=1/v

 

[haruspex]

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?

 

[mdavies23]

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

 

[haruspex]

We know t, s, and a

One more.

 

[mdavies23]

One more.

v

 

[haruspex]

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?

 

[mdavies23]

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

 

[haruspex]

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?

 

[mdavies23]

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

 

[jbriggs444]

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.

 

[haruspex]

The initial speed is the same as the elevator

Yes.

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?