Velocity of a particle varies as `v=2t^3-3t^2` in `(km)/(hr)` If `t=0` is taken at 12:00 noon
Q. What is the velocity of the particle at 12:00 noon?

Options

A. `0.5 (km)/(hr)`
B. `zero
C. 1km `(km)/(hr)`
D. 2 `(km)/(hr)`

Explanation

To find the velocity of the particle at 12:00 noon, we will follow these steps:

### Step 1: Understand the given equation for velocity
The velocity of the particle is given by the equation:
\[ v = 2t^3 - 3t^2 \]
where \( v \) is in kilometers per hour (km/hr) and \( t \) is the time in hours since 12:00 noon.

### Step 2: Determine the value of \( t \) at 12:00 noon
Since \( t = 0 \) is taken at 12:00 noon, we can substitute \( t = 0 \) into the equation.

### Step 3: Substitute \( t = 0 \) into the velocity equation
Now, we substitute \( t = 0 \) into the velocity equation:
\[ v = 2(0)^3 - 3(0)^2 \]

### Step 4: Calculate the velocity
Calculating the above expression:
\[ v = 2 \cdot 0 - 3 \cdot 0 \]
\[ v = 0 - 0 \]
\[ v = 0 \]

### Step 5: Conclusion
Thus, the velocity of the particle at 12:00 noon is:
\[ v = 0 \text{ km/hr} \]