Velocity of a particle is given as `v = (2t^(2) - 3)m//s`. The acceleration of particle at t = 3s will be :

Options

A. `18 ms^(-2)`
B. `12 ms^(-2)`
C. `15 ms^(-2)`
D. zero

Answer

To find the acceleration of the particle at \( t = 3 \) seconds, we start with the given velocity function:

\[ v(t) = 2t^2 - 3 \, \text{m/s} \]

### Step 1: Differentiate the velocity function to find acceleration

Acceleration \( a(t) \) is defined as the rate of change of velocity with respect to time. Mathematically, this is expressed as:

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

Now we differentiate the velocity function \( v(t) \):

\[ a(t) = \frac{d}{dt}(2t^2 - 3) \]

### Step 2: Perform the differentiation

Using the power rule of differentiation, we differentiate each term:

1. The derivative of \( 2t^2 \) is \( 4t \).
2. The derivative of the constant \( -3 \) is \( 0 \).

Thus, we have:

\[ a(t) = 4t \]

### Step 3: Substitute \( t = 3 \) seconds into the acceleration function

Now, we need to find the acceleration at \( t = 3 \) seconds:

\[ a(3) = 4 \times 3 \]

### Step 4: Calculate the value

Calculating this gives:

\[ a(3) = 12 \, \text{m/s}^2 \]

### Final Answer

The acceleration of the particle at \( t = 3 \) seconds is:

\[ \boxed{12 \, \text{m/s}^2} \]

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