If `s=4t+(1)/(t)` is the equation o motion of a particle, then the acceleration when velocity vanishes, is

Options

A. 0
B. 16
C. 8
D. 1

Explanation

To solve the problem, we need to find the acceleration of a particle described by the equation of motion \( s = 4t + \frac{1}{t} \) when the velocity is zero. We will follow these steps:

### Step 1: Find the velocity
The velocity \( v \) is the first derivative of the displacement \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt} \left( 4t + \frac{1}{t} \right) \]
Calculating the derivative:
\[ v = 4 - \frac{1}{t^2} \]

### Step 2: Find the time when velocity is zero
To find when the velocity vanishes, we set \( v = 0 \):
\[ 4 - \frac{1}{t^2} = 0 \]
Rearranging gives:
\[ \frac{1}{t^2} = 4 \]
Taking the reciprocal:
\[ t^2 = \frac{1}{4} \]
Taking the square root:
\[ t = \frac{1}{2} \]

### Step 3: Find the acceleration
The acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( 4 - \frac{1}{t^2} \right) \]
Calculating the derivative:
\[ a = 0 + \frac{2}{t^3} = \frac{2}{t^3} \]

### Step 4: Substitute \( t = \frac{1}{2} \) into the acceleration formula
Now, we substitute \( t = \frac{1}{2} \) into the acceleration equation:
\[ a = \frac{2}{\left(\frac{1}{2}\right)^3} = \frac{2}{\frac{1}{8}} = 2 \times 8 = 16 \]

### Conclusion
The acceleration when the velocity vanishes is \( 16 \).

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