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

Options

A. 12:00 noon
B. 0.54166666666667
C. 0.45833333333333
D. 0.58333333333333

Explanation

To solve the problem, we need to find the time at which the speed of the particle is minimum given the velocity function \( v(t) = 2t^3 - 3t^2 \).

### Step-by-Step Solution:

1. **Identify the Velocity Function**:
The velocity of the particle is given by:
\[ v(t) = 2t^3 - 3t^2 \]

2. **Find the First Derivative**:
To find the critical points where the speed might be minimum, we need to take the derivative of the velocity function with respect to time \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(2t^3 - 3t^2) = 6t^2 - 6t \]

3. **Set the First Derivative to Zero**:
We set the first derivative equal to zero to find the critical points:
\[ 6t^2 - 6t = 0 \]
Factoring gives:
\[ 6t(t - 1) = 0 \]
This results in:
\[ t = 0 \quad \text{or} \quad t = 1 \]

4. **Find the Second Derivative**:
To determine whether these critical points correspond to a minimum or maximum, we calculate the second derivative:
\[ \frac{d^2v}{dt^2} = \frac{d}{dt}(6t^2 - 6t) = 12t - 6 \]

5. **Evaluate the Second Derivative at Critical Points**:
- For \( t = 0 \):
\[ \frac{d^2v}{dt^2} \bigg|_{t=0} = 12(0) - 6 = -6 \quad (\text{not a minimum}) \]
- For \( t = 1 \):
\[ \frac{d^2v}{dt^2} \bigg|_{t=1} = 12(1) - 6 = 6 \quad (\text{a minimum}) \]

6. **Determine the Time Corresponding to \( t = 1 \)**:
Since \( t = 0 \) corresponds to 12:00 noon, \( t = 1 \) corresponds to:
\[ 12:00 \, \text{noon} + 1 \, \text{hour} = 1:00 \, \text{PM} \]

### Conclusion:
The speed of the particle is minimum at **1:00 PM**.