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Let's analyze the problem step by step.

The position of the particle is given by \( s(t) = 3t^3 + 5t + 9 \).

### (a) Find the velocity at time \( t \).

The velocity is the first derivative of the position function \( s(t) \) with respect to time \( t \):
\[
v(t) = \frac{ds(t)}{dt}
\]
Given \( s(t) = 3t^3 + 5t + 9 \), let's differentiate:

\[
v(t) = \frac{d}{dt} \left( 3t^3 + 5t + 9 \right)
= 9t^2 + 5
\]

So, the velocity function is:
\[
v(t) = 9t^2 + 5 \, \frac{m}{s}
\]

### (b) Find the velocity at \( t = 3 \) seconds.

We substitute \( t = 3 \) into the velocity function:
\[
v(3) = 9(3)^2 + 5 = 9(9) + 5 = 81 + 5 = 86 \, \frac{m}{s}
\]

### (c) Find the acceleration at time \( t \).

The acceleration is the derivative of the velocity function \( v(t) \) with respect to time \( t \):
\[
a(t) = \frac{dv(t)}{dt}
\]
Since \( v(t) = 9t^2 + 5 \), let's differentiate:
\[
a(t) = \frac{d}{dt} \left( 9t^2 + 5 \right)
= 18t
\]

So, the acceleration function is:
\[
a(t) = 18t \, \frac{m}{s^2}
\]

### (d) Find the acceleration at \( t = 3 \) seconds.

We substitute \( t = 3 \) into the acceleration function:
\[
a(3) = 18(3) = 54 \, \frac{m}{s^2}
\]

### Summary of Answers:
- (a) \( v(t) = 9t^2 + 5 \, \frac{m}{s} \)
- (b) \( v(3) = 86 \, \frac{m}{s} \)
- (c) \( a(t) = 18t \, \frac{m}{s^2} \)
- (d) \( a(3) = 54 \, \frac{m}{s^2} \)

Would you like any further explanation or details on these steps?

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Here are 5 related questions to practice:
1. What would the velocity be at \( t = 5 \) seconds?
2. How would the position function look if the velocity was \( v(t) = 6t^2 + 4t \)?
3. How do you find the jerk (third derivative of position) in this case?
4. What is the significance of the constant 9 in the position equation?
5. How would you determine when the particle is at rest?

**Tip**: Always remember to differentiate the position function to find velocity and differentiate the velocity function to find acceleration!

Suppose that the position of a particle is given by s(t) = 3t^3 + 5t + 9. (a) Find the velocity at time t. (b) Find the velocity at time t = 3 seconds. (c) Find the acceleration at time t. (d) Find the acceleration at time t = 3 seconds.

This problem involves finding the velocity and acceleration of a particle given its position function s(t) = 3t^3 + 5t + 9. The solution uses calculus principles like differentiation to determine the velocity and acceleration at specific times.

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