A particle, initially at rest, starts moving in a straight line with an acceleration `a=6t+4 m//s^(2)`. The distance covered by it in 3 s is

Options

A. 15 m
B. 30 m
C. 45 m
D. 60 m

Answer

To solve the problem, we need to find the distance covered by a particle that starts from rest and has an acceleration given by the equation \( a(t) = 6t + 4 \, \text{m/s}^2 \).

### Step-by-Step Solution:

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

2. **Set up the equation for acceleration**:
- Given \( a(t) = 6t + 4 \), we can write:
\[ \frac{dv}{dt} = 6t + 4 \]

3. **Integrate to find velocity**:
- To find the velocity \( v(t) \), integrate the acceleration:
\[ v(t) = \int (6t + 4) \, dt \]
- Performing the integration:
\[ v(t) = 3t^2 + 4t + C \]
- Since the particle starts from rest, we have \( v(0) = 0 \):
\[ 0 = 3(0)^2 + 4(0) + C \implies C = 0 \]
- Therefore, the velocity function is:
\[ v(t) = 3t^2 + 4t \]

4. **Integrate to find displacement**:
- Displacement \( s(t) \) is the integral of velocity:
\[ s(t) = \int v(t) \, dt = \int (3t^2 + 4t) \, dt \]
- Performing the integration:
\[ s(t) = t^3 + 2t^2 + D \]
- Since the initial displacement is zero, \( s(0) = 0 \):
\[ 0 = (0)^3 + 2(0)^2 + D \implies D = 0 \]
- Therefore, the displacement function is:
\[ s(t) = t^3 + 2t^2 \]

5. **Calculate the distance covered in 3 seconds**:
- Substitute \( t = 3 \) into the displacement equation:
\[ s(3) = (3)^3 + 2(3)^2 \]
- Calculate:
\[ s(3) = 27 + 2 \times 9 = 27 + 18 = 45 \, \text{meters} \]

### Final Answer:
The distance covered by the particle in 3 seconds is **45 meters**.