The velocity-time relation of an electron starting from rest is given by u=kt, where `k=2m//s^(2)`. The distance traversed in 3 sec is:-

Options

A. 9m
B. 16 m
C. 27 m
D. 36 m

Answer

To solve the problem, we need to find the distance traversed by the electron in 3 seconds given the velocity-time relation \( u = kt \), where \( k = 2 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Identify the Velocity Function:**
The velocity of the electron is given by:
\[ u = kt \]
Substituting the value of \( k \):
\[ u = 2t \, \text{m/s} \]

2. **Set Up the Relationship Between Velocity and Distance:**
The velocity \( u \) is the derivative of displacement \( s \) with respect to time \( t \):
\[ \frac{ds}{dt} = u = 2t \]

3. **Integrate to Find Displacement:**
To find the displacement \( s \), we need to integrate the velocity function with respect to time:
\[ ds = 2t \, dt \]
Integrating both sides from \( t = 0 \) to \( t = 3 \):
\[ s = \int_0^3 2t \, dt \]

4. **Perform the Integration:**
The integral of \( 2t \) is:
\[ s = 2 \cdot \frac{t^2}{2} \bigg|_0^3 = t^2 \bigg|_0^3 = 3^2 - 0^2 = 9 \, \text{m} \]

5. **Conclusion:**
The distance traversed by the electron in 3 seconds is:
\[ s = 9 \, \text{m} \]

### Final Answer:
The distance traversed in 3 seconds is **9 meters**.

---