A stone is thrown from the top of a tall cliff.  Its acceleration is a constant​ -32 

StartFraction ft Over secant squared EndFractionftsec2  

​(So A(t)​ = -32).  Its velocity after 

33 

seconds is 

negative 12−12 

StartFraction ft Over secant EndFractionftsec​, 

and its height after 

33 

seconds is 

216216 

ft.

Given the acceleration \( A(t) = -32 \) ft/s\(^2\), we want to find the velocity \( V(t) \) and the position \( S(t) \) of the stone as functions of time \( t \).

### 1. Finding the velocity function \( V(t) \)

The acceleration \( A(t) \) is the derivative of the velocity \( V(t) \). Therefore, we can find \( V(t) \) by integrating \( A(t) \):

\[ A(t) = \frac{dV(t)}{dt} = -32 \]

Integrate both sides with respect to \( t \):

\[ V(t) = \int -32 \, dt = -32t + C_1 \]

We know the velocity after 3 seconds is -12 ft/s. Hence,

\[ V(3) = -12 \]
\[ -32(3) + C_1 = -12 \]
\[ -96 + C_1 = -12 \]
\[ C_1 = -12 + 96 \]
\[ C_1 = 84 \]

Thus, the velocity function is:

\[ V(t) = -32t + 84 \]

### 2. Finding the position function \( S(t) \)

The velocity \( V(t) \) is the derivative of the position \( S(t) \). Therefore, we can find \( S(t) \) by integrating \( V(t) \):

\[ V(t) = \frac{dS(t)}{dt} = -32t + 84 \]

Integrate both sides with respect to \( t \):

\[ S(t) = \int (-32t + 84) \, dt = -16t^2 + 84t + C_2 \]

We know the height after 3 seconds is 216 ft. Hence,

\[ S(3) = 216 \]
\[ -16(3)^2 + 84(3) + C_2 = 216 \]
\[ -16(9) + 252 + C_2 = 216 \]
\[ -144 + 252 + C_2 = 216 \]
\[ 108 + C_2 = 216 \]
\[ C_2 = 216 - 108 \]
\[ C_2 = 108 \]

Thus, the position function is:

\[ S(t) = -16t^2 + 84t + 108 \]

### Summary

- Acceleration: \( A(t) = -32 \) ft/s\(^2\)
- Velocity: \( V(t) = -32t + 84 \) ft/s
- Position: \( S(t) = -16t^2 + 84t + 108 \) ft

Do you want more details or have any questions?

Here are five related questions you might consider:

1. What is the initial velocity of the stone?
2. What is the initial height of the stone?
3. At what time does the stone reach its maximum height?
4. When will the stone hit the ground?
5. How long does it take for the stone to reach its maximum height?

**Tip:** Always check initial conditions and constants of integration when solving differential equations to ensure the solution fits the given physical scenario.

A stone is thrown from the top of a tall cliff. Its acceleration is a constant -32 ft/sec^2 (So A(t) = -32). Its velocity after 3 seconds is -12 ft/sec, and its height after 3 seconds is 216 ft.

Explore how to find the velocity and position functions of a stone thrown from a cliff with constant acceleration (-32 ft/sec^2). Learn to apply integration to determine velocity and height, using initial conditions such as velocity after 3 seconds (-12 ft/sec) and height (216 ft). Suitable for advanced high school students in Grades 11-12.

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