- Scenario: A car is traveling along a straight road. Its position (in meters) at time t (in seconds) is given by s(t) = t³ - 6t² + 9t. Find the car's velocity and acceleration at time t = 2 seconds

To find the car's velocity and acceleration at time \( t = 2 \) seconds, we start with the position function:

\[
s(t) = t^3 - 6t^2 + 9t
\]

### Step 1: Find the Velocity

The velocity \( v(t) \) is the first derivative of the position function \( s(t) \):

\[
v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t)
\]

Calculating the derivative:

\[
v(t) = 3t^2 - 12t + 9
\]

Now, substituting \( t = 2 \):

\[
v(2) = 3(2)^2 - 12(2) + 9 = 3(4) - 24 + 9 = 12 - 24 + 9 = -3 \text{ m/s}
\]

### Step 2: Find the Acceleration

The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \):

\[
a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 9)
\]

Calculating the derivative:

\[
a(t) = 6t - 12
\]

Now, substituting \( t = 2 \):

\[
a(2) = 6(2) - 12 = 12 - 12 = 0 \text{ m/s}^2
\]

### Summary

- **Velocity at \( t = 2 \) seconds**: \( -3 \) m/s
- **Acceleration at \( t = 2 \) seconds**: \( 0 \) m/s²

If you have any questions or need further details, let me know! Here are some related questions to consider:

1. What is the significance of the negative velocity in this scenario?
2. How would the position function change if the coefficients were different?
3. What does an acceleration of zero indicate about the car's motion?
4. How can we find the maximum or minimum position of the car using this function?
5. What is the impact of different initial conditions on the car's motion?

**Tip**: To analyze motion further, consider using a graph of the position, velocity, and acceleration functions to visualize the changes over time.

A car is traveling along a straight road. Its position (in meters) at time t (in seconds) is given by s(t) = t³ - 6t² + 9t. Find the car's velocity and acceleration at time t = 2 seconds.

Discover how to find the velocity and acceleration of a car given its position function s(t) = t³ - 6t² + 9t. At t = 2 seconds, the car's velocity is -3 m/s and its acceleration is 0 m/s². This explanation covers differentiation and kinematics concepts suitable for Grades 11-12.

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