If velocity of a particle is given by `v=2t^(2)-2` then find the acceleration of particle at t = 2 s.

Options

A. 0
B. 2
C. 4
D. 8

Answer

To find the acceleration of the particle at \( t = 2 \) seconds, we start with the given velocity function:

\[ v(t) = 2t^2 - 2 \]

### Step 1: Differentiate the velocity function to find acceleration
Acceleration \( a(t) \) is defined as the derivative of velocity with respect to time:

\[ a(t) = \frac{dv}{dt} \]

We will differentiate \( v(t) \):

\[ a(t) = \frac{d}{dt}(2t^2 - 2) \]

### Step 2: Apply the power rule of differentiation
Using the power rule, we differentiate each term:

\[ \frac{d}{dt}(2t^2) = 2 \cdot 2t^{2-1} = 4t \]
\[ \frac{d}{dt}(-2) = 0 \]

Thus, the acceleration function becomes:

\[ a(t) = 4t \]

### Step 3: Substitute \( t = 2 \) seconds into the acceleration function
Now we will find the acceleration at \( t = 2 \) seconds:

\[ a(2) = 4 \cdot 2 = 8 \, \text{m/s}^2 \]

### Final Result
The acceleration of the particle at \( t = 2 \) seconds is:

\[ \boxed{8 \, \text{m/s}^2} \]