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 find the final velocity \( y \) of a particle after time \( t \) given its initial velocity \( x \) and acceleration \( a = 6t \), we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and time
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, mathematically expressed as:
\[ a = \frac{dv}{dt} \]
Given that \( a = 6t \), we can write:
\[ \frac{dv}{dt} = 6t \]

### Step 2: Rearrange the equation
To find the change in velocity, we can rearrange the equation:
\[ dv = 6t \, dt \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side will be integrated with respect to \( v \) from the initial velocity \( x \) to the final velocity \( y \), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_{x}^{y} dv = \int_{0}^{t} 6t \, dt \]

### Step 4: Perform the integration
The left side integrates to:
\[ y - x \]
The right side integrates as follows:
\[ \int 6t \, dt = 6 \cdot \frac{t^2}{2} = 3t^2 \]
So we have:
\[ y - x = 3t^2 \]

### Step 5: Solve for final velocity \( y \)
Now, we can solve for \( y \):
\[ y = x + 3t^2 \]

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