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### Problem 18: A ball is thrown straight down from the top of a 220-foot building.

#### (a) Write the position function for the ball.
The given equation is:
\[
s(t) = -16t^2 + v_0 t + s_0
\]
- The initial velocity \( v_0 \) is given as \(-22\) feet per second.
- The initial position \( s_0 \) is \(220\) feet (the height of the building).

Substituting into the equation, we get the position function:
\[
s(t) = -16t^2 - 22t + 220
\]

#### (b) Find the average velocity of the ball from \( t=0 \) to \( t=1 \) seconds.
The formula for average velocity is:
\[
\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]
Here, \( t_1 = 0 \) and \( t_2 = 1 \), and we use the position function from part (a).

Calculate:
- \( s(0) = -16(0)^2 - 22(0) + 220 = 220 \)
- \( s(1) = -16(1)^2 - 22(1) + 220 = -16 - 22 + 220 = 182 \)

Thus, the average velocity is:
\[
\text{Average velocity} = \frac{182 - 220}{1 - 0} = \frac{-38}{1} = -38 \text{ feet per second}
\]

#### (c) Find the velocity function for the ball.
The velocity function is the derivative of the position function \( s(t) \). 
\[
v(t) = \frac{d}{dt}[-16t^2 - 22t + 220] = -32t - 22
\]

#### (d) Find the acceleration function for the ball.
The acceleration function is the derivative of the velocity function \( v(t) \). 
\[
a(t) = \frac{d}{dt}[-32t - 22] = -32 \text{ feet per second squared}
\]

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### Problem 19: A car is traveling on a straight road, and its velocity is modeled by a piecewise function based on the graph.

#### (a) What is the car's velocity at \( t = 8 \) seconds?
From the graph, at \( t = 8 \) seconds, the velocity is 20 meters per second (read directly from the graph).

#### (b) What is the car's acceleration at \( t = 8 \) seconds?
The acceleration is the slope of the velocity graph at \( t = 8 \) seconds. Between \( t = 0 \) and \( t = 8 \), the velocity is constant at 20 m/s, so the acceleration is 0 meters per second squared.

#### (c) What is the car's average acceleration from \( t = 16 \) seconds to \( t = 20 \) seconds?
Between \( t = 16 \) and \( t = 20 \), the velocity decreases from 20 m/s to 0 m/s. The average acceleration is:
\[
\text{Average acceleration} = \frac{v(20) - v(16)}{20 - 16} = \frac{0 - 20}{4} = -5 \text{ meters per second squared}
\]

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Would you like more details on any of the parts? Here are some questions to consider:

1. How is the position function derived from kinematic equations?
2. What does negative velocity indicate about the ball's motion?
3. How would the acceleration change if the initial velocity were positive?
4. Can you explain how to interpret a velocity-time graph for acceleration?
5. How would average velocity differ if the time interval was larger in problem 18(b)?

**Tip:** Remember, the slope of a velocity-time graph gives the acceleration, and a flat line means zero acceleration.

Problem 18: A ball is thrown straight down from the top of a 220-foot building with an initial velocity of -22 feet per second. s(t) = -16t^2 + v_0 * t + s_0. (a) Write the position function for the ball. (b) Find the average velocity of the ball from t = 0 to t = 1 second. (c) Find the velocity function for the ball. (d) Find the acceleration function for the ball.

Problem 19: A car is traveling on a straight road, and for 0 ≤ t ≤ 24 seconds, its velocity v(t) (in meters per second) is modeled by a piecewise graph. (a) What is the car's velocity at time t = 8 seconds? (b) What is the car's acceleration at time t = 8 seconds? (c) What is the car's average acceleration from t = 16 to t = 20 seconds?

Detailed solutions to AP Calculus kinematics problems involving a falling object and a car's velocity-time graph. Learn how to derive position, velocity, and acceleration functions, and interpret piecewise velocity graphs. Suitable for Grades 11-12.

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