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

Options

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

Explanation

To solve the problem of finding the displacement of a particle after 2 seconds when its acceleration is given by the expression \( a = 3t^2 + 2t + 1 \, \text{m/s}^2 \), we will follow these steps:

### Step 1: Integrate the acceleration to find velocity
The acceleration \( a(t) \) is given as:
\[ a(t) = 3t^2 + 2t + 1 \]
To find the velocity \( v(t) \), we integrate the acceleration with respect to time \( t \):
\[ v(t) = \int a(t) \, dt = \int (3t^2 + 2t + 1) \, dt \]

### Step 2: Perform the integration
Calculating the integral:
\[ v(t) = \left( \frac{3t^3}{3} + \frac{2t^2}{2} + t \right) + C = t^3 + t^2 + t + C \]
where \( C \) is the constant of integration.

### Step 3: Apply initial conditions to find \( C \)
We know that the initial velocity \( v(0) = 0 \) (the particle starts from rest):
\[ v(0) = 0^3 + 0^2 + 0 + C = 0 \implies C = 0 \]
Thus, the velocity function simplifies to:
\[ v(t) = t^3 + t^2 + t \]

### Step 4: Integrate the velocity to find displacement
Next, we find the displacement \( x(t) \) by integrating the velocity:
\[ x(t) = \int v(t) \, dt = \int (t^3 + t^2 + t) \, dt \]

### Step 5: Perform the integration for displacement
Calculating the integral:
\[ x(t) = \left( \frac{t^4}{4} + \frac{t^3}{3} + \frac{t^2}{2} \right) + D \]
where \( D \) is another constant of integration.

### Step 6: Apply initial conditions to find \( D \)
Assuming the initial displacement \( x(0) = 0 \):
\[ x(0) = \frac{0^4}{4} + \frac{0^3}{3} + \frac{0^2}{2} + D = 0 \implies D = 0 \]
Thus, the displacement function simplifies to:
\[ x(t) = \frac{t^4}{4} + \frac{t^3}{3} + \frac{t^2}{2} \]

### Step 7: Calculate displacement after 2 seconds
Now, we need to find the displacement after 2 seconds:
\[ x(2) = \frac{2^4}{4} + \frac{2^3}{3} + \frac{2^2}{2} \]
Calculating each term:
- \( \frac{2^4}{4} = \frac{16}{4} = 4 \)
- \( \frac{2^3}{3} = \frac{8}{3} \)
- \( \frac{2^2}{2} = \frac{4}{2} = 2 \)

Combining these:
\[ x(2) = 4 + \frac{8}{3} + 2 = 6 + \frac{8}{3} = \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \, \text{meters} \]

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