The acceleration of a'particle starting from rest, varies with time according to the relation `a=kt+c`. The velocity of the particle after time `t` will be :

Options

A. `kt^2+1/2ct`
B. `1/2kt^2+ct`
C. `1/2(kt^2+c)`
D. `kt^2+ ct`

Explanation

To solve the problem, we need to find the velocity of a particle whose acceleration varies with time according to the equation \( a = kt + c \), where \( k \) and \( c \) are constants. The particle starts from rest, which means its initial velocity is zero.

### Step-by-Step Solution:

1. **Understand the relationship between acceleration and velocity**:
- Acceleration is defined as the rate of change of velocity with respect to time. Therefore, we can express this relationship mathematically as:
\[ a = \frac{dv}{dt} \]

2. **Substitute the given acceleration equation**:
- We know from the problem that:
\[ a = kt + c \]
- Thus, we can rewrite the equation as:
\[ \frac{dv}{dt} = kt + c \]

3. **Integrate both sides**:
- To find the velocity \( v \), we need to integrate the acceleration with respect to time \( t \):
\[ v = \int (kt + c) \, dt \]

4. **Perform the integration**:
- The integral can be calculated as follows:
\[ v = \int (kt + c) \, dt = \int kt \, dt + \int c \, dt \]
- The first integral \( \int kt \, dt \) gives:
\[ \frac{k}{2} t^2 \]
- The second integral \( \int c \, dt \) gives:
\[ ct \]
- Therefore, combining these results, we have:
\[ v = \frac{k}{2} t^2 + ct + C \]
- Here, \( C \) is the constant of integration.

5. **Apply the initial condition**:
- Since the particle starts from rest, the initial velocity at \( t = 0 \) is \( v(0) = 0 \):
\[ v(0) = \frac{k}{2} (0)^2 + c(0) + C = 0 \implies C = 0 \]

6. **Final expression for velocity**:
- Thus, the velocity of the particle after time \( t \) is:
\[ v = \frac{k}{2} t^2 + ct \]

### Conclusion:
The velocity of the particle after time \( t \) is given by:
\[ v = \frac{kt^2}{2} + ct \]