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, we need to follow these steps:

### Step 1: Write down the equation of motion
The equation of motion given is:
\[ s = 12t + 3t^2 - 2t^3 \]
where \( s \) is the displacement in meters and \( t \) is the time in seconds.

### Step 2: Differentiate the equation of motion to find the velocity
The velocity \( v \) is the first derivative of the displacement \( s \) with respect to time \( t \). We differentiate \( s \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}(12t + 3t^2 - 2t^3) \]
Using the power rule of differentiation, we get:
\[ v = 12 + 6t - 6t^2 \]

### Step 3: Calculate the velocity at the start (when \( t = 0 \))
To find the velocity at the start, we substitute \( t = 0 \) into the velocity equation:
\[ v(0) = 12 + 6(0) - 6(0)^2 \]
This simplifies to:
\[ v(0) = 12 + 0 - 0 = 12 \text{ m/s} \]

### Conclusion
The velocity of the car at the start is:
\[ \boxed{12 \text{ m/s}} \]

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