Can you please check my answers and help me fix the one I don't or or did wron .

The height in feet of a free falling object t seconds after release is s(t)=-16t^2+ v_0t+s_0, where s_0 is the height(in feet) at which the object is realsed, and v_0 is the initial velocity (in feet per second). Suppose the coin is dropped from a height of 1454 feet.

A. Determine the position and velocity functions
P(t)=-16t^2+1454
P'(t)=-32t
B.Determine the average velocity on the interval [1,3]
How would you go about this I tried to find the slope of the secant line and subtracted s(3)-s(2)/3-2 and got -64ft/sec
C.Find the instaneous velocity when t=1 and t=3
T=3 -96 ft/sec
T=1 -32 ft/sec
D.At what time is the instaneous velocity of the coin equal to the average velocity of the coin found in part B?
Need help here
E. What is the name of the theorwm that says there must be at least one solution to part D?
Mean Value thm
F. Find the velocity of the coin just before it hits the ground.
T=9.53284 ft/sec

A .

correct

B.
for the average velocity no Calculus is needed
P(1) = -16(1) + 1454 = 1438
P(3) = -16(9) + 1454 = 1310

average velocity = (1310 - 1438)/(3-1) = -64 ft/s

C. now you want to set
-32t = -64
t = 2

E. correct
F. I disagree.

you want P(t) = 0
-16t^2 + 1454 = 0
t^2 = 90.875
t = 9.532 , you had that, but gave it the wrong units.
You want the velocity when t = 9.532
velocity = -32(9.532) = -305.05 ft/s

A. Your position function, P(t), is correct: P(t) = -16t^2 + 1454. However, your velocity function, P'(t), is incorrect. To find the velocity function, you need to take the derivative of the position function with respect to time.

Taking the derivative of P(t) = -16t^2 + 1454, we get P'(t) = -32t. So the correct velocity function is P'(t) = -32t.

B. To find the average velocity on the interval [1, 3], you need to find the change in position over the change in time. In this case, the change in time is 3 - 1 = 2 seconds.

Plugging the values into the position function, we get P(3) = -16(3)^2 + 1454 = -144. Similarly, P(1) = -16(1)^2 + 1454 = 1438.

The change in position is P(3) - P(1), and the change in time is 2. So, the average velocity is (P(3) - P(1)) / (3 - 1) = (-144 - 1438) / 2 = -791 ft/sec.

Your answer of -64 ft/sec is incorrect. Please check your calculation.

C. To find the instantaneous velocity when t = 1 and t = 3, you need to evaluate the velocity function, P'(t), at those specific times.

For t = 1, plug 1 into the velocity function P'(t) = -32t: P'(1) = -32(1) = -32 ft/sec.

For t = 3, plug 3 into the velocity function: P'(3) = -32(3) = -96 ft/sec.

Your answers of -32 ft/sec and -96 ft/sec are correct.

D. To find the time at which the instantaneous velocity of the coin is equal to the average velocity found in part B, you need to set the velocity function, P'(t), equal to the average velocity (-791 ft/sec) and solve for t.

-32t = -791
t = -791 / -32
t ≈ 24.72 seconds

So, at approximately t = 24.72 seconds, the instantaneous velocity of the coin will be equal to the average velocity found in part B.

E. The correct name of the theorem that guarantees there must be at least one solution to part D is the Intermediate Value Theorem, not the Mean Value Theorem.

F. To find the velocity of the coin just before it hits the ground, you need to find the instantaneous velocity when t = 9.53284. Substitute this value into the velocity function: P'(9.53284) = -32(9.53284) ≈ -304.49 ft/sec.

So, the velocity of the coin just before it hits the ground is approximately -304.49 ft/sec.