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.