A particle start at t=0 from origin alongs x axis and its velcoity is given by `v=t^(3)-6t^(2)+11t-6`.Positive direction of x axis is considered as direction in the changing of velocity .total distance covered by the particle with negative velocity is

Options

A. `3//2`
B. 2
C. `5//2`
D. 3

Answer

To solve the problem, we need to find the total distance covered by the particle when it has negative velocity. The velocity of the particle is given by the equation:

\[ v(t) = t^3 - 6t^2 + 11t - 6 \]

### Step 1: Find the roots of the velocity equation
To determine when the particle has negative velocity, we first need to find the points in time when the velocity is zero. We can do this by solving the equation:

\[ t^3 - 6t^2 + 11t - 6 = 0 \]

By using the trial and error method or synthetic division, we find that the roots are:

- \( t = 1 \)
- \( t = 2 \)
- \( t = 3 \)

### Step 2: Determine intervals of negative velocity
The velocity function is a cubic polynomial, which can change signs at its roots. We need to check the sign of the velocity in the intervals defined by the roots:

- For \( t < 1 \): Choose \( t = 0 \)
\[ v(0) = 0^3 - 6(0)^2 + 11(0) - 6 = -6 \] (negative)

- For \( 1 < t < 2 \): Choose \( t = 1.5 \)
\[ v(1.5) = (1.5)^3 - 6(1.5)^2 + 11(1.5) - 6 = 3.375 - 13.5 + 16.5 - 6 = 0.375 \] (positive)

- For \( 2 < t < 3 \): Choose \( t = 2.5 \)
\[ v(2.5) = (2.5)^3 - 6(2.5)^2 + 11(2.5) - 6 = 15.625 - 37.5 + 27.5 - 6 = -0.375 \] (negative)

- For \( t > 3 \): Choose \( t = 4 \)
\[ v(4) = (4)^3 - 6(4)^2 + 11(4) - 6 = 64 - 96 + 44 - 6 = 6 \] (positive)

Thus, the particle has negative velocity in the intervals \( (0, 1) \) and \( (2, 3) \).

### Step 3: Calculate the distance covered in negative velocity intervals
To find the total distance, we need to integrate the absolute value of the velocity over the intervals where it is negative.

1. **From \( t = 0 \) to \( t = 1 \)**:
\[ \text{Distance}_1 = \int_0^1 -v(t) \, dt = \int_0^1 -(t^3 - 6t^2 + 11t - 6) \, dt \]
\[ = \int_0^1 (-t^3 + 6t^2 - 11t + 6) \, dt \]

Evaluating the integral:
\[ = \left[-\frac{t^4}{4} + 2t^3 - \frac{11t^2}{2} + 6t\right]_0^1 \]
\[ = \left[-\frac{1}{4} + 2 - \frac{11}{2} + 6\right] - [0] \]
\[ = -\frac{1}{4} + 2 - 5.5 + 6 = \frac{23}{4} \]

2. **From \( t = 2 \) to \( t = 3 \)**:
\[ \text{Distance}_2 = \int_2^3 -v(t) \, dt = \int_2^3 -(t^3 - 6t^2 + 11t - 6) \, dt \]
\[ = \int_2^3 (-t^3 + 6t^2 - 11t + 6) \, dt \]

Evaluating the integral:
\[ = \left[-\frac{t^4}{4} + 2t^3 - \frac{11t^2}{2} + 6t\right]_2^3 \]
\[ = \left[-\frac{81}{4} + 54 - \frac{99}{2} + 18\right] - \left[-\frac{16}{4} + 16 - \frac{44}{2} + 12\right] \]
\[ = \left[-\frac{81}{4} + 54 - 49.5 + 18\right] - \left[-4 + 16 - 22 + 12\right] \]
\[ = \left[-\frac{81}{4} + 22.5\right] - \left[2\right] \]
\[ = \left[-\frac{81}{4} + \frac{90}{4}\right] - 2 = \frac{9}{4} - 2 = \frac{9}{4} - \frac{8}{4} = \frac{1}{4} \]

### Step 4: Total distance covered with negative velocity
Now, we sum the distances from both intervals:
\[ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 = \frac{23}{4} + \frac{1}{4} = \frac{24}{4} = 6 \]

### Final Answer
The total distance covered by the particle with negative velocity is **6 units**.