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)+ct`
B. `1.2 kt^(2)+ct`
C. `1/2 (kt^(2)+ct)`
D. `kt^(2)+1/2 ct`

Answer

To find the velocity of a particle after time \( t \) given that its acceleration varies with time according to the relation \( a = kt + c \), we can follow these steps:

### Step 1: Understand the relationship between acceleration and velocity
The acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]

### Step 2: Substitute the given acceleration equation
From the problem, we know that:
\[ a = kt + c \]
Thus, we can write:
\[ \frac{dv}{dt} = kt + c \]

### Step 3: Rearrange the equation for integration
We can rearrange this equation to separate the variables:
\[ dv = (kt + c) dt \]

### Step 4: Integrate both sides
Now, we will integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):
\[ \int dv = \int (kt + c) dt \]
The left side integrates to \( v \), and the right side can be integrated as follows:
\[ v = \int (kt + c) dt = \frac{kt^2}{2} + ct + C \]
where \( C \) is the constant of integration.

### Step 5: Apply initial conditions
Given that the particle starts from rest, the initial velocity \( u \) at \( t = 0 \) is:
\[ u = 0 \]
Substituting \( t = 0 \) into the velocity equation:
\[ 0 = \frac{k(0)^2}{2} + c(0) + C \implies C = 0 \]

### Step 6: Write the final expression for velocity
Now substituting \( C \) back into our velocity equation, we get:
\[ v = \frac{kt^2}{2} + ct \]

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

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