The momentum of a particle moving in straight line is given by
`p = t+(1)/(t)("in "m//s)`
find the (time `t gt0`) at which the net force aciting on particle is 0 and it's momentum at that time.

Options

A. ` 1 kg m//sec `
B. `2 kgm//sec`
C. `3kg m//sec`
D. `4kg m//sec`

Answer

To solve the problem step by step, we will follow the outlined process:

1. **Given Momentum Expression**:
The momentum \( p \) of the particle is given by the equation:
\[ p = t + \frac{1}{t} \]

2. **Finding the Force**:
The force acting on the particle can be found by differentiating the momentum with respect to time \( t \). The formula for force \( F \) is given by:
\[ F = \frac{dp}{dt} \]
We differentiate \( p \):
\[ \frac{dp}{dt} = \frac{d}{dt}\left(t + \frac{1}{t}\right) \]
Using the rules of differentiation:
- The derivative of \( t \) is \( 1 \).
- The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \).

Therefore, we have:
\[ \frac{dp}{dt} = 1 - \frac{1}{t^2} \]

3. **Setting Force to Zero**:
To find the time at which the net force is zero, we set the derivative equal to zero:
\[ 1 - \frac{1}{t^2} = 0 \]
Rearranging gives:
\[ \frac{1}{t^2} = 1 \]
Taking the reciprocal:
\[ t^2 = 1 \]
Thus, we find:
\[ t = \pm 1 \]
Since time must be greater than zero, we take:
\[ t = 1 \text{ second} \]

4. **Finding the Momentum at that Time**:
Now we substitute \( t = 1 \) back into the momentum equation to find the momentum at that time:
\[ p = 1 + \frac{1}{1} = 1 + 1 = 2 \text{ kg m/s} \]

5. **Final Answers**:
- The time at which the net force acting on the particle is zero is \( t = 1 \) second.
- The momentum at that time is \( p = 2 \) kg m/s.

### Summary of the Solution:
- The time \( t \) at which the net force is zero is \( 1 \) second.
- The momentum at that time is \( 2 \) kg m/s.