The problem provides the following information:

• Acceleration: $a(t) = -42$
• Initial velocity: $v(0) = 40$
• Initial position: $s(0) = 30$

We need to find the expressions for velocity $v(t)$ and position $s(t)$.

Step 1: Find the velocity function $v(t)$

Since acceleration $a(t)$ is the derivative of velocity, we can integrate $a(t)$ to find $v(t)$.

$v(t) = \int a(t) \, dt = \int -42 \, dt = -42t + C_1$

We are given that $v(0) = 40$. Using this to solve for $C_1$:

$v(0) = -42(0) + C_1 = 40 \quad \Rightarrow \quad C_1 = 40$

Thus, the velocity function is:

$v(t) = -42t + 40$

Step 2: Find the position function $s(t)$

Since velocity $v(t)$ is the derivative of position $s(t)$, we can integrate $v(t)$ to find $s(t)$.

$s(t) = \int v(t) \, dt = \int (-42t + 40) \, dt = -21t^2 + 40t + C_2$

We are given that $s(0) = 30$. Using this to solve for $C_2$:

$s(0) = -21(0)^2 + 40(0) + C_2 = 30 \quad \Rightarrow \quad C_2 = 30$

Thus, the position function is:

$s(t) = -21t^2 + 40t + 30$

Final results:

• Velocity: $v(t) = -42t + 40$
• Position: $s(t) = -21t^2 + 40t + 30$

Would you like further clarification or details on this process?

Here are 5 related questions to expand your understanding:

1. What is the relationship between acceleration, velocity, and position in general?
2. How does constant acceleration affect the velocity and position of an object?
3. What would happen to the position and velocity functions if the initial conditions changed?
4. Can you derive the velocity function if the acceleration were time-dependent, say $a(t) = -42t$?
5. What is the significance of the constants $C_1$ and $C_2$ in these integrations?

Tip: When integrating to find velocity or position, always remember to account for the constant of integration using the given initial conditions.