A car moves along a straight line whose equation of motion is given by
`s=12t+3t^(2)-2t^(3)`
where s is in metres and t is in seconds. The velocity of the car at start will be :-

Options

A. 7 m/s
B. 9 m/s
C. 12 m/s
D. 16 m/s

Answer

To find the velocity of the car at the start (when \( t = 0 \)), we need to follow these steps:

### Step 1: Write down the equation of motion
The equation of motion given is:
\[ s = 12t + 3t^2 - 2t^3 \]

### Step 2: Differentiate the equation with respect to time
To find the velocity, we need to differentiate the displacement \( s \) with respect to time \( t \). The velocity \( v \) is given by:
\[ v = \frac{ds}{dt} \]

Differentiating the equation:
\[ \frac{ds}{dt} = \frac{d}{dt}(12t) + \frac{d}{dt}(3t^2) - \frac{d}{dt}(2t^3) \]

Calculating each term:
- The derivative of \( 12t \) is \( 12 \).
- The derivative of \( 3t^2 \) is \( 6t \).
- The derivative of \( -2t^3 \) is \( -6t^2 \).

Putting it all together:
\[ v = 12 + 6t - 6t^2 \]

### Step 3: Substitute \( t = 0 \) to find the initial velocity
Now, we substitute \( t = 0 \) into the velocity equation to find the initial velocity:
\[ v(0) = 12 + 6(0) - 6(0)^2 \]
\[ v(0) = 12 + 0 - 0 = 12 \, \text{m/s} \]

### Conclusion
The velocity of the car at the start (when \( t = 0 \)) is:
\[ \boxed{12 \, \text{m/s}} \]