If `a=(3t^2+2t+1)` `m/s^2` is the expression according to which the acceleration of a particle varies. Then-
Q. The expression for instantaneous velocity at any time `t` will be (if the particle was initially at rest)-

Options

A. `t^(3)+2t+1`
B. `t^(3)+t+1`
C. `t^(3)+t^(2)+t`
D. `t^(3)+t^(2)+t+C`

Explanation

To find the instantaneous velocity of a particle given its acceleration as a function of time, we can follow these steps:

### Step 1: Understand the relationship between acceleration and velocity
Acceleration \( a \) is defined as the rate of change of velocity \( v \) with respect to time \( t \). Mathematically, this can be expressed as:
\[ a = \frac{dv}{dt} \]

### Step 2: Substitute the given expression for acceleration
We are given that:
\[ a = 3t^2 + 2t + 1 \quad \text{(in m/s}^2\text{)} \]
Thus, we can write:
\[ \frac{dv}{dt} = 3t^2 + 2t + 1 \]

### Step 3: Rearrange the equation for integration
To find the velocity, we can rearrange the equation:
\[ dv = (3t^2 + 2t + 1) dt \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side integrates with respect to \( v \) and the right side integrates with respect to \( t \):
\[ \int dv = \int (3t^2 + 2t + 1) dt \]

### Step 5: Perform the integration
The left side gives:
\[ v = \int (3t^2 + 2t + 1) dt \]
The right side can be integrated term by term:
\[ \int 3t^2 dt = t^3, \quad \int 2t dt = t^2, \quad \int 1 dt = t \]
Thus, we have:
\[ v = t^3 + t^2 + t + C \]
where \( C \) is the constant of integration.

### Step 6: Apply initial conditions
Since the particle was initially at rest, we know that when \( t = 0 \), \( v = 0 \):
\[ 0 = 0^3 + 0^2 + 0 + C \implies C = 0 \]
So, the equation for velocity simplifies to:
\[ v = t^3 + t^2 + t \]

### Final Expression
The expression for instantaneous velocity at any time \( t \) is:
\[ v(t) = t^3 + t^2 + t \quad \text{(in m/s)} \]