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 of finding the total distance covered by the particle with negative velocity, we will follow these steps:

### Step 1: Determine the velocity function
The velocity of the particle is given by:
\[ v(t) = t^3 - 6t^2 + 11t - 6 \]

### Step 2: Find the points where the velocity is zero
To find when the particle changes direction (from negative to positive velocity or vice versa), we need to find the roots of the velocity equation:
\[ v(t) = 0 \]
This leads to solving:
\[ t^3 - 6t^2 + 11t - 6 = 0 \]

### Step 3: Factor the polynomial
We can factor the polynomial:
\[ v(t) = (t - 1)(t^2 - 5t + 6) \]
Now, we can further factor the quadratic:
\[ t^2 - 5t + 6 = (t - 2)(t - 3) \]
Thus, the complete factorization is:
\[ v(t) = (t - 1)(t - 2)(t - 3) \]

### Step 4: Identify the roots
The roots of the equation are:
- \( t = 1 \)
- \( t = 2 \)
- \( t = 3 \)

### Step 5: Analyze the intervals of velocity
We will check the sign of the velocity in the intervals determined by the roots:
1. For \( t < 1 \): Choose \( t = 0 \) → \( v(0) = -6 \) (negative)
2. For \( 1 < t < 2 \): Choose \( t = 1.5 \) → \( v(1.5) = 1.875 \) (positive)
3. For \( 2 < t < 3 \): Choose \( t = 2.5 \) → \( v(2.5) = -1.875 \) (negative)
4. For \( t > 3 \): Choose \( t = 4 \) → \( v(4) = 6 \) (positive)

### Step 6: Determine intervals with negative velocity
The particle has negative velocity in the intervals:
- From \( t = 0 \) to \( t = 1 \)
- From \( t = 2 \) to \( t = 3 \)

### Step 7: Calculate the distance traveled in these intervals
To find the distance, we need to calculate the integral of the absolute value of the velocity over these intervals. Since the velocity is negative in these intervals, we can directly integrate the velocity function.

#### Distance from \( t = 0 \) to \( t = 1 \):
\[ s_1 = \int_{0}^{1} (t^3 - 6t^2 + 11t - 6) \, dt \]

Calculating the integral:
\[ s_1 = \left[ \frac{t^4}{4} - 2t^3 + \frac{11t^2}{2} - 6t \right]_{0}^{1} \]
Evaluating at the limits:
\[ s_1 = \left( \frac{1}{4} - 2 + \frac{11}{2} - 6 \right) - 0 \]
\[ s_1 = \frac{1}{4} - 2 + 5.5 - 6 = \frac{1}{4} - 2 - 0.5 = \frac{1}{4} - 2.5 = \frac{1 - 10}{4} = -\frac{9}{4} \]
Since we take absolute value, \( s_1 = \frac{9}{4} \).

#### Distance from \( t = 2 \) to \( t = 3 \):
\[ s_2 = \int_{2}^{3} (t^3 - 6t^2 + 11t - 6) \, dt \]
Calculating the integral:
\[ s_2 = \left[ \frac{t^4}{4} - 2t^3 + \frac{11t^2}{2} - 6t \right]_{2}^{3} \]
Evaluating at the limits:
\[ s_2 = \left( \frac{81}{4} - 54 + \frac{99}{2} - 18 \right) - \left( \frac{16}{4} - 16 + \frac{44}{2} - 12 \right) \]
Calculating both parts:
\[ s_2 = \left( \frac{81}{4} - 54 + \frac{198}{4} - \frac{72}{4} \right) - \left( 4 - 16 + 22 - 12 \right) \]
\[ s_2 = \left( \frac{81 + 198 - 216}{4} \right) - (-2) \]
\[ s_2 = \frac{63}{4} + 2 = \frac{63}{4} + \frac{8}{4} = \frac{71}{4} \]

### Step 8: Total distance covered with negative velocity
The total distance covered by the particle with negative velocity is:
\[ \text{Total Distance} = s_1 + s_2 = \frac{9}{4} + \frac{71}{4} = \frac{80}{4} = 20 \]

### Final Answer
The total distance covered by the particle with negative velocity is \( \frac{5}{2} \).