The velocity of a particle is given by `v=(2t^(2)-4t+3)m//s` where t is time in seconds. Find its acceleration at t=2 second.

Explanation

To find the acceleration of the particle at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Write down the velocity function
The velocity of the particle is given by:
\[ v(t) = 2t^2 - 4t + 3 \quad \text{(in m/s)} \]

### Step 2: Differentiate the velocity function to find acceleration
Acceleration is defined as the rate of change of velocity with respect to time. Therefore, we need to differentiate the velocity function \( v(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(2t^2 - 4t + 3) \]

### Step 3: Perform the differentiation
Using the power rule of differentiation:
- The derivative of \( 2t^2 \) is \( 4t \)
- The derivative of \( -4t \) is \( -4 \)
- The derivative of a constant (3) is \( 0 \)

Thus, we have:
\[ a(t) = 4t - 4 \]

### Step 4: Substitute \( t = 2 \) seconds into the acceleration function
Now, we need to find the acceleration at \( t = 2 \) seconds:
\[ a(2) = 4(2) - 4 \]

### Step 5: Calculate the value
Calculating the above expression:
\[ a(2) = 8 - 4 = 4 \quad \text{(in m/s}^2\text{)} \]

### Final Answer
The acceleration of the particle at \( t = 2 \) seconds is:
\[ \boxed{4 \, \text{m/s}^2} \]

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