A parrot flies in a straight line for 6s. Velocity of the parrot is given by v=`|t-3|`. Time (t) is measured in seconds and velocity in m/s. displacement of the parrot in 6s is

Options

A. 0
B. 5m
C. 6m
D. 9m

Explanation

To solve the problem, we need to find the displacement of the parrot over a time interval of 6 seconds, given that its velocity is defined as \( v = |t - 3| \).

### Step-by-Step Solution:

1. **Understand the Velocity Function**:
The velocity of the parrot is given by \( v = |t - 3| \). This means we need to consider two cases based on the value of \( t \):
- For \( t < 3 \), \( v = -(t - 3) = 3 - t \)
- For \( t \geq 3 \), \( v = t - 3 \)

2. **Break the Time Interval**:
Since we are interested in the displacement from \( t = 0 \) to \( t = 6 \), we will break this interval into two parts:
- From \( t = 0 \) to \( t = 3 \)
- From \( t = 3 \) to \( t = 6 \)

3. **Calculate Displacement from \( t = 0 \) to \( t = 3 \)**:
For \( t \) in this interval, the velocity is \( v = 3 - t \).
- The displacement \( ds \) can be expressed as \( ds = v dt = (3 - t) dt \).
- Now we integrate this from \( t = 0 \) to \( t = 3 \):
\[ s_1 = \int_0^3 (3 - t) dt \]
- Calculating the integral:
\[ s_1 = \left[ 3t - \frac{t^2}{2} \right]_0^3 = \left( 3(3) - \frac{3^2}{2} \right) - \left( 0 - 0 \right) = 9 - \frac{9}{2} = \frac{18}{2} - \frac{9}{2} = \frac{9}{2} \text{ meters} \]

4. **Calculate Displacement from \( t = 3 \) to \( t = 6 \)**:
For \( t \) in this interval, the velocity is \( v = t - 3 \).
- The displacement \( ds \) can be expressed as \( ds = v dt = (t - 3) dt \).
- Now we integrate this from \( t = 3 \) to \( t = 6 \):
\[ s_2 = \int_3^6 (t - 3) dt \]
- Calculating the integral:
\[ s_2 = \left[ \frac{t^2}{2} - 3t \right]_3^6 = \left( \frac{6^2}{2} - 3(6) \right) - \left( \frac{3^2}{2} - 3(3) \right) \]
\[ = \left( \frac{36}{2} - 18 \right) - \left( \frac{9}{2} - 9 \right) = (18 - 18) - \left( \frac{9}{2} - \frac{18}{2} \right) = 0 + \frac{9}{2} = \frac{9}{2} \text{ meters} \]

5. **Total Displacement**:
The total displacement \( s \) is the sum of the displacements from both intervals:
\[ s = s_1 + s_2 = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9 \text{ meters} \]

### Final Answer:
The displacement of the parrot in 6 seconds is **9 meters**.