a book accidentally falls from a shelf 4.2 m high.a librarian is standing nearby and moves .80m,starting from rest, to catch the book.what must be his average acceleration if he catches the book when it is 1.8m above the floor?
ok, when is the book at the ending position?
hf=hi-1/2 g t^2
t=sqrt(2.4*5/9.8) sec
Now, acceration.
d=1/2 a t^2
a=2d/t^2=1.6/t^2
320 metres per second
To find the average acceleration of the librarian, we can use the equation of motion:
d = v₀t + (1/2)at²
Where:
- d is the distance traveled by the librarian,
- v₀ is the initial velocity (which is 0 m/s since the librarian starts from rest),
- t is the time taken to catch the book,
- a is the average acceleration of the librarian.
In this scenario, the librarian moves a distance of 0.80 m to catch the book, starting from rest. We need to find the time it takes to catch the book and the average acceleration.
To determine the time taken, we first need to calculate the initial velocity of the book when it falls from a height of 4.2 m. We can use the equation of motion:
v² = u² + 2as
Where:
- v is the final velocity of the book (0 m/s when it is caught),
- u is the initial velocity of the book (also 0 m/s since it starts from rest),
- a is the acceleration due to gravity (approximately -9.8 m/s²),
- s is the displacement or distance traveled by the book (4.2 m).
Plugging these values into the equation, we get:
0 = 0 + 2(-9.8)(4.2)
Simplifying, we get:
0 = -2(9.8)(4.2)
0 = -82.32
Since there is no real solution, it means there is an error in the problem statement or the calculations. It is not physically possible for the book to be caught when it is 1.8 m above the floor if it falls from a height of 4.2 m.