Finding Distance and Velocity of a Loose Bolt from Moving Elevator

AI Thread Summary
A bolt falls from an elevator moving upward at 6.0 m/s and takes 3.0 seconds to reach the bottom of the shaft. The calculations show that the initial height of the elevator when the bolt came loose is approximately 26.1 meters, with the negative value indicating the distance fallen. The final speed of the bolt upon impact is calculated to be 23.4 m/s. It's crucial to define the direction of vectors correctly, with upward as positive and downward as negative, to avoid confusion in calculations. Understanding these vector components is essential for solving similar physics problems accurately.
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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


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

<br /> v^2 = v_0^2 + 2 a \Delta x<br />


The Attempt at a Solution



I first used <br /> x = x_0 + v_0 t + (1/2) a t^2<br /> and plugged in V0 = 6 m/s , t = 3 sec and a = -9.8 m/s2 since the object is going down. I ended up getting a \Deltay of -26.1 m. Is it ok for this value to be negative?

Then I used <br /> v^2 = v_0^2 + 2 a \Delta x<br /> and plugged in my -26.1 m to get a Vf of 23.4 m/s


I'm confused about the when to use a negative or positive velocity or \Deltay.
 
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xr250h said:
<br /> x = x_0 + v_0 t + (1/2) a t^2<br />

First you need to know what is x in your equation. In this case it would be the distance from the bottom of the shaft to the bolt (at the certain time t). So, you actually want to know x_0, because it's the initial height. After time interval t, x is zero (bolt has reached the bottom).

That's why it is important to consider such equations in vector form and define which vector points in which direction. In this case, it would be

<br /> \vec{x}=\vec{x_0}+\vec{v_0}t+\frac{1}{2}\vec{g}t^2<br />

x point upwards to the bolt, x_0 also points upwards and denotes the initial height, v_0 also points upwards (because the elevator goes up) and g, of course, points downwards. So, it's best to take up as positive direction (we actually set the bottom of the shaft as the origin of coordinate system). Now, everything that points upwards is positive and everything else is negative.

<br /> \vec{x}=(x_0+v_0 t-\frac{1}{2}g t^2)\hat{x}<br />

\hat{x} is unit vector pointing upwards (it's used to denote direction and mesurement unit - its length is 1).

I hope all of this is not too much confusing, but these details are very important.
 
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