A particle is moving along the x-axis whose position is given by ` x= 4 - 9t + t^3/3`. Mark the correct statement(s) in relation to its motion.

Options

A. a. The direction of motion is not changing at any of the instants.
B. b. The direction of the motion is changing at `t = 3s`.
C. c. For `0lttlt3 s`, the particle is slowing down.
D. d. For `0lttlt3 s`, the particle is speeding up.

Explanation

To analyze the motion of the particle given by the position function \( x(t) = 4 - 9t + \frac{t^3}{3} \), we need to find the velocity and acceleration of the particle and determine the behavior of the particle in terms of direction and speed.

### Step 1: Find the velocity
The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(4 - 9t + \frac{t^3}{3}\right) \]

Calculating the derivative:

\[ v(t) = 0 - 9 + t^2 = t^2 - 9 \]

### Step 2: Find when the velocity is zero
To find when the particle changes direction, we set the velocity equal to zero:

\[ t^2 - 9 = 0 \]

Solving for \( t \):

\[ t^2 = 9 \implies t = 3 \quad (\text{considering } t \geq 0) \]

### Step 3: Determine the acceleration
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \):

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

### Step 4: Analyze the intervals of motion
1. **For \( 0 < t < 3 \)**:
- Velocity: \( v(t) = t^2 - 9 < 0 \) (negative)
- Acceleration: \( a(t) = 2t > 0 \) (positive)
- The particle is slowing down because the velocity is negative and the acceleration is positive.

2. **At \( t = 3 \)**:
- The velocity is zero, indicating a change in direction.

3. **For \( t > 3 \)**:
- Velocity: \( v(t) = t^2 - 9 > 0 \) (positive)
- Acceleration: \( a(t) = 2t > 0 \) (positive)
- The particle is speeding up because both velocity and acceleration are positive.

### Conclusion
Based on the analysis, we can summarize the findings:
- The direction of motion is changing at \( t = 3 \) seconds.
- The particle is slowing down for \( 0 < t < 3 \) seconds.
- The particle is speeding up for \( t > 3 \) seconds.

### Correct Statements
- **B**: The direction of the motion is changing at \( t = 3 \) seconds.
- **C**: For \( 0 < t < 3 \) seconds, the particle is slowing down.
- **D**: For \( t > 3 \) seconds, the particle is speeding up.