The distance travelled ‘S’ by an accelerated particle of mass M is given by the following relation (in MKS units)
`S = 6t + 3t^2`
The velocity of the particle after 2 seconds is

Options

A. 6
B. 12
C. 18
D. 24

Explanation

To find the velocity of the particle after 2 seconds given the distance traveled \( S = 6t + 3t^2 \), we can follow these steps:

### Step 1: Understand the relationship between distance and velocity
The velocity \( V \) of an object is defined as the rate of change of distance with respect to time. Mathematically, this is expressed as:
\[ V = \frac{dS}{dt} \]

### Step 2: Differentiate the distance function
Given the distance function:
\[ S = 6t + 3t^2 \]
we need to differentiate \( S \) with respect to \( t \):
\[ V = \frac{dS}{dt} = \frac{d}{dt}(6t + 3t^2) \]

### Step 3: Apply differentiation rules
Using the rules of differentiation:
- The derivative of \( 6t \) is \( 6 \).
- The derivative of \( 3t^2 \) is \( 6t \) (using the power rule).

Thus, we have:
\[ V = 6 + 6t \]

### Step 4: Substitute \( t = 2 \) seconds into the velocity equation
Now we need to find the velocity at \( t = 2 \) seconds:
\[ V(2) = 6 + 6(2) \]
Calculating this gives:
\[ V(2) = 6 + 12 = 18 \, \text{m/s} \]

### Final Answer
The velocity of the particle after 2 seconds is:
\[ \boxed{18 \, \text{m/s}} \]

---