The velocity `'v'` of a particle moving along straight line is given in terms of time `t` as `v=3(t^(2)-t)` where `t` is in seconds and `v` is in `m//s`.
The speed is minimum after `t=0` second at instant of time

Options

A. `0.5 sec`
B. `1 sec`
C. `2 sec`
D. None of these

Explanation

To find the time at which the speed of the particle is minimum, we start with the given velocity function:

\[ v(t) = 3(t^2 - t) \]

### Step 1: Find the first derivative of the velocity function
To find the critical points where the speed could be minimum, we first differentiate the velocity function with respect to time \( t \).

\[ \frac{dv}{dt} = \frac{d}{dt}[3(t^2 - t)] \]

Using the power rule for differentiation:

\[ \frac{dv}{dt} = 3(2t - 1) = 6t - 3 \]

### Step 2: Set the first derivative equal to zero
To find the critical points, we set the first derivative equal to zero:

\[ 6t - 3 = 0 \]

Solving for \( t \):

\[ 6t = 3 \]
\[ t = \frac{3}{6} = 0.5 \text{ seconds} \]

### Step 3: Find the second derivative of the velocity function
Next, we find the second derivative to determine whether this critical point is a minimum or maximum.

\[ \frac{d^2v}{dt^2} = \frac{d}{dt}[6t - 3] \]

Differentiating again:

\[ \frac{d^2v}{dt^2} = 6 \]

### Step 4: Analyze the second derivative
Since the second derivative \( \frac{d^2v}{dt^2} = 6 \) is positive, this indicates that the critical point at \( t = 0.5 \) seconds is indeed a minimum point for the velocity.

### Step 5: Determine the speed at the critical point
Now, we need to check the speed at \( t = 0.5 \) seconds to ensure it is indeed the minimum speed after \( t = 0 \).

Substituting \( t = 0.5 \) into the velocity function:

\[ v(0.5) = 3((0.5)^2 - 0.5) \]
\[ = 3(0.25 - 0.5) \]
\[ = 3(-0.25) \]
\[ = -0.75 \text{ m/s} \]

Since speed is the absolute value of velocity, we take:

\[ \text{Speed} = |v(0.5)| = 0.75 \text{ m/s} \]

### Step 6: Check for other critical points
Next, we should check if there are any other points where the velocity could be zero, as speed cannot be negative. Setting the velocity function to zero:

\[ 3(t^2 - t) = 0 \]
\[ 3t(t - 1) = 0 \]

This gives us:

\[ t = 0 \text{ or } t = 1 \text{ second} \]

### Step 7: Analyze the speed at these points
1. At \( t = 0 \):
\[ v(0) = 3(0^2 - 0) = 0 \text{ m/s} \]
Speed = 0 m/s

2. At \( t = 1 \):
\[ v(1) = 3(1^2 - 1) = 0 \text{ m/s} \]
Speed = 0 m/s

### Conclusion
The minimum speed occurs at \( t = 1 \) second, where the speed is 0 m/s.

Thus, the time at which the speed is minimum after \( t = 0 \) seconds is:

**Final Answer: \( t = 1 \text{ second} \)**