Relative motion: Understanding acceleration and velocity in an elevator

[asdf1]

If an elevator moves at an acceleration of 5 ft per second squared going up and when the velocity is 8 ft/sec going up, a screw falls from the ceiling of the elevator down to the floor of the elevator. The height of the elevator is 9 ft. Find the length the screw dropped.

My problem:
length= 8t-0.5*(32-4)$$t^2$$

but the worked out solution to the problem says that the acceleration is 32 ft/$$sec^2$$...
Why is that?

 

[neutrino]

Because you are using units in which g has a magnitude of 32.

 

[HallsofIvy]

If an elevator moves at an acceleration of 5 ft per second squared going up and when the velocity is 8 ft/sec going up, a screw falls from the ceiling of the elevator down to the floor of the elevator. The height of the elevator is 9 ft. Find the length the screw dropped.

My problem:
length= 8t-0.5*(32-4)$$t^2$$

but the worked out solution to the problem says that the acceleration is 32 ft/$$sec^2$$...
Why is that?

I presume you actually meant the at the elevator was accelerating at 4 ft/sec². Or did you mean to write $8t- 0.5(32- 5)t^2$?
Obviously, relative to the elevator, the the screw drops 9 ft. ! The question is then the distance the screw falls relative to some frame of reference in which the elevator is moving at 8 ft/sec at the instant the screw drops.

Here's how I would argue: while the screw is falling downward, with acceleration, d₁= -32 ft/sec², initial velocity 8 ft/sec, so that the distance it falls in t seconds is -(8t- 16t²)= 16t²- 8t (since the distance it falls is positive), the elevator is accelerating upward at 4 (or is it 5?), also initial velocity 8 ft/sec, so that the distance it rises in t seconds is d₂= 8t+ 2t². The distance the screw falls, relative to the elevator is
d1- d2= (16- 2)t². The coefficient of t² is precisely your 0.5*(32-4) but notice that the two "8t" terms have canceled out! Since both elevator and screw had that same upward initial velocity, it is irrelevant, relative to the elevator. Since the elevator is 9 ft high, obviously 0.5*(32-4)t²= (16-2)t²= 14t²= 9. Solve that for t.

But that is not what was asked! You are asked for the distance the screw "actually" falls (relative to some "non-moving" frame of reference- obviously the distance the screw falls relative to the elevator is 9 ft.). To answer that, plug the t you got into d1= 16t²- 8t.

Presumably, that will be less than 9 ft.

 

[Office_Shredder]

The screw is accelerating with magnitude 32 + 5 relative to the floor

 

[asdf1]

Sorry for the misprint! It's accelerating at 4 ft/sec[tex]^2[tex]...
thank you very much for clearing up my problem! ^_^