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

Options

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

Explanation

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. **Understand the Given Relation**:
The velocity \( u \) of the electron is given by the equation:
\[ u = kt \]
where \( k = 2 \, \text{m/s}^2 \). This means that the velocity of the electron increases linearly with time.

2. **Express Velocity in Terms of Time**:
Substitute the value of \( k \):
\[ u = 2t \]

3. **Relate Velocity to Displacement**:
The velocity \( u \) is the derivative of displacement \( s \) with respect to time \( t \):
\[ u = \frac{ds}{dt} \]
Therefore, we can write:
\[ \frac{ds}{dt} = 2t \]

4. **Separate Variables and Integrate**:
Rearranging gives:
\[ ds = 2t \, dt \]
Now, we integrate both sides. The integral of \( ds \) gives \( s \), and the integral of \( 2t \, dt \) is:
\[ s = \int 2t \, dt = 2 \cdot \frac{t^2}{2} = t^2 + C \]
Since the electron starts from rest (initial displacement \( s = 0 \) when \( t = 0 \)), we can find the constant \( C \):
\[ s(0) = 0^2 + C \implies C = 0 \]
Thus, the equation simplifies to:
\[ s = t^2 \]

5. **Calculate the Distance at \( t = 3 \) seconds**:
Now, we can find the distance traversed in 3 seconds:
\[ s(3) = (3)^2 = 9 \, \text{m} \]

### Final Answer:
The distance traversed by the electron in 3 seconds is:
\[ \boxed{9 \, \text{m}} \]