If displacement S at time t is `S=-t^(3)+3t^(2)+5`, then velocity at time `t=2` sec is

Options

A. 3 units/ sec
B. 6 units/sec
C. 12 units/sec
D. 0

Answer

To find the velocity at time \( t = 2 \) seconds given the displacement function \( S = -t^3 + 3t^2 + 5 \), we will follow these steps:

### Step 1: Understand the relationship between displacement and velocity
Velocity is defined as the rate of change of displacement with respect to time. Mathematically, this is represented as:
\[ v = \frac{dS}{dt} \]

### Step 2: Differentiate the displacement function
We need to differentiate the displacement function \( S(t) = -t^3 + 3t^2 + 5 \) with respect to \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt}(-t^3) + \frac{d}{dt}(3t^2) + \frac{d}{dt}(5) \]
Calculating each term:
- The derivative of \( -t^3 \) is \( -3t^2 \).
- The derivative of \( 3t^2 \) is \( 6t \).
- The derivative of the constant \( 5 \) is \( 0 \).

Combining these results, we have:
\[ \frac{dS}{dt} = -3t^2 + 6t \]

### Step 3: Substitute \( t = 2 \) into the velocity function
Now we will substitute \( t = 2 \) seconds into the velocity function:
\[ v(2) = -3(2^2) + 6(2) \]
Calculating \( 2^2 \):
\[ 2^2 = 4 \]
Now substituting this value:
\[ v(2) = -3(4) + 6(2) = -12 + 12 \]
Simplifying this gives:
\[ v(2) = 0 \]

### Conclusion
The velocity at time \( t = 2 \) seconds is:
\[ \boxed{0} \text{ units per second} \]