Why Is the Constant Speed of an Elevator Ignored in a Free Fall Calculation?

[godkills]

While riding on an elevator descending with a constant speed of 3.0 m/s, you accidentally drop a book from under your arm.
How long does it take for the book to reach the elevator floor, 1.2 m below your arm?
What is the book’s speed when it hits the elevator floor?My question is why is 3.0 m/s in constant speed ignored in this question?

We are using the equation x = x₀ + v₀ + 1/2(a)(t²)

Why is that v₀ is not plugged in? When 3.0 m/s is v₀ (velocity).

Since velocity here is descending then the velocity in this question is -3.0 m/s correct?

also i am having problem solving part b in which i don't know where to begin

 

[tvavanasd]

Both the elevator floor and the book are within the same inertial frame of reference (both are moving at a constant speed in a straight line). You are interested in the relationship between them, not that with objects outside that frame.
Vo of the book relative to the floor is zero.
The sign that you give the velocity must agree with that for position and acceleration (if you call down positive, acceleration down is also positive).
Once you know the time for the book to fall 1.2 m, can you not calculate speed (or velocity) based on other constant acceleration equations?

 

[godkills]

Vo of the book relative to the floor is zero.

Is that because the book is in free fall meaning that only gravity is affecting it and that the book has no speed only acceleration which is gravity?

Free fall (free from any effects other then gravity so speed is one of the effects correct?)

For part (b) I keep getting 7.9 but the answer is 4.8

 

[tvavanasd]

Remember to show units in your answers.

If you were inside the elevator, you would not be able to tell whether it was at rest, or moving up or down at a constant speed (assuming it's motion is smooth).

All motion of interest in this question is with respect to the inside of the elevator (the inertial frame), not the building.

In this case, you should completely ignore the velocity of the elevator. True, the velocity of the book when it hits the floor relative to the building is about 7.85 m/s, but the book will actually hit the floor of the elevator at approximately 4.85 m/s (relative to the floor). This is representative of how hard it will hit the elevator floor (7.85 m/s is not).

I get:
t = sqrt (2 * 1.2 / 9.8)
t = 0.495 s

v = a * t
v = 9.8 * 0.495
v = 4.85 m/s