A particle moves according to law `s=t^(3)-6t^(2)+9t+15`. Find the velocity when t = 0.

Answer

To find the velocity of the particle at \( t = 0 \), we first need to determine the expression for velocity by differentiating the displacement function \( s(t) \).

### Step 1: Write down the displacement function
The displacement function is given by:
\[ s(t) = t^3 - 6t^2 + 9t + 15 \]

### Step 2: Differentiate the displacement function to find velocity
The velocity \( v(t) \) is the derivative of the displacement function \( s(t) \) with respect to time \( t \):
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t + 15) \]

### Step 3: Apply the power rule to differentiate
Using the power rule of differentiation:
- The derivative of \( t^3 \) is \( 3t^2 \)
- The derivative of \( -6t^2 \) is \( -12t \)
- The derivative of \( 9t \) is \( 9 \)
- The derivative of a constant (15) is \( 0 \)

Thus, we have:
\[ v(t) = 3t^2 - 12t + 9 \]

### Step 4: Substitute \( t = 0 \) into the velocity function
Now, we need to find the velocity when \( t = 0 \):
\[ v(0) = 3(0)^2 - 12(0) + 9 \]
\[ v(0) = 0 - 0 + 9 = 9 \]

### Final Answer
The velocity of the particle when \( t = 0 \) is:
\[ \boxed{9} \]