# Relative velocity of bolt

1. Mar 3, 2013

### Toranc3

1. The problem statement, all variables and given/known data

An elevator is moving upward at a constant speed of 2.50 m/s. A bolt in the elevator ceiling 3.00m above the elevator floor works loose and falls.

A. How long will it take the bolt to fall to the elevator floor?

I am not sure if I should do this relative to elevator or earth.

2. Relevant equations

Vb/g=Vb/e+Ve/g

Yb/g=Yb/e+Ye/g

3.The attempt at a solution
Vb/g=Vb/e+Ve/g
Vb/g= velocity of bolt relative to ground
Vb/e= velocity of bolt relative to earth
Ve/g= velocity of elevator relative to ground

Yb/g=Yb/e+Ye/g
Yb/g=position of bolt relative to ground
Yb/e= position of bolt relative to earth
Ye/g=position of elevator relative to ground

Vb/g=Vb/e+Ve/g
Vb/g=0+2.5m/s

Yb/g=Yb/e+Ye/g
Yb/g=3m+ 2.5m/s*t

Is this right so far? Where do I go from here? Thanks!

2. Mar 3, 2013

### haruspex

You can make this question much easier by recognising that since the elevator is moving at constant speed it constitutes an inertial frame.

3. Mar 3, 2013

### Toranc3

Can you clarify what you mean? :/

4. Mar 3, 2013

### haruspex

Take the elevator as the reference frame. What are the initial height, initial velocity, final height and acceleration of the bolt in that frame?

5. Mar 3, 2013

### Toranc3

3m, 0m/s, 0m, -9.81m/s^(2)?

I think I see what you mean. What if the elevator was not traveling at a constant speed but rather a constant acceleration. How would that change?

6. Mar 3, 2013

### tms

He means that the laws of physics are the same in every inertial reference frame. You can think of an inertial frame as a frame that is not accelerating (that is, one that is moving at constant speed relative to another inertial frame). It's a little more complicated than that, but you can ignore those complications for present purposes.