A particla starting from rest undergoes acceleration given by `a=|t-2|" m/s"^(2)` where t is time in sec. Find velocity of particle after 4 sec. is :-

Answer

To solve the problem, we need to find the velocity of a particle after 4 seconds given that its acceleration is defined as \( a = |t - 2| \) m/s². The particle starts from rest, which means its initial velocity \( v_0 = 0 \) m/s at \( t = 0 \).

### Step-by-Step Solution:

1. **Understand the Acceleration Function**:
The acceleration is given by \( a = |t - 2| \). This means we need to consider the absolute value, which will change the expression based on the value of \( t \).

2. **Determine the Intervals for \( t \)**:
- For \( t < 2 \): \( a = -(t - 2) = 2 - t \)
- For \( t \geq 2 \): \( a = t - 2 \)

3. **Set Up the Integral for Velocity**:
The change in velocity \( dv \) can be expressed as:
\[ dv = a \, dt \]
We will integrate this to find the velocity.

4. **Split the Integral**:
We need to calculate the velocity from \( t = 0 \) to \( t = 4 \), which requires splitting the integral at \( t = 2 \):
\[ v = \int_0^2 (2 - t) \, dt + \int_2^4 (t - 2) \, dt \]

5. **Calculate the First Integral**:
\[ \int_0^2 (2 - t) \, dt = \left[ 2t - \frac{t^2}{2} \right]_0^2 = \left( 2(2) - \frac{2^2}{2} \right) - (0) = 4 - 2 = 2 \]

6. **Calculate the Second Integral**:
\[ \int_2^4 (t - 2) \, dt = \left[ \frac{t^2}{2} - 2t \right]_2^4 = \left( \frac{4^2}{2} - 2(4) \right) - \left( \frac{2^2}{2} - 2(2) \right) \]
- Upper limit: \( \frac{16}{2} - 8 = 8 - 8 = 0 \)
- Lower limit: \( 2 - 4 = -2 \)
- Therefore, the second integral gives us \( 0 - (-2) = 2 \).

7. **Combine the Results**:
Now, add the results of both integrals to find the total change in velocity:
\[ v = 2 + 2 = 4 \text{ m/s} \]

### Final Answer:
The velocity of the particle after 4 seconds is \( \mathbf{4 \, m/s} \).