A particle moves along `x-` axis. It's velocity is a function of time according to relation `V=(3t^(2)-18t+24)m//s` assume at `t=0` particle is at origin.
Time interval in which particle speed continuous decreases?

Options

A. `0-3 sec`
B. `0-2 sec`
C. `2-4 sec`
D. `2-3 sec`

Explanation

To solve the problem, we need to determine the time interval during which the speed of the particle is continuously decreasing. We start with the given velocity function:

\[ V(t) = 3t^2 - 18t + 24 \]

### Step 1: Find the acceleration
Acceleration \( A(t) \) is the derivative of velocity with respect to time. We differentiate the velocity function:

\[ A(t) = \frac{dV}{dt} = \frac{d}{dt}(3t^2 - 18t + 24) \]

Calculating the derivative:

\[ A(t) = 6t - 18 \]

### Step 2: Determine when the velocity is zero
To find when the speed is decreasing, we need to find the points where the velocity changes sign. We set the velocity function equal to zero and solve for \( t \):

\[ 3t^2 - 18t + 24 = 0 \]

Dividing the entire equation by 3:

\[ t^2 - 6t + 8 = 0 \]

Now, we factor the quadratic:

\[ (t - 2)(t - 4) = 0 \]

Thus, the roots are:

\[ t = 2 \quad \text{and} \quad t = 4 \]

### Step 3: Analyze intervals
The velocity function changes sign at \( t = 2 \) and \( t = 4 \). We need to analyze the sign of the velocity and acceleration in the intervals \( (-\infty, 2) \), \( (2, 4) \), and \( (4, \infty) \).

1. **Interval \( (0, 2) \)**:
- Choose \( t = 1 \):
\[ V(1) = 3(1)^2 - 18(1) + 24 = 3 - 18 + 24 = 9 \quad (\text{positive}) \]
- Choose \( t = 1 \) for acceleration:
\[ A(1) = 6(1) - 18 = 6 - 18 = -12 \quad (\text{negative}) \]
- Here, velocity is positive and acceleration is negative. Therefore, speed is decreasing.

2. **Interval \( (2, 4) \)**:
- Choose \( t = 3 \):
\[ V(3) = 3(3)^2 - 18(3) + 24 = 27 - 54 + 24 = -3 \quad (\text{negative}) \]
- Choose \( t = 3 \) for acceleration:
\[ A(3) = 6(3) - 18 = 18 - 18 = 0 \quad (\text{zero}) \]
- Here, velocity is negative and acceleration is zero. Therefore, speed is not decreasing continuously.

3. **Interval \( (4, \infty) \)**:
- Choose \( t = 5 \):
\[ V(5) = 3(5)^2 - 18(5) + 24 = 75 - 90 + 24 = 9 \quad (\text{positive}) \]
- Choose \( t = 5 \) for acceleration:
\[ A(5) = 6(5) - 18 = 30 - 18 = 12 \quad (\text{positive}) \]
- Here, both velocity and acceleration are positive. Therefore, speed is not decreasing.

### Conclusion
The only interval where the speed of the particle is continuously decreasing is:

\[ \text{Time interval: } (0, 2) \text{ seconds} \]