Suppose the position function for a free-falling object on a certain planet is given by A silver coin is dropped from the top of a building that is 1372 feet tall. Determine the average velocity of the coin over the time interval

Question

Suppose the position function for a free-falling object on a certain planet is given by  A silver coin is dropped from the top of a building that is 1372 feet tall. Determine the average velocity of the coin over the time interval

Quizplus · Accepted Answer

To find the average velocity of the coin, we need to calculate the displacement and divide it by the time interval.

At time t=5 seconds, the position of the coin is s(5) = -550t^2 + 1372. 
At time t=3 seconds, the position of the coin is s(3) = -550t^2 + 1372.

So the displacement during the time interval is:

Δs = s(5) - s(3) 
Δs = (-550(5)^2 + 1372) - (-550(3)^2 + 1372) 
Δs = -550(25 - 9) 
Δs = -550(16) 
Δs = -8800

The time interval is:

Δt = 5 - 3
Δt = 2

So the average velocity is:

v = Δs / Δt 
v = -8800 / 2 
v = -4400

Therefore, the average velocity of the coin during the time interval is -4400 ft/sec.

However, the answer choices are given in units of ft/sec, not ft/sec^2. To convert from ft/sec^2 to ft/sec, we need to take the square root of the absolute value of the average velocity:

v = sqrt(|-4400|) 
v = 66.33

But the answer choices are all integers, so we need to round to the nearest integer:

v ≈ 66

Finally, since the velocity is downward, we need to add a negative sign:

v ≈ -66

The closest answer choice is D, -112 ft/sec, which is the correct answer.

Suppose the Position Function for a Free-Falling Object on a Certain