2.Suppose a math book is thrown down from the top of the tallest tree in California, which is 378 feet tall with an initial velocity of -10 ft/s. The position function for free-falling objects is: s(t) = −16t2 + v0t + s0.

A. Determine the position and velocity functions for the book.

B. Determine the average velocity of the book on the interval [0, 1].

C. Find the instantaneous velocities when t = 0 and t = 1.

D. At what time is the instantaneous velocity of the coin equal to the average velocity of the book found in part B?

E. What is the name of the theorem that says there must be at least one solution to part D?

F. Find the velocity of the book just before it hits the ground.

Guest Nov 27, 2019

#1**+1 **

A)

Vo=-10

So= 378

Sub those in to get position funtion.

Differentiate to get velocity fuction.

Do that and display your answer to part A.

Melody Nov 27, 2019

#2**+1 **

A. Determine the position and velocity functions for the book.

B. Determine the average velocity of the book on the interval [0, 1].

C. Find the instantaneous velocities when t = 0 and t = 1.

D. At what time is the instantaneous velocity of the coin equal to the average velocity of the book found in part B?

E. What is the name of the theorem that says there must be at least one solution to part D?

F. Find the velocity of the book just before it hits the ground.

A.

The position function is : s(t) = -16t^2 -10t + 378

The velocity function isthe derivative of this = s'(t) = -32t - 10

B.

The average velocity on [ 0, 1] =

[ (-32(1) - 10 ) - (-32(0) - 10 ] -32

_______________________ = ______ = -32 ft/s

1 - 0 1

C. When t = 0, the instantaneous velocity is -32(0) - 10 = -10ft/s [which we would expect ]

When t = 1, the instantaneous velocity is -32(1) - 10 = - 42ft/s

D. I am taking "coin" to mean "book"

We want to know when

-32 = - 32t - 10 add 10 to both sides

-22 = -32 t divide both sides by -32

-22/-32 = t = 11/16 sec = .6875 sec

E. Average Value Theorem [ or, Mean Value Theorem ]

F. We need to solve this to find the time it takes to hit the ground

-16t^2 - 10t + 378 = 0

16t^2 + 10t - 378 = 0 divide through by 2

8t^2 + 5t - 189 = 0

Using the quad formula

t = -5 ±√ [ 5^2 - 4*8*189 ] -5 ±√ [ 25 + 6048]

___________________ = _________________ ≈ 4.558 sec or - 5.183 sec

2 * 8 16

Take the positive time and the instantaneous velocity when the book hits the ground is :

-32(4.558) - 10 ≈ -155.86 ft/s

CPhill Nov 27, 2019