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

Answer

To find the velocity of a particle at \( t = 4 \) seconds given the acceleration function \( a(t) = 2 + |t - 2| \), we will break the problem into two parts based on the value of \( t \).

### Step 1: Define the acceleration function for different time intervals
The acceleration function can be expressed as:
- For \( t < 2 \): \( a(t) = 2 + (2 - t) = 4 - t \)
- For \( t \geq 2 \): \( a(t) = 2 + (t - 2) = t \)

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

We will integrate the acceleration to find the velocity:
\[ \frac{dv}{dt} = 4 - t \]
Integrating from \( t = 0 \) to \( t = 2 \):
\[ \int_{0}^{v_2} dv = \int_{0}^{2} (4 - t) dt \]
Calculating the right-hand side:
\[ v_2 - v(0) = \left[ 4t - \frac{t^2}{2} \right]_{0}^{2} \]
\[ v_2 = \left(4 \cdot 2 - \frac{2^2}{2}\right) - 0 = 8 - 2 = 6 \text{ m/s} \]

### Step 3: Calculate the velocity for \( 2 < t \leq 4 \)
Now we will calculate the velocity from \( t = 2 \) to \( t = 4 \):
\[ \frac{dv}{dt} = t \]
Integrating from \( t = 2 \) to \( t = 4 \):
\[ \int_{v_2}^{v_4} dv = \int_{2}^{4} t \, dt \]
Calculating the right-hand side:
\[ v_4 - v_2 = \left[\frac{t^2}{2}\right]_{2}^{4} \]
Substituting the limits:
\[ v_4 - 6 = \left(\frac{4^2}{2} - \frac{2^2}{2}\right) \]
\[ v_4 - 6 = \left(\frac{16}{2} - \frac{4}{2}\right) = 8 - 2 = 6 \]
Thus,
\[ v_4 = 6 + 6 = 12 \text{ m/s} \]

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

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