If `a=(3t^2+2t+1)` `m/s^2` is the expression according to which the acceleration of a particle varies. Then-
Find displacement of the particle after 2 seconds of start.

Options

A. 26 m
B. `(26)/(3)`m
C. `(30)/(7)`m
D. `(26)/(7)`m

Explanation

To find the displacement of the particle after 2 seconds given the acceleration \( a = 3t^2 + 2t + 1 \, \text{m/s}^2 \), we can follow these steps:

### Step 1: Integrate acceleration to find velocity
The acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can express this as:
\[ dv = a \, dt \]
Substituting the expression for acceleration:
\[ dv = (3t^2 + 2t + 1) \, dt \]
Now, we integrate both sides. The limits for \( v \) will be from \( 0 \) to \( v \) (initial velocity is \( 0 \)), and for \( t \) from \( 0 \) to \( t \):
\[ \int_0^v dv = \int_0^t (3t^2 + 2t + 1) \, dt \]
This gives:
\[ v = \int_0^t (3t^2 + 2t + 1) \, dt \]

### Step 2: Calculate the integral
Now we calculate the integral on the right-hand side:
\[ \int (3t^2) \, dt = t^3, \quad \int (2t) \, dt = t^2, \quad \int (1) \, dt = t \]
Thus, we have:
\[ v = \left[ t^3 + t^2 + t \right]_0^t = t^3 + t^2 + t \]

### Step 3: Integrate velocity to find displacement
Now, we use the velocity to find displacement:
\[ v = \frac{ds}{dt} \]
This can be rewritten as:
\[ ds = v \, dt \]
Substituting the expression for velocity:
\[ ds = (t^3 + t^2 + t) \, dt \]
Integrating both sides from \( 0 \) to \( s \) and \( 0 \) to \( t \):
\[ \int_0^s ds = \int_0^t (t^3 + t^2 + t) \, dt \]
This gives:
\[ s = \int_0^t (t^3 + t^2 + t) \, dt \]

### Step 4: Calculate the integral for displacement
Calculating the integral:
\[ \int (t^3) \, dt = \frac{t^4}{4}, \quad \int (t^2) \, dt = \frac{t^3}{3}, \quad \int (t) \, dt = \frac{t^2}{2} \]
Thus:
\[ s = \left[ \frac{t^4}{4} + \frac{t^3}{3} + \frac{t^2}{2} \right]_0^t = \frac{t^4}{4} + \frac{t^3}{3} + \frac{t^2}{2} \]

### Step 5: Substitute \( t = 2 \) seconds
Now, we substitute \( t = 2 \):
\[ s = \frac{(2)^4}{4} + \frac{(2)^3}{3} + \frac{(2)^2}{2} \]
Calculating each term:
\[ s = \frac{16}{4} + \frac{8}{3} + \frac{4}{2} = 4 + \frac{8}{3} + 2 \]
Combining the terms:
\[ s = 6 + \frac{8}{3} = \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \, \text{meters} \]

### Final Answer
The displacement of the particle after 2 seconds is:
\[ \boxed{\frac{26}{3} \, \text{meters}} \]