The velocity of particle at time t is given by the relation `v=6t-(t^(2))/(6)`. The distance traveled in 3 s is, if s=0 at t=0

Options

A. `(39)/(2)`
B. `(57)/(2)`
C. `(51)/(2)`
D. `(33)/(2)`

Explanation

To find the distance traveled by the particle in 3 seconds given the velocity function \( v = 6t - \frac{t^2}{6} \), we can follow these steps:

### Step 1: Write the relationship between velocity and distance
The velocity \( v \) is defined as the rate of change of distance \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} \]
Thus, we can rewrite the equation as:
\[ ds = v \, dt \]

### Step 2: Substitute the expression for velocity
Substituting the given velocity expression into the equation:
\[ ds = \left(6t - \frac{t^2}{6}\right) dt \]

### Step 3: Integrate both sides
Now, we integrate both sides to find the distance \( s \):
\[ s = \int \left(6t - \frac{t^2}{6}\right) dt \]

### Step 4: Perform the integration
We can break the integral into two parts:
\[ s = \int 6t \, dt - \int \frac{t^2}{6} \, dt \]
Calculating these integrals:
\[ s = 6 \cdot \frac{t^2}{2} - \frac{1}{6} \cdot \frac{t^3}{3} + C \]
\[ s = 3t^2 - \frac{t^3}{18} + C \]

### Step 5: Determine the constant of integration
We know that at \( t = 0 \), the distance \( s = 0 \):
\[ s(0) = 3(0)^2 - \frac{(0)^3}{18} + C = 0 \]
This implies \( C = 0 \). Therefore, the equation for distance simplifies to:
\[ s = 3t^2 - \frac{t^3}{18} \]

### Step 6: Calculate the distance traveled in 3 seconds
Now, we substitute \( t = 3 \) into the distance equation:
\[ s(3) = 3(3^2) - \frac{(3)^3}{18} \]
Calculating this:
\[ s(3) = 3(9) - \frac{27}{18} \]
\[ s(3) = 27 - \frac{3}{2} \]
Converting \( 27 \) to a fraction:
\[ s(3) = \frac{54}{2} - \frac{3}{2} = \frac{51}{2} \]

### Final Answer
The distance traveled in 3 seconds is:
\[ \boxed{\frac{51}{2}} \]