The acceleration of a particle which starts from rest varies with time according to relation a=2t+3. The velocity of the particle at time t would be

Options

A. `2t^3+3`
B. `t^2+3t`
C. `2t^2+3t`
D. zero

Answer

To find the velocity of a particle whose acceleration varies with time according to the relation \( a = 2t + 3 \), 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 with respect to time. Mathematically, this is expressed as:
\[ a = \frac{dv}{dt} \]

### Step 2: Substitute the given acceleration into the equation
From the problem, we have:
\[ a = 2t + 3 \]
So we can write:
\[ \frac{dv}{dt} = 2t + 3 \]

### Step 3: Rearrange the equation for integration
We can rearrange this equation to isolate \( dv \):
\[ dv = (2t + 3) dt \]

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

The left side gives:
\[ v \]

The right side can be integrated as follows:
\[ \int (2t + 3) dt = \int 2t \, dt + \int 3 \, dt = t^2 + 3t + C \]
where \( C \) is the constant of integration.

### Step 5: Determine the constant of integration
Since the particle starts from rest, we know that at \( t = 0 \), the velocity \( v = 0 \):
\[ v(0) = 0 = 0^2 + 3(0) + C \implies C = 0 \]

### Step 6: Write the final expression for velocity
Substituting \( C \) back into the equation gives us:
\[ v = t^2 + 3t \]

### Conclusion
Thus, the velocity of the particle at time \( t \) is:
\[ v(t) = t^2 + 3t \]

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