The velocity of the particle at any time t is given by `vu = 2t(3 - t) m s^(-1)`. At what time is its velocity maximum?

Options

A. 2 s
B. 3 s
C. 2/3 s
D. 3/2 s

Explanation

To find the time at which the velocity of the particle is maximum, we will follow these steps:

### Step 1: Write the given velocity equation
The velocity of the particle at any time \( t \) is given by:
\[ v(t) = 2t(3 - t) \]

### Step 2: Expand the equation
We can expand the equation to make differentiation easier:
\[ v(t) = 2t(3 - t) = 6t - 2t^2 \]

### Step 3: Differentiate the velocity equation
To find the maximum velocity, we need to differentiate the velocity function with respect to time \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(6t - 2t^2) = 6 - 4t \]

### Step 4: Set the derivative equal to zero
To find the critical points where the velocity could be maximum or minimum, we set the derivative equal to zero:
\[ 6 - 4t = 0 \]

### Step 5: Solve for \( t \)
Now, we solve for \( t \):
\[ 4t = 6 \implies t = \frac{6}{4} = \frac{3}{2} \text{ seconds} \]

### Step 6: Conclusion
The time at which the velocity is maximum is:
\[ t = \frac{3}{2} \text{ seconds} \]