A particle starts from rest and moves with acceleration a which varies with time `t` as `a=kt` where `k` is a costant. The displacement `s` of the particle at time `t` is

Options

A. `1/2 kt^(2)`
B. `1/2 at^(2)`
C. `1/6 at^(2)`
D. None

Answer

To solve the problem, we need to find the displacement \( s \) of a particle that starts from rest and moves with an acceleration that varies with time as \( a = kt \), where \( k \) is a constant.

### Step-by-Step Solution:

1. **Understand the relationship between acceleration, velocity, and displacement**:
- We know that acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} \]

2. **Substitute the given acceleration**:
- Given \( a = kt \), we can write:
\[ \frac{dv}{dt} = kt \]

3. **Integrate to find velocity**:
- To find the velocity \( v \), we integrate both sides with respect to \( t \):
\[ dv = kt \, dt \]
- Integrating gives:
\[ v = \int kt \, dt = \frac{kt^2}{2} + C \]
- Since the particle starts from rest, the initial velocity \( v(0) = 0 \), so \( C = 0 \). Thus, we have:
\[ v = \frac{kt^2}{2} \]

4. **Relate velocity to displacement**:
- We know that velocity \( v \) is also the derivative of displacement \( s \) with respect to time:
\[ v = \frac{ds}{dt} \]
- Substituting for \( v \):
\[ \frac{ds}{dt} = \frac{kt^2}{2} \]

5. **Integrate to find displacement**:
- Now we integrate to find the displacement \( s \):
\[ ds = \frac{kt^2}{2} \, dt \]
- Integrating gives:
\[ s = \int \frac{kt^2}{2} \, dt = \frac{k}{2} \cdot \frac{t^3}{3} + C \]
- Again, since the particle starts from rest, the initial displacement \( s(0) = 0 \), so \( C = 0 \). Thus, we have:
\[ s = \frac{kt^3}{6} \]

### Final Answer:
The displacement \( s \) of the particle at time \( t \) is:
\[ s = \frac{kt^3}{6} \]