A point initially at rest moves along x-axis. Its acceleration varies with time as `a = (6t + 5) m//s^(2)`. If it starts from origin, the distance covered in 2 s is:

Options

A. 20m
B. 18m
C. 16m
D. 25m

Explanation

To solve the problem step by step, we will follow the process of integrating the acceleration to find the velocity and then integrating the velocity to find the distance.

### Step 1: Write down the acceleration equation
The acceleration of the point is given by:
\[ a(t) = 6t + 5 \, \text{m/s}^2 \]

### Step 2: Relate acceleration to velocity
We know that acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = 6t + 5 \]

### Step 3: Integrate to find velocity
To find the velocity, we integrate both sides with respect to time \( t \):
\[ \int dv = \int (6t + 5) dt \]

This gives us:
\[ v = 3t^2 + 5t + C \]
Since the point starts from rest, the initial velocity \( v(0) = 0 \). Therefore, we can find the constant \( C \):
\[ 0 = 3(0)^2 + 5(0) + C \Rightarrow C = 0 \]
Thus, the velocity equation simplifies to:
\[ v(t) = 3t^2 + 5t \]

### Step 4: Relate velocity to distance
Now, we know that velocity is the derivative of distance with respect to time:
\[ v = \frac{ds}{dt} \]
So we can write:
\[ \frac{ds}{dt} = 3t^2 + 5t \]

### Step 5: Integrate to find distance
We integrate both sides with respect to time \( t \):
\[ \int ds = \int (3t^2 + 5t) dt \]

This gives us:
\[ s = t^3 + \frac{5}{2}t^2 + C' \]
Again, since the point starts from the origin, the initial distance \( s(0) = 0 \). Therefore, we find the constant \( C' \):
\[ 0 = (0)^3 + \frac{5}{2}(0)^2 + C' \Rightarrow C' = 0 \]
Thus, the distance equation simplifies to:
\[ s(t) = t^3 + \frac{5}{2}t^2 \]

### Step 6: Calculate distance at \( t = 2 \) seconds
Now, we substitute \( t = 2 \) seconds into the distance equation:
\[ s(2) = (2)^3 + \frac{5}{2}(2)^2 \]
Calculating this gives:
\[ s(2) = 8 + \frac{5}{2} \cdot 4 \]
\[ s(2) = 8 + 10 = 18 \, \text{meters} \]

### Final Answer
The distance covered in 2 seconds is:
\[ \boxed{18 \, \text{meters}} \]

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