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

Explanation

To solve the problem of finding the distance covered by a particle moving with an acceleration given by \( a = 6t + 4 \, \text{m/s}^2 \) over a time interval of 3 seconds, we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and displacement.
Acceleration is the rate of change of velocity, and velocity is the rate of change of displacement. Therefore, we can express these relationships mathematically:
- \( a = \frac{dv}{dt} \)
- \( v = \frac{dx}{dt} \)

### Step 2: Integrate the acceleration to find the velocity.
Given the acceleration \( a = 6t + 4 \), we can write:
\[ \frac{dv}{dt} = 6t + 4 \]
Now, we integrate both sides with respect to time \( t \):
\[ \int dv = \int (6t + 4) dt \]
This gives us:
\[ v = 3t^2 + 4t + C \]
Since the particle is initially at rest (at \( t = 0 \), \( v = 0 \)), we can find the constant \( C \):
\[ 0 = 3(0)^2 + 4(0) + C \implies C = 0 \]
Thus, the velocity function is:
\[ v = 3t^2 + 4t \]

### Step 3: Integrate the velocity to find the displacement.
Next, we need to find the displacement \( x \) by integrating the velocity:
\[ \frac{dx}{dt} = 3t^2 + 4t \]
Integrating both sides gives:
\[ \int dx = \int (3t^2 + 4t) dt \]
This results in:
\[ x = t^3 + 2t^2 + C \]
Again, since the particle starts from rest at \( t = 0 \) (and thus \( x = 0 \)), we find:
\[ 0 = (0)^3 + 2(0)^2 + C \implies C = 0 \]
So, the displacement function is:
\[ x = t^3 + 2t^2 \]

### Step 4: Calculate the distance covered in 3 seconds.
Now, we can find the distance covered in 3 seconds by substituting \( t = 3 \) into the displacement equation:
\[ x(3) = (3)^3 + 2(3)^2 \]
Calculating this gives:
\[ x(3) = 27 + 2 \times 9 = 27 + 18 = 45 \, \text{meters} \]

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

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