Velocity is given by `v=4t(1-2t),` then find time at which velocity is maximum

Options

A. 0.25 s
B. 1 s
C. 0.45 s
D. 4s

Answer

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

### Step 1: Write down the given velocity equation.
The velocity \( v \) is given by:
\[ v = 4t(1 - 2t) \]

### Step 2: Expand the velocity equation.
To make differentiation easier, we can expand the equation:
\[ v = 4t - 8t^2 \]

### Step 3: Differentiate the velocity with respect to time \( t \).
We need to find the derivative of \( v \) with respect to \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(4t - 8t^2) \]
Using the power rule of differentiation:
\[ \frac{dv}{dt} = 4 - 16t \]

### Step 4: Set the derivative equal to zero to find critical points.
To find the time when the velocity is maximum, we set the derivative equal to zero:
\[ 4 - 16t = 0 \]

### Step 5: Solve for \( t \).
Rearranging the equation gives:
\[ 16t = 4 \implies t = \frac{4}{16} = \frac{1}{4} \]
Thus,
\[ t = 0.25 \text{ seconds} \]

### Conclusion
The time at which the velocity is maximum is \( t = 0.25 \) seconds.

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