The initial velocity of a particle is x `ms^(-1)` (at t=0) and acceleration a is function for time, given by, a= 6t. Which of the following relation is correct for final velocity y after time t?

Options

A. `y =x+1/2(6t)^2`
B. `y =x^2+(6t)`
C. `y=x+3t^2`
D. `y=x+at`

Answer

To solve the problem, we need to find the final velocity \( y \) of a particle after time \( t \), given that the initial velocity is \( x \, \text{ms}^{-1} \) and the acceleration \( a \) is a function of time defined as \( a = 6t \).

### Step-by-Step Solution:

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

2. **Substitute the given acceleration**:
- From the problem, we know:
\[ a = 6t \]
- Therefore, we can write:
\[ \frac{dv}{dt} = 6t \]

3. **Rearranging the equation**:
- Rearranging gives:
\[ 6t \, dt = dv \]

4. **Integrate both sides**:
- Now, we will integrate both sides. The left side will be integrated with respect to \( t \) and the right side with respect to \( v \):
\[ \int 6t \, dt = \int dv \]

5. **Set the limits for integration**:
- The limits for the left side (time) will be from \( 0 \) to \( t \), and for the right side (velocity) from \( x \) to \( y \):
\[ \int_0^t 6t \, dt = \int_x^y dv \]

6. **Perform the integration**:
- The left side integrates to:
\[ 6 \left( \frac{t^2}{2} \right) \bigg|_0^t = 3t^2 \]
- The right side integrates to:
\[ v \bigg|_x^y = y - x \]

7. **Combine the results**:
- Setting the results of the integrals equal gives:
\[ 3t^2 = y - x \]

8. **Solve for final velocity \( y \)**:
- Rearranging the equation to solve for \( y \):
\[ y = x + 3t^2 \]

### Final Answer:
The relation for the final velocity \( y \) after time \( t \) is:
\[ y = x + 3t^2 \]