A particle starts from rest and moves with an acceleration of `a={2+|t-2|}m//s^(2)` The velocity of the particle at `t=4` sec is

Options

A. ` 16 m//s`
B. ` 20 m//s`
C. ` 8m//s `
D. ` 12 m//s`

Explanation

To find the velocity of the particle at \( t = 4 \) seconds given the acceleration function \( a(t) = 2 + |t - 2| \), we will follow these steps:

### Step 1: Define the acceleration function based on the value of \( t \)
The acceleration function \( a(t) = 2 + |t - 2| \) can be broken down into two cases based on the value of \( t \):

- **Case 1:** For \( t < 2 \)
\[ a(t) = 2 + (2 - t) = 4 - t \]

- **Case 2:** For \( t \geq 2 \)
\[ a(t) = 2 + (t - 2) = t \]

### Step 2: Calculate the velocity from \( t = 0 \) to \( t = 2 \)
Since the particle starts from rest, the initial velocity \( v(0) = 0 \).

For \( 0 \leq t < 2 \), we will integrate the acceleration function:
\[ v(t) = \int a(t) \, dt = \int (4 - t) \, dt \]
Calculating the integral:
\[ v(t) = 4t - \frac{t^2}{2} + C \]
Since \( v(0) = 0 \), we find \( C = 0 \):
\[ v(t) = 4t - \frac{t^2}{2} \]

Now, we evaluate \( v(t) \) at \( t = 2 \):
\[ v(2) = 4(2) - \frac{(2)^2}{2} = 8 - 2 = 6 \, \text{m/s} \]

### Step 3: Calculate the velocity from \( t = 2 \) to \( t = 4 \)
For \( t \geq 2 \), we use the acceleration function \( a(t) = t \):
\[ v(t) = v(2) + \int_{2}^{t} a(t) \, dt = 6 + \int_{2}^{t} t \, dt \]
Calculating the integral:
\[ \int t \, dt = \frac{t^2}{2} \]
Thus,
\[ v(t) = 6 + \left[ \frac{t^2}{2} \right]_{2}^{t} = 6 + \left( \frac{t^2}{2} - \frac{(2)^2}{2} \right) \]
Evaluating at \( t = 4 \):
\[ v(4) = 6 + \left( \frac{(4)^2}{2} - \frac{(2)^2}{2} \right) = 6 + \left( \frac{16}{2} - \frac{4}{2} \right) = 6 + (8 - 2) = 6 + 6 = 12 \, \text{m/s} \]

### Final Answer
The velocity of the particle at \( t = 4 \) seconds is \( 12 \, \text{m/s} \).

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