Position of a point object is given by `x = (2)/(3)t^(3) - 4t^(2)+6t+7` (x is in meter and t is in seconds). What would be the distance travelled by object from t = 0 to instant when particle its direction of velocity for last time.

Options

A. `(16)/(3)m`
B. `(32)/(3)m`
C. `(8)/(3)m`
D. `0`

Explanation

To solve the problem, we need to find the distance traveled by the object from \( t = 0 \) to the instant when the particle changes its direction of velocity for the last time. We can follow these steps:

### Step 1: Find the expression for velocity
The position of the object is given by:
\[ x(t) = \frac{2}{3}t^3 - 4t^2 + 6t + 7 \]
To find the velocity, we differentiate the position function with respect to time:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{2}{3}t^3 - 4t^2 + 6t + 7 \right) \]
Calculating the derivative:
\[ v(t) = 2t^2 - 8t + 6 \]

### Step 2: Find when the velocity is zero
To find when the particle changes its direction of velocity, we set the velocity equation to zero:
\[ 2t^2 - 8t + 6 = 0 \]
Dividing the entire equation by 2:
\[ t^2 - 4t + 3 = 0 \]
Now, we can factor the quadratic:
\[ (t - 1)(t - 3) = 0 \]
Thus, the solutions are:
\[ t = 1 \quad \text{and} \quad t = 3 \]

### Step 3: Identify the last time the particle changes direction
The particle changes direction at \( t = 1 \) seconds and \( t = 3 \) seconds. We are interested in the last time it changes direction, which is at \( t = 3 \) seconds.

### Step 4: Calculate the distance traveled from \( t = 0 \) to \( t = 3 \)
We need to calculate the position at \( t = 0 \), \( t = 1 \), and \( t = 3 \).

1. **Position at \( t = 0 \)**:
\[ x(0) = \frac{2}{3}(0)^3 - 4(0)^2 + 6(0) + 7 = 7 \, \text{m} \]

2. **Position at \( t = 1 \)**:
\[ x(1) = \frac{2}{3}(1)^3 - 4(1)^2 + 6(1) + 7 = \frac{2}{3} - 4 + 6 + 7 = \frac{2}{3} + 9 = \frac{29}{3} \, \text{m} \]

3. **Position at \( t = 3 \)**:
\[ x(3) = \frac{2}{3}(3)^3 - 4(3)^2 + 6(3) + 7 = \frac{2}{3}(27) - 4(9) + 18 + 7 \]
\[ = 18 - 36 + 18 + 7 = 7 \, \text{m} \]

### Step 5: Calculate the total distance traveled
The distance traveled from \( t = 0 \) to \( t = 1 \):
\[ d_1 = x(1) - x(0) = \frac{29}{3} - 7 = \frac{29}{3} - \frac{21}{3} = \frac{8}{3} \, \text{m} \]

The distance traveled from \( t = 1 \) to \( t = 3 \):
\[ d_2 = x(3) - x(1) = 7 - \frac{29}{3} = \frac{21}{3} - \frac{29}{3} = -\frac{8}{3} \, \text{m} \]
Since we are calculating distance, we take the absolute value:
\[ |d_2| = \frac{8}{3} \, \text{m} \]

### Step 6: Total distance traveled
The total distance traveled from \( t = 0 \) to \( t = 3 \) is:
\[ \text{Total Distance} = d_1 + |d_2| = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \, \text{m} \]

### Final Answer
The distance traveled by the object from \( t = 0 \) to the instant when the particle changes its direction of velocity for the last time is:
\[ \boxed{\frac{16}{3} \, \text{m}} \]