The initial velocity of a particle is u (at t = 0) and the acceleration f is given by f = at. Which of the following relation is valid?

Options

A. `v=u+at^(2)`
B. `v=u+(at^(2))/(2)`
C. `v=u+at`
D. v=u

Answer

To solve the problem, we need to analyze the motion of a particle with an initial velocity \( u \) and an acceleration that varies with time, given by \( f = at \).

### Step-by-Step Solution:

1. **Understand the given information**:
- Initial velocity of the particle at \( t = 0 \) is \( u \).
- The acceleration \( f \) is given by \( f = at \), where \( a \) is a constant.

2. **Relate acceleration to velocity**:
- Acceleration is defined as the rate of change of velocity. Mathematically, this can be expressed as:
\[ f = \frac{dv}{dt} \]
- Since \( f = at \), we can write:
\[ \frac{dv}{dt} = at \]

3. **Separate variables and integrate**:
- Rearranging gives:
\[ dv = at \, dt \]
- Now we integrate both sides. The left side integrates from the initial velocity \( u \) to the final velocity \( v \), and the right side integrates from \( t = 0 \) to \( t \):
\[ \int_{u}^{v} dv = \int_{0}^{t} at \, dt \]

4. **Perform the integration**:
- The left side gives:
\[ v - u \]
- The right side can be integrated as follows:
\[ \int_{0}^{t} at \, dt = a \int_{0}^{t} t \, dt = a \left[ \frac{t^2}{2} \right]_{0}^{t} = a \frac{t^2}{2} \]

5. **Combine results**:
- Setting the results of the integrations equal gives:
\[ v - u = a \frac{t^2}{2} \]
- Rearranging this equation, we find:
\[ v = u + a \frac{t^2}{2} \]

6. **Identify the valid relation**:
- The derived equation \( v = u + a \frac{t^2}{2} \) matches with one of the provided options. Therefore, the valid relation is:
\[ v = u + \frac{at^2}{2} \]

### Conclusion:
The correct answer is option **B**: \( v = u + \frac{at^2}{2} \).