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## Homework Statement

A bolt comes loose from underneath an elevator that is moving upward at a constant speed of 6.0 m/s. The bolt reaches the bottom of the elevator shaft in 3.0 s.

a.) How high above the bottom of the shaft was the elevator when the bolt came loose?

b.) What is the speed of the bolt when it hits the bottom of the shaft?

## Homework Equations

[tex]

x = x_0 + v_0 t + (1/2) a t^2

[/tex]

[tex]

v^2 = v_0^2 + 2 a \Delta x

[/tex]

## The Attempt at a Solution

I first used [tex]

x = x_0 + v_0 t + (1/2) a t^2

[/tex] and plugged in V

_{0}= 6 m/s , t = 3 sec and a = -9.8 m/s

^{2}since the object is going down. I ended up getting a [tex]\Delta[/tex]y of -26.1 m. Is it ok for this value to be negative?

Then I used [tex]

v^2 = v_0^2 + 2 a \Delta x

[/tex] and plugged in my -26.1 m to get a V

_{f}of 23.4 m/s

I'm confused about the when to use a negative or positive velocity or [tex]\Delta[/tex]y.